Q1: Evaluate the determinants in Exercises 1 and 2. 1. left right

To Find: The value of the determinant left right. Formula: For a determinant of order 2, left right = ad - bc. Solution: Let the given determinant be Δ. Δ = left right Using the formula, we have: Δ = (2)(-1) - (4)(-5) Δ = -2 - (-20) Δ = -2 + 20 Δ = 18 Final Answer: The value of the determinant is…

(i) To Find: The value of the determinant left right. Solution: Let the determinant be Δ_1. Δ_1 = left right Δ_1 = (cos θ)(cos θ) - (-sin θ)(sin θ) Δ_1 = cos^2 θ + sin^2 θ Using the trigonometric identity cos^2 θ + sin^2 θ = 1, we get: Δ_1 = 1 Final Answer for (i): 1 (ii) To Find: The value of the…

Q3: If A=left[ right], then show that 2 ~A =4 ~A.

Given: The matrix A=left[ right]. To Show: 2 ~A =4 ~A. Proof: First, we calculate A. A = left right = (1)(2) - (2)(4) = 2 - 8 = -6 Next, we find the matrix 2A. 2A = 2 left[ right] = left[ right] = left[ right] Now, we calculate the determinant of 2A. 2A = left right = (2)(4) - (4)(8) = 8 - 32 = -24…

Q4: If A=left[ right], then show that 3 ~A =27 ~A.

Given: The matrix A=left[ right]. To Show: 3 ~A =27 ~A. Proof: First, we calculate A. We expand along the first column ( C_1 ) for simplicity. A = 1 left right - 0 left right + 0 left right A = 1((1)(4) - (2)(0)) - 0 + 0 = 4 Next, we find the matrix 3A. 3A = 3 left[ right] = left[ right] Now, we…

(i) To Evaluate: Δ = left right Solution: Expanding along the second row ( R_2 ) as it contains two zeros. Δ = -0 left right + 0 left right - (-1) left right Δ = 0 + 0 + 1((3)(-5) - (-1)(3)) Δ = -15 - (-3) = -15 + 3 = -12 Final Answer for (i): -12 (ii) To Evaluate: Δ = left right Solution:…

Q1 (i): (1,0),(6,0),(4,3)

Formula: The area of a triangle with vertices (x_1, y_1), (x_2, y_2), and (x_3, y_3) is given by: Δ = 1/2 left left right right where the outer absolute value ensures the area is positive. (i) Given vertices: (1,0), (6,0), (4,3) Solution: Let (x_1, y_1) = (1,0), (x_2, y_2) = (6,0), and (x_3, y_3) =…

Q1 (ii): (2,7),(1,1),(10,8)

Formula: The area of a triangle with vertices (x_1, y_1), (x_2, y_2), and (x_3, y_3) is given by: Δ = 1/2 left left right right where the outer absolute value ensures the area is positive. (i) Given vertices: (1,0), (6,0), (4,3) Solution: Let (x_1, y_1) = (1,0), (x_2, y_2) = (6,0), and (x_3, y_3) =…

Q1 (iii): (-2,-3),(3,2),(-1,-8)

Formula: The area of a triangle with vertices (x_1, y_1), (x_2, y_2), and (x_3, y_3) is given by: Δ = 1/2 left left right right where the outer absolute value ensures the area is positive. (i) Given vertices: (1,0), (6,0), (4,3) Solution: Let (x_1, y_1) = (1,0), (x_2, y_2) = (6,0), and (x_3, y_3) =…

Q3 (i): (k, 0),(4,0),(0,2)

Formula: Area of a triangle Δ = 1/2 left right. When the area is given, we must consider both positive and negative values of the determinant, i.e., 1/2 det = ± Area. (i) Given: Area = 4 sq. units, vertices are (k, 0), (4,0), (0,2). Solution: 1/2 left right = ± 4 left right = ± 8 Expanding along…

Q3 (ii): (-2,0),(0,4),(0, k)

Formula: Area of a triangle Δ = 1/2 left right. When the area is given, we must consider both positive and negative values of the determinant, i.e., 1/2 det = ± Area. (i) Given: Area = 4 sq. units, vertices are (k, 0), (4,0), (0,2). Solution: 1/2 left right = ± 4 left right = ± 8 Expanding along…

Determinants Class 12 Mathematics NCERT Solutions

Q1EXERCISE 4.1

Evaluate the determinants in Exercises 1 and 2. $\left|\begin{array}{rr}2 & 4 \ -5 & -1\end{array}\right|$

Solution

To Find: The value of the determinant

\left|\begin{array}{rr}2 & 4 \ -5 & -1\end{array}\right|

.

Formula: For a determinant of order 2,

\left|\begin{array}{ll}a & b \ c & d\end{array}\right| = ad - bc

.

Solution: Let the given determinant be

\Delta

.

\Delta = \left|\begin{array}{rr}2 & 4 \ -5 & -1\end{array}\right|

Using the formula, we have:

\Delta = (2)(-1) - (4)(-5)

\Delta = -2 - (-20)

\Delta = -2 + 20

\Delta = 18

Final Answer: The value of the determinant is 18.

Q2EXERCISE 4.1

(i) left right (ii) left right

(i)

$\left|\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}\right|$

(ii)

$\left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}\right|$

Solution

(i) To Find: The value of the determinant

\left|\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}\right|

.

Solution: Let the determinant be

\Delta_1

.

\Delta_1 = \left|\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}\right|

\Delta_1 = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta)

\Delta_1 = \cos^2 \theta + \sin^2 \theta

Using the trigonometric identity

\cos^2 \theta + \sin^2 \theta = 1

, we get:

\Delta_1 = 1

Final Answer for (i): 1

(ii) To Find: The value of the determinant

\left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}\right|

.

Solution: Let the determinant be

\Delta_2

.

\Delta_2 = \left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}\right|

\Delta_2 = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)

We know the identity

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

. So,

(x+1)(x^2-x+1) = x^3 + 1^3 = x^3 + 1

. We also know the identity

a^2 - b^2 = (a-b)(a+b)

. So,

(x-1)(x+1) = x^2 - 1

. Substituting these into the expression:

\Delta_2 = (x^3 + 1) - (x^2 - 1)

\Delta_2 = x^3 + 1 - x^2 + 1

\Delta_2 = x^3 - x^2 + 2

Final Answer for (ii):

x^3 - x^2 + 2

Q3EXERCISE 4.1

If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}\right]$ , then show that $|2 \mathrm{~A}|=4|\mathrm{~A}|$ .

Solution

Given: The matrix

\mathrm{A}=\left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}\right]

.

To Show:

|2 \mathrm{~A}|=4|\mathrm{~A}|

.

Proof: First, we calculate

|\mathrm{A}|

.

|\mathrm{A}| = \left|\begin{array}{ll}1 & 2 \ 4 & 2\end{array}\right| = (1)(2) - (2)(4) = 2 - 8 = -6

Next, we find the matrix

2\mathrm{A}

.

2\mathrm{A} = 2 \left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}\right] = \left[\begin{array}{ll}2 \times 1 & 2 \times 2 \ 2 \times 4 & 2 \times 2\end{array}\right] = \left[\begin{array}{ll}2 & 4 \ 8 & 4\end{array}\right]

Now, we calculate the determinant of

2\mathrm{A}

.

|2\mathrm{A}| = \left|\begin{array}{ll}2 & 4 \ 8 & 4\end{array}\right| = (2)(4) - (4)(8) = 8 - 32 = -24

Now we evaluate the Right Hand Side (RHS) of the equation to be shown. RHS =

4|\mathrm{A}| = 4 \times (-6) = -24

.

We have LHS =

|2\mathrm{A}| = -24

and RHS =

4|\mathrm{A}| = -24

. Since LHS = RHS, the statement is proved.

Hence Proved.

Q4EXERCISE 4.1

If $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4\end{array}\right]$ , then show that $|3 \mathrm{~A}|=27|\mathrm{~A}|$ .

Solution

Given: The matrix

\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4\end{array}\right]

.

To Show:

|3 \mathrm{~A}|=27|\mathrm{~A}|

.

Proof: First, we calculate

|\mathrm{A}|

. We expand along the first column (

C_1

) for simplicity.

|\mathrm{A}| = 1 \left|\begin{array}{ll}1 & 2 \ 0 & 4\end{array}\right| - 0 \left|\begin{array}{ll}0 & 1 \ 0 & 4\end{array}\right| + 0 \left|\begin{array}{ll}0 & 1 \ 1 & 2\end{array}\right|

|\mathrm{A}| = 1((1)(4) - (2)(0)) - 0 + 0 = 4

Next, we find the matrix

3\mathrm{A}

.

3\mathrm{A} = 3 \left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4\end{array}\right] = \left[\begin{array}{ccc}3 & 0 & 3 \ 0 & 3 & 6 \ 0 & 0 & 12\end{array}\right]

Now, we calculate the determinant of

3\mathrm{A}

. We expand along the first column (

C_1

).

|3\mathrm{A}| = 3 \left|\begin{array}{cc}3 & 6 \ 0 & 12\end{array}\right| - 0 \left|\begin{array}{cc}0 & 3 \ 0 & 12\end{array}\right| + 0 \left|\begin{array}{cc}0 & 3 \ 3 & 6\end{array}\right|

|3\mathrm{A}| = 3((3)(12) - (6)(0)) - 0 + 0 = 3(36) = 108

Now we evaluate the Right Hand Side (RHS) of the equation to be shown. RHS =

27|\mathrm{A}| = 27 \times 4 = 108

.

We have LHS =

|3\mathrm{A}| = 108

and RHS =

27|\mathrm{A}| = 108

. Since LHS = RHS, the statement is proved.

Hence Proved.

Q5EXERCISE 4.1

Evaluate the determinants

(i)

$\left|\begin{array}{rrr}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}\right|$

(ii)

$\left|\begin{array}{rrr}3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1\end{array}\right|$

(iii)

$\left|\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0\end{array}\right|$

(iv)

$\left|\begin{array}{rrr}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}\right|$

Solution

(i) To Evaluate:

\Delta = \left|\begin{array}{rrr}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}\right|

Solution: Expanding along the second row (

R_2

) as it contains two zeros.

\Delta = -0 \left|\begin{array}{rr}-1 & -2 \ -5 & 0\end{array}\right| + 0 \left|\begin{array}{rr}3 & -2 \ 3 & 0\end{array}\right| - (-1) \left|\begin{array}{rr}3 & -1 \ 3 & -5\end{array}\right|

\Delta = 0 + 0 + 1((3)(-5) - (-1)(3))

\Delta = -15 - (-3) = -15 + 3 = -12

Final Answer for (i): -12

(ii) To Evaluate:

\Delta = \left|\begin{array}{rrr}3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1\end{array}\right|

Solution: Expanding along the first row (

R_1

).

\Delta = 3 \left|\begin{array}{rr}1 & -2 \ 3 & 1\end{array}\right| - (-4) \left|\begin{array}{rr}1 & -2 \ 2 & 1\end{array}\right| + 5 \left|\begin{array}{rr}1 & 1 \ 2 & 3\end{array}\right|

\Delta = 3((1)(1) - (-2)(3)) + 4((1)(1) - (-2)(2)) + 5((1)(3) - (1)(2))

\Delta = 3(1 + 6) + 4(1 + 4) + 5(3 - 2)

\Delta = 3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46

Final Answer for (ii): 46

(iii) To Evaluate:

\Delta = \left|\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0\end{array}\right|

Solution: Expanding along the first row (

R_1

).

\Delta = 0 \left|\begin{array}{rr}0 & -3 \ 3 & 0\end{array}\right| - 1 \left|\begin{array}{rr}-1 & -3 \ -2 & 0\end{array}\right| + 2 \left|\begin{array}{rr}-1 & 0 \ -2 & 3\end{array}\right|

\Delta = 0 - 1((-1)(0) - (-3)(-2)) + 2((-1)(3) - (0)(-2))

\Delta = -1(0 - 6) + 2(-3 - 0) = -1(-6) + 2(-3) = 6 - 6 = 0

Final Answer for (iii): 0

(iv) To Evaluate:

\Delta = \left|\begin{array}{rrr}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}\right|

Solution: Expanding along the first column (

C_1

) as it contains one zero.

\Delta = 2 \left|\begin{array}{rr}2 & -1 \ -5 & 0\end{array}\right| - 0 \left|\begin{array}{rr}-1 & -2 \ -5 & 0\end{array}\right| + 3 \left|\begin{array}{rr}-1 & -2 \ 2 & -1\end{array}\right|

\Delta = 2((2)(0) - (-1)(-5)) - 0 + 3((-1)(-1) - (-2)(2))

\Delta = 2(0 - 5) + 3(1 - (-4)) = 2(-5) + 3(1 + 4) = -10 + 3(5) = -10 + 15 = 5

Final Answer for (iv): 5

Q6EXERCISE 4.1

If $\mathrm{A}=\left[\begin{array}{lll}1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9\end{array}\right]$ , find $|\mathrm{A}|$ .

Solution

Given: The matrix

\mathrm{A}=\left[\begin{array}{lll}1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9\end{array}\right]

.

To Find: The value of the determinant

|\mathrm{A}|

.

Solution: We expand the determinant along the first row (

R_1

).

|\mathrm{A}| = 1 \left|\begin{array}{ll}1 & -3 \ 4 & -9\end{array}\right| - 1 \left|\begin{array}{ll}2 & -3 \ 5 & -9\end{array}\right| + (-2) \left|\begin{array}{ll}2 & 1 \ 5 & 4\end{array}\right|

|\mathrm{A}| = 1((1)(-9) - (-3)(4)) - 1((2)(-9) - (-3)(5)) - 2((2)(4) - (1)(5))

|\mathrm{A}| = 1(-9 - (-12)) - 1(-18 - (-15)) - 2(8 - 5)

|\mathrm{A}| = 1(-9 + 12) - 1(-18 + 15) - 2(3)

|\mathrm{A}| = 1(3) - 1(-3) - 6

|\mathrm{A}| = 3 + 3 - 6 = 0

Final Answer:

|\mathrm{A}| = 0

Q7EXERCISE 4.1

Find values of $x$ , if

(i)

$\left|\begin{array}{ll}2 & 4 \ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \ 6 & x\end{array}\right|$

(ii)

$\left|\begin{array}{ll}2 & 3 \ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \ 2 x & 5\end{array}\right|$

Solution

(i) Given:

\left|\begin{array}{ll}2 & 4 \ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \ 6 & x\end{array}\right|

To Find: The value of

x

.

Solution: First, evaluate the determinant on the LHS.

LHS = (2)(1) - (4)(5) = 2 - 20 = -18

Next, evaluate the determinant on the RHS.

RHS = (2x)(x) - (4)(6) = 2x^2 - 24

Now, equate the LHS and RHS.

-18 = 2x^2 - 24

2x^2 = 24 - 18

2x^2 = 6

x^2 = 3

x = \pm\sqrt{3}

Final Answer for (i):

x = \pm\sqrt{3}

(ii) Given:

\left|\begin{array}{ll}2 & 3 \ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \ 2 x & 5\end{array}\right|

To Find: The value of

x

.

Solution: First, evaluate the determinant on the LHS.

LHS = (2)(5) - (3)(4) = 10 - 12 = -2

Next, evaluate the determinant on the RHS.

RHS = (x)(5) - (3)(2x) = 5x - 6x = -x

Now, equate the LHS and RHS.

-2 = -x

x = 2

Final Answer for (ii):

x = 2

Q8EXERCISE 4.1

If $\left|\begin{array}{cc}x & 2 \ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \ 18 & 6\end{array}\right|$ , then $x$ is equal to

(A)

6

(B)

\pm 6

(C)

-6

(D)

0

Solution

Given: The equation

\left|\begin{array}{cc}x & 2 \ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \ 18 & 6\end{array}\right|

.

To Find: The value of

x

.

Solution: First, evaluate the determinant on the LHS.

LHS = (x)(x) - (2)(18) = x^2 - 36

Next, evaluate the determinant on the RHS.

RHS = (6)(6) - (2)(18) = 36 - 36 = 0

Now, equate the LHS and RHS.

x^2 - 36 = 0

x^2 = 36

x = \pm\sqrt{36}

x = \pm 6

This corresponds to option (B).

Final Answer: The correct option is (B).

Q1EXERCISE 4.2

Find area of the triangle with vertices at the point given in each of the following:

(i)

$(1,0),(6,0),(4,3)$

(ii)

$(2,7),(1,1),(10,8)$

(iii)

$(-2,-3),(3,2),(-1,-8)$

Solution

Formula: The area of a triangle with vertices

(x_1, y_1), (x_2, y_2),

and

(x_3, y_3)

is given by:

\Delta = \frac{1}{2} \left| \left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}\right| \right|

where the outer absolute value ensures the area is positive.

(i) Given vertices:

(1,0), (6,0), (4,3)

Solution: Let

(x_1, y_1) = (1,0)

,

(x_2, y_2) = (6,0)

, and

(x_3, y_3) = (4,3)

.

\text{Area} = \frac{1}{2} \left| \left|\begin{array}{lll}1 & 0 & 1 \ 6 & 0 & 1 \ 4 & 3 & 1\end{array}\right| \right|

Expanding along the second column (

C_2

):

= \frac{1}{2} | -0 + 0 - 3\left|\begin{array}{ll}1 & 1 \ 6 & 1\end{array}\right| |

= \frac{1}{2} | -3((1)(1) - (1)(6)) | = \frac{1}{2} | -3(1-6) | = \frac{1}{2} | -3(-5) | = \frac{1}{2} |15| = \frac{15}{2}

Final Answer for (i):

\frac{15}{2}

sq. units

(ii) Given vertices:

(2,7), (1,1), (10,8)

Solution: Let

(x_1, y_1) = (2,7)

,

(x_2, y_2) = (1,1)

, and

(x_3, y_3) = (10,8)

.

\text{Area} = \frac{1}{2} \left| \left|\begin{array}{ccc}2 & 7 & 1 \ 1 & 1 & 1 \ 10 & 8 & 1\end{array}\right| \right|

Expanding along the first row (

R_1

):

= \frac{1}{2} | 2\left|\begin{array}{ll}1 & 1 \ 8 & 1\end{array}\right| - 7\left|\begin{array}{cc}1 & 1 \ 10 & 1\end{array}\right| + 1\left|\begin{array}{cc}1 & 1 \ 10 & 8\end{array}\right| |

= \frac{1}{2} | 2(1-8) - 7(1-10) + 1(8-10) |

= \frac{1}{2} | 2(-7) - 7(-9) + 1(-2) | = \frac{1}{2} | -14 + 63 - 2 | = \frac{1}{2} | 47 | = \frac{47}{2}

Final Answer for (ii):

\frac{47}{2}

sq. units

(iii) Given vertices:

(-2,-3), (3,2), (-1,-8)

Solution: Let

(x_1, y_1) = (-2,-3)

,

(x_2, y_2) = (3,2)

, and

(x_3, y_3) = (-1,-8)

.

\text{Area} = \frac{1}{2} \left| \left|\begin{array}{rrr}-2 & -3 & 1 \ 3 & 2 & 1 \ -1 & -8 & 1\end{array}\right| \right|

Expanding along the first row (

R_1

):

= \frac{1}{2} | -2\left|\begin{array}{rr}2 & 1 \ -8 & 1\end{array}\right| - (-3)\left|\begin{array}{rr}3 & 1 \ -1 & 1\end{array}\right| + 1\left|\begin{array}{rr}3 & 2 \ -1 & -8\end{array}\right| |

= \frac{1}{2} | -2(2 - (-8)) + 3(3 - (-1)) + 1(-24 - (-2)) |

= \frac{1}{2} | -2(10) + 3(4) + 1(-22) | = \frac{1}{2} | -20 + 12 - 22 | = \frac{1}{2} | -30 | = 15

Final Answer for (iii): 15 sq. units

Q2EXERCISE 4.2

Show that points $\mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b)$ are collinear.

Solution

Given: Points

\mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b)

.

To Show: The points A, B, and C are collinear.

Condition for Collinearity: Three points are collinear if the area of the triangle formed by them is zero.

Proof: We calculate the area of triangle ABC using the determinant formula.

\text{Area}(\triangle ABC) = \frac{1}{2} \left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}\right| = \frac{1}{2} \left|\begin{array}{lll}a & b+c & 1 \ b & c+a & 1 \ c & a+b & 1\end{array}\right|

Let

\Delta

be the value of the determinant. We apply the column operation

C_1 \rightarrow C_1 + C_2

.

\Delta = \left|\begin{array}{lll}a+(b+c) & b+c & 1 \ b+(c+a) & c+a & 1 \ c+(a+b) & a+b & 1\end{array}\right| = \left|\begin{array}{lll}a+b+c & b+c & 1 \ a+b+c & c+a & 1 \ a+b+c & a+b & 1\end{array}\right|

Now, we take

(a+b+c)

common from the first column (

C_1

).

\Delta = (a+b+c) \left|\begin{array}{lll}1 & b+c & 1 \ 1 & c+a & 1 \ 1 & a+b & 1\end{array}\right|

Since the first column (

C_1

) and the third column (

C_3

) are identical, the value of the determinant is zero.

\Delta = (a+b+c) \times 0 = 0

Since the area of the triangle formed by points A, B, and C is zero, the points are collinear.

Hence Proved.

Q3EXERCISE 4.2

Find values of $k$ if area of triangle is 4 sq . units and vertices are

(i)

$(k, 0),(4,0),(0,2)$

(ii)

$(-2,0),(0,4),(0, k)$

Solution

Formula: Area of a triangle

\Delta = \frac{1}{2} \left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}\right|

. When the area is given, we must consider both positive and negative values of the determinant, i.e.,

\frac{1}{2} \det = \pm \text{Area}

.

(i) Given: Area = 4 sq. units, vertices are

(k, 0), (4,0), (0,2)

.

Solution:

\frac{1}{2} \left|\begin{array}{lll}k & 0 & 1 \ 4 & 0 & 1 \ 0 & 2 & 1\end{array}\right| = \pm 4

\left|\begin{array}{lll}k & 0 & 1 \ 4 & 0 & 1 \ 0 & 2 & 1\end{array}\right| = \pm 8

Expanding along the second column (

C_2

):

-0 + 0 - 2\left|\begin{array}{ll}k & 1 \ 4 & 1\end{array}\right| = \pm 8

-2(k(1) - 1(4)) = \pm 8

-2(k-4) = \pm 8

k-4 = \frac{\pm 8}{-2}

k-4 = \mp 4

Case 1:

k-4 = -4 \implies k = 0

. Case 2:

k-4 = 4 \implies k = 8

.

Final Answer for (i):

k = 0, 8

(ii) Given: Area = 4 sq. units, vertices are

(-2,0), (0,4), (0,k)

.

Solution:

\frac{1}{2} \left|\begin{array}{rrr}-2 & 0 & 1 \ 0 & 4 & 1 \ 0 & k & 1\end{array}\right| = \pm 4

\left|\begin{array}{rrr}-2 & 0 & 1 \ 0 & 4 & 1 \ 0 & k & 1\end{array}\right| = \pm 8

Expanding along the first column (

C_1

):

-2\left|\begin{array}{ll}4 & 1 \ k & 1\end{array}\right| - 0 + 0 = \pm 8

-2(4(1) - 1(k)) = \pm 8

-2(4-k) = \pm 8

4-k = \frac{\pm 8}{-2}

4-k = \mp 4

Case 1:

4-k = -4 \implies k = 8

. Case 2:

4-k = 4 \implies k = 0

.

Final Answer for (ii):

k = 0, 8

Q4EXERCISE 4.2

(i) Find equation of line joining (1,2) and (3,6) using determinants. (ii) Find equation of line joining…

(i)

Find equation of line joining $(1,2)$ and $(3,6)$ using determinants.

(ii)

Find equation of line joining $(3,1)$ and $(9,3)$ using determinants.

Solution

Concept: If three points A, B, and P are on the same line (collinear), the area of the triangle formed by them is zero. Let the given points be A

(x_1, y_1)

and B

(x_2, y_2)

, and let P

(x,y)

be any point on the line joining A and B. Then Area(

\triangle

ABP) = 0.

\frac{1}{2} \left|\begin{array}{lll}x & y & 1 \ x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1\end{array}\right| = 0

(i) Given points:

(1,2)

and

(3,6)

.

Solution: Let P

(x,y)

be any point on the line joining A

(1,2)

and B

(3,6)

. Then the points P, A, and B are collinear. So, the area of

\triangle

PAB is 0.

\frac{1}{2} \left|\begin{array}{lll}x & y & 1 \ 1 & 2 & 1 \ 3 & 6 & 1\end{array}\right| = 0

\left|\begin{array}{lll}x & y & 1 \ 1 & 2 & 1 \ 3 & 6 & 1\end{array}\right| = 0

Expanding along the first row (

R_1

):

x\left|\begin{array}{ll}2 & 1 \ 6 & 1\end{array}\right| - y\left|\begin{array}{ll}1 & 1 \ 3 & 1\end{array}\right| + 1\left|\begin{array}{ll}1 & 2 \ 3 & 6\end{array}\right| = 0

x(2-6) - y(1-3) + 1(6-6) = 0

x(-4) - y(-2) + 1(0) = 0

-4x + 2y = 0

2y = 4x

y = 2x

Final Answer for (i): The equation of the line is

y = 2x

.

(ii) Given points:

(3,1)

and

(9,3)

.

Solution: Let P

(x,y)

be any point on the line joining A

(3,1)

and B

(9,3)

. Then the points P, A, and B are collinear. So, the area of

\triangle

PAB is 0.

\frac{1}{2} \left|\begin{array}{lll}x & y & 1 \ 3 & 1 & 1 \ 9 & 3 & 1\end{array}\right| = 0

\left|\begin{array}{lll}x & y & 1 \ 3 & 1 & 1 \ 9 & 3 & 1\end{array}\right| = 0

Expanding along the first row (

R_1

):

x\left|\begin{array}{ll}1 & 1 \ 3 & 1\end{array}\right| - y\left|\begin{array}{ll}3 & 1 \ 9 & 1\end{array}\right| + 1\left|\begin{array}{ll}3 & 1 \ 9 & 3\end{array}\right| = 0

x(1-3) - y(3-9) + 1(9-9) = 0

x(-2) - y(-6) + 1(0) = 0

-2x + 6y = 0

6y = 2x

3y = x

x - 3y = 0

Final Answer for (ii): The equation of the line is

x - 3y = 0

.

Q5EXERCISE 4.2

If area of triangle is 35 sq units with vertices $(2,-6),(5,4)$ and $(k, 4)$ . Then $k$ is

(A)

12

(B)

-2

(C)

-12,-2

(D)

12,-2

Solution

Given: Area = 35 sq. units, vertices are

(2,-6), (5,4), (k,4)

.

To Find: The value of

k

.

Solution: Using the area of triangle formula:

\frac{1}{2} \left|\begin{array}{rrr}2 & -6 & 1 \ 5 & 4 & 1 \ k & 4 & 1\end{array}\right| = \pm 35

\left|\begin{array}{rrr}2 & -6 & 1 \ 5 & 4 & 1 \ k & 4 & 1\end{array}\right| = \pm 70

Expanding along the first row (

R_1

):

2\left|\begin{array}{ll}4 & 1 \ 4 & 1\end{array}\right| - (-6)\left|\begin{array}{ll}5 & 1 \ k & 1\end{array}\right| + 1\left|\begin{array}{ll}5 & 4 \ k & 4\end{array}\right| = \pm 70

2(4-4) + 6(5-k) + 1(20-4k) = \pm 70

2(0) + 30 - 6k + 20 - 4k = \pm 70

50 - 10k = \pm 70

Case 1:

50 - 10k = 70

-10k = 70 - 50

-10k = 20

k = -2

Case 2:

50 - 10k = -70

-10k = -70 - 50

-10k = -120

k = 12

So, the values of

k

are 12 and -2. This corresponds to option (D).

Final Answer: The correct option is (D).

Q1EXERCISE 4.3

Write Minors and Cofactors of the elements of following determinants:

(i)

$\left|\begin{array}{rr}2 & -4 \ 0 & 3\end{array}\right|$

(ii)

$\left|\begin{array}{ll}a & c \ b & d\end{array}\right|$

Solution

Definitions:

Minor of an element $a_{ij}$ , denoted by $M_{ij}$ , is the determinant obtained by deleting the $i$ -th row and $j$ -th column.
Cofactor of an element $a_{ij}$ , denoted by $A_{ij}$ , is defined by $A_{ij} = (-1)^{i+j} M_{ij}$ .

(i) Determinant:

\left|\begin{array}{rr}2 & -4 \ 0 & 3\end{array}\right|

Here,

a_{11}=2, a_{12}=-4, a_{21}=0, a_{22}=3

.

Minors:

M_{11}

(Minor of

a_{11}=2

) = 3

M_{12}

(Minor of

a_{12}=-4

) = 0

M_{21}

(Minor of

a_{21}=0

) = -4

M_{22}

(Minor of

a_{22}=3

) = 2

Cofactors:

A_{11} = (-1)^{1+1} M_{11} = (1)(3) = 3

A_{12} = (-1)^{1+2} M_{12} = (-1)(0) = 0

A_{21} = (-1)^{2+1} M_{21} = (-1)(-4) = 4

A_{22} = (-1)^{2+2} M_{22} = (1)(2) = 2

(ii) Determinant:

\left|\begin{array}{ll}a & c \ b & d\end{array}\right|

Here,

a_{11}=a, a_{12}=c, a_{21}=b, a_{22}=d

.

Minors:

M_{11}

(Minor of

a_{11}=a

) = d

M_{12}

(Minor of

a_{12}=c

) = b

M_{21}

(Minor of

a_{21}=b

) = c

M_{22}

(Minor of

a_{22}=d

) = a

Cofactors:

A_{11} = (-1)^{1+1} M_{11} = (1)(d) = d

A_{12} = (-1)^{1+2} M_{12} = (-1)(b) = -b

A_{21} = (-1)^{2+1} M_{21} = (-1)(c) = -c

A_{22} = (-1)^{2+2} M_{22} = (1)(a) = a

Q2EXERCISE 4.3

Write Minors and Cofactors of the elements of following determinants:

(i)

$\left|\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right|$

(ii)

$\left|\begin{array}{rrr}1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2\end{array}\right|$

Solution

(i) Determinant:

\Delta = \left|\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right|

Minors:

M_{11} = \left|\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right| = 1

M_{12} = \left|\begin{array}{ll}0 & 0 \ 0 & 1\end{array}\right| = 0

M_{13} = \left|\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right| = 0

M_{21} = \left|\begin{array}{ll}0 & 0 \ 0 & 1\end{array}\right| = 0

M_{22} = \left|\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right| = 1

M_{23} = \left|\begin{array}{ll}1 & 0 \ 0 & 0\end{array}\right| = 0

M_{31} = \left|\begin{array}{ll}0 & 0 \ 1 & 0\end{array}\right| = 0

M_{32} = \left|\begin{array}{ll}1 & 0 \ 0 & 0\end{array}\right| = 0

M_{33} = \left|\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right| = 1

Cofactors:

A_{11} = (-1)^{1+1}M_{11} = 1

A_{12} = (-1)^{1+2}M_{12} = 0

A_{13} = (-1)^{1+3}M_{13} = 0

A_{21} = (-1)^{2+1}M_{21} = 0

A_{22} = (-1)^{2+2}M_{22} = 1

A_{23} = (-1)^{2+3}M_{23} = 0

A_{31} = (-1)^{3+1}M_{31} = 0

A_{32} = (-1)^{3+2}M_{32} = 0

A_{33} = (-1)^{3+3}M_{33} = 1

(ii) Determinant:

\Delta = \left|\begin{array}{rrr}1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2\end{array}\right|

Minors:

M_{11} = \left|\begin{array}{rr}5 & -1 \ 1 & 2\end{array}\right| = 10 - (-1) = 11

M_{12} = \left|\begin{array}{rr}3 & -1 \ 0 & 2\end{array}\right| = 6 - 0 = 6

M_{13} = \left|\begin{array}{ll}3 & 5 \ 0 & 1\end{array}\right| = 3 - 0 = 3

M_{21} = \left|\begin{array}{ll}0 & 4 \ 1 & 2\end{array}\right| = 0 - 4 = -4

M_{22} = \left|\begin{array}{ll}1 & 4 \ 0 & 2\end{array}\right| = 2 - 0 = 2

M_{23} = \left|\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right| = 1 - 0 = 1

M_{31} = \left|\begin{array}{rr}0 & 4 \ 5 & -1\end{array}\right| = 0 - 20 = -20

M_{32} = \left|\begin{array}{rr}1 & 4 \ 3 & -1\end{array}\right| = -1 - 12 = -13

M_{33} = \left|\begin{array}{ll}1 & 0 \ 3 & 5\end{array}\right| = 5 - 0 = 5

Cofactors:

A_{11} = (-1)^{1+1}M_{11} = 11

A_{12} = (-1)^{1+2}M_{12} = -6

A_{13} = (-1)^{1+3}M_{13} = 3

A_{21} = (-1)^{2+1}M_{21} = -(-4) = 4

A_{22} = (-1)^{2+2}M_{22} = 2

A_{23} = (-1)^{2+3}M_{23} = -1

A_{31} = (-1)^{3+1}M_{31} = -20

A_{32} = (-1)^{3+2}M_{32} = -(-13) = 13

A_{33} = (-1)^{3+3}M_{33} = 5

Q3EXERCISE 4.3

Using Cofactors of elements of second row, evaluate $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3\end{array}\right|$ .

Solution

Given: The determinant

\Delta=\left|\begin{array}{lll}5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3\end{array}\right|

.

To Find: The value of

\Delta

using cofactors of the elements of the second row.

Formula: The value of a determinant can be found by summing the products of the elements of any row with their corresponding cofactors. For the second row (

R_2

):

\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}

Solution: The elements of the second row are

a_{21}=2, a_{22}=0, a_{23}=1

. We need to find their cofactors

A_{21}, A_{22}, A_{23}

.

Cofactor of $a_{21}$ :

M_{21} = \left|\begin{array}{ll}3 & 8 \ 2 & 3\end{array}\right| = (3)(3) - (8)(2) = 9 - 16 = -7

A_{21} = (-1)^{2+1} M_{21} = (-1)(-7) = 7

Cofactor of $a_{22}$ :

M_{22} = \left|\begin{array}{ll}5 & 8 \ 1 & 3\end{array}\right| = (5)(3) - (8)(1) = 15 - 8 = 7

A_{22} = (-1)^{2+2} M_{22} = (1)(7) = 7

Cofactor of $a_{23}$ :

M_{23} = \left|\begin{array}{ll}5 & 3 \ 1 & 2\end{array}\right| = (5)(2) - (3)(1) = 10 - 3 = 7

A_{23} = (-1)^{2+3} M_{23} = (-1)(7) = -7

Now, we use the formula:

\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}

\Delta = (2)(7) + (0)(7) + (1)(-7)

\Delta = 14 + 0 - 7

\Delta = 7

Final Answer: The value of the determinant is 7.

Q4EXERCISE 4.3

Using Cofactors of elements of third column, evaluate $\Delta=\left|\begin{array}{lll}1 & x & y z \ 1 & y & z x \ 1 & z & x y\end{array}\right|$ .

Solution

Given: The determinant

\Delta=\left|\begin{array}{lll}1 & x & y z \ 1 & y & z x \ 1 & z & x y\end{array}\right|

.

To Find: The value of

\Delta

using cofactors of the elements of the third column.

Formula: For the third column (

C_3

):

\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}

Solution: The elements of the third column are

a_{13}=yz, a_{23}=zx, a_{33}=xy

. We need to find their cofactors

A_{13}, A_{23}, A_{33}

.

Cofactor of $a_{13}$ :

M_{13} = \left|\begin{array}{ll}1 & y \ 1 & z\end{array}\right| = z - y

A_{13} = (-1)^{1+3} M_{13} = (1)(z - y) = z-y

Cofactor of $a_{23}$ :

M_{23} = \left|\begin{array}{ll}1 & x \ 1 & z\end{array}\right| = z - x

A_{23} = (-1)^{2+3} M_{23} = (-1)(z - x) = x-z

Cofactor of $a_{33}$ :

M_{33} = \left|\begin{array}{ll}1 & x \ 1 & y\end{array}\right| = y - x

A_{33} = (-1)^{3+3} M_{33} = (1)(y - x) = y-x

Now, we use the formula:

\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}

\Delta = yz(z-y) + zx(x-z) + xy(y-x)

\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y

Rearranging the terms:

\Delta = x^2z - x^2y - z^2x + y^2z + xy^2 - yz^2

\Delta = x^2(z-y) - zx(z-y) - yz(z-y) + xy^2 - yz^2 + yz^2 - y^2z

\Delta = x^2(z-y) - x(z^2-y^2) + yz(y-z)

\Delta = x^2(z-y) - x(z-y)(z+y) - yz(z-y)

Take

(z-y)

as a common factor:

\Delta = (z-y)[x^2 - x(z+y) - yz]

\Delta = (z-y)[x^2 - xz - xy - yz]

This doesn't seem to simplify well. Let's try another factorization from the expanded form.

\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y

Let's group by powers of x:

\Delta = x^2(z-y) - x(z^2-y^2) + (yz^2 - y^2z)

\Delta = x^2(z-y) - x(z-y)(z+y) + yz(z-y)

Factor out

(z-y)

:

\Delta = (z-y)[x^2 - x(z+y) + yz]

\Delta = (z-y)[x^2 - xz - xy + yz]

Factor by grouping inside the bracket:

\Delta = (z-y)[x(x-z) - y(x-z)]

\Delta = (z-y)(x-z)(x-y)

It can also be written as

(x-y)(y-z)(z-x)

.

Final Answer: The value of the determinant is

(x-y)(y-z)(z-x)

.

Q5EXERCISE 4.3

If $\Delta=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33}\end{array}\right|$ and $\mathrm{A}_{i j}$ is Cofactors of $a_{i j}$ , then value of $\Delta$ is given by

(A)

a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33}

(B)

a_{11} \mathrm{~A}_{11}+a_{12} \mathrm{~A}_{21}+a_{13} \mathrm{~A}_{31}

(C)

a_{21} \mathrm{~A}_{11}+a_{22} \mathrm{~A}_{12}+a_{23} \mathrm{~A}_{13}

(D)

a_{11} \mathrm{~A}_{11}+a_{21} \mathrm{~A}_{21}+a_{31} \mathrm{~A}_{31}

Solution

Concept: The value of a determinant is the sum of the products of the elements of any one row or any one column with their corresponding cofactors.

Let's analyze the options: (A)

a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33}

: This is the sum of products of elements of the first row (

a_{11}, a_{12}, a_{13}

) with the cofactors of the third row (

A_{31}, A_{32}, A_{33}

). The value of this expression is 0.

(B)

a_{11} \mathrm{~A}_{11}+a_{12} \mathrm{~A}_{21}+a_{13} \mathrm{~A}_{31}

: This expression does not follow the rule. It mixes elements from the first row with cofactors from different elements.

(C)

a_{21} \mathrm{~A}_{11}+a_{22} \mathrm{~A}_{12}+a_{23} \mathrm{~A}_{13}

: This is the sum of products of elements of the second row (

a_{21}, a_{22}, a_{23}

) with the cofactors of the first row (

A_{11}, A_{12}, A_{13}

). The value of this expression is 0.

(D)

a_{11} \mathrm{~A}_{11}+a_{21} \mathrm{~A}_{21}+a_{31} \mathrm{~A}_{31}

: This is the sum of the products of the elements of the first column (

a_{11}, a_{21}, a_{31}

) with their corresponding cofactors (

A_{11}, A_{21}, A_{31}

). According to the definition, this gives the value of the determinant

\Delta

.

Another valid expression would be for the first row:

a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}

. This is not an option.

Final Answer: The correct option is (D).

Q1EXERCISE 4.4

Find adjoint of each of the matrices in Exercises 1 and 2. $\begin{array}{ll}1 & 2 \ 3 & 4\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]

.

To Find: The adjoint of matrix A, denoted as adj A.

Definition: The adjoint of a matrix is the transpose of its cofactor matrix. For a 2x2 matrix

\mathrm{A} = \left[\begin{array}{ll}a & b \ c & d\end{array}\right]

, the adjoint is

\text{adj A} = \left[\begin{array}{cc}d & -b \ -c & a\end{array}\right]

.

Solution using Cofactors: First, find the cofactors of all elements.

a_{11} = 1, a_{12} = 2, a_{21} = 3, a_{22} = 4

.

A_{11} = (-1)^{1+1}(4) = 4

A_{12} = (-1)^{1+2}(3) = -3

A_{21} = (-1)^{2+1}(2) = -2

A_{22} = (-1)^{2+2}(1) = 1

The cofactor matrix is

\left[\begin{array}{cc}A_{11} & A_{12} \ A_{21} & A_{22}\end{array}\right] = \left[\begin{array}{cc}4 & -3 \ -2 & 1\end{array}\right]

.

The adjoint of A is the transpose of the cofactor matrix:

\text{adj A} = \left[\begin{array}{cc}4 & -3 \ -2 & 1\end{array}\right]^T = \left[\begin{array}{cc}4 & -2 \ -3 & 1\end{array}\right]

Alternative method (Shortcut for 2x2): For

\mathrm{A} = \left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]

, we interchange the diagonal elements (1 and 4) and change the sign of the off-diagonal elements (2 and 3).

\text{adj A} = \left[\begin{array}{cc}4 & -2 \ -3 & 1\end{array}\right]

Final Answer: The adjoint of the matrix is

\left[\begin{array}{cc}4 & -2 \ -3 & 1\end{array}\right]

.

Q2EXERCISE 4.4

$\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}\right]

.

To Find: The adjoint of matrix A, adj A.

Solution: We need to find the cofactor of each element.

A_{11} = (-1)^{1+1} \left|\begin{array}{ll}3 & 5 \ 0 & 1\end{array}\right| = 1(3-0) = 3

A_{12} = (-1)^{1+2} \left|\begin{array}{cc}2 & 5 \ -2 & 1\end{array}\right| = -1(2 - (-10)) = -12

A_{13} = (-1)^{1+3} \left|\begin{array}{cc}2 & 3 \ -2 & 0\end{array}\right| = 1(0 - (-6)) = 6

A_{21} = (-1)^{2+1} \left|\begin{array}{cc}-1 & 2 \ 0 & 1\end{array}\right| = -1(-1 - 0) = 1

A_{22} = (-1)^{2+2} \left|\begin{array}{cc}1 & 2 \ -2 & 1\end{array}\right| = 1(1 - (-4)) = 5

A_{23} = (-1)^{2+3} \left|\begin{array}{cc}1 & -1 \ -2 & 0\end{array}\right| = -1(0 - 2) = 2

A_{31} = (-1)^{3+1} \left|\begin{array}{cc}-1 & 2 \ 3 & 5\end{array}\right| = 1(-5 - 6) = -11

A_{32} = (-1)^{3+2} \left|\begin{array}{cc}1 & 2 \ 2 & 5\end{array}\right| = -1(5 - 4) = -1

A_{33} = (-1)^{3+3} \left|\begin{array}{cc}1 & -1 \ 2 & 3\end{array}\right| = 1(3 - (-2)) = 5

The cofactor matrix is:

C = \left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33}\end{array}\right] = \left[\begin{array}{rrr}3 & -12 & 6 \ 1 & 5 & 2 \ -11 & -1 & 5\end{array}\right]

The adjoint of A is the transpose of the cofactor matrix:

\text{adj A} = C^T = \left[\begin{array}{rrr}3 & 1 & -11 \ -12 & 5 & -1 \ 6 & 2 & 5\end{array}\right]

Final Answer:

\text{adj A} = \left[\begin{array}{rrr}3 & 1 & -11 \ -12 & 5 & -1 \ 6 & 2 & 5\end{array}\right]

.

Q3EXERCISE 4.4

Verify $\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}$ in Exercises 3 and 4 3. $\begin{array}{cc}2 & 3 \ -4 & -6\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{cc}2 & 3 \ -4 & -6\end{array}\right]

.

To Verify:

\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}

.

Solution: Step 1: Calculate the determinant

|\mathrm{A}|

.

|\mathrm{A}| = (2)(-6) - (3)(-4) = -12 - (-12) = -12 + 12 = 0

Step 2: Find the adjoint of A, adj A. For a 2x2 matrix

\left[\begin{array}{ll}a & b \ c & d\end{array}\right]

, adj A =

\left[\begin{array}{cc}d & -b \ -c & a\end{array}\right]

.

\text{adj A} = \left[\begin{array}{cc}-6 & -3 \ 4 & 2\end{array}\right]

Step 3: Calculate

\mathrm{A}(\operatorname{adj} \mathrm{A})

.

\mathrm{A}(\operatorname{adj} \mathrm{A}) = \left[\begin{array}{cc}2 & 3 \ -4 & -6\end{array}\right] \left[\begin{array}{cc}-6 & -3 \ 4 & 2\end{array}\right]

= \left[\begin{array}{cc}(2)(-6)+(3)(4) & (2)(-3)+(3)(2) \ (-4)(-6)+(-6)(4) & (-4)(-3)+(-6)(2)\end{array}\right]

= \left[\begin{array}{cc}-12+12 & -6+6 \ 24-24 & 12-12\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

Step 4: Calculate

(\operatorname{adj} \mathrm{A}) \mathrm{A}

.

(\operatorname{adj} \mathrm{A}) \mathrm{A} = \left[\begin{array}{cc}-6 & -3 \ 4 & 2\end{array}\right] \left[\begin{array}{cc}2 & 3 \ -4 & -6\end{array}\right]

= \left[\begin{array}{cc}(-6)(2)+(-3)(-4) & (-6)(3)+(-3)(-6) \ (4)(2)+(2)(-4) & (4)(3)+(2)(-6)\end{array}\right]

= \left[\begin{array}{cc}-12+12 & -18+18 \ 8-8 & 12-12\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

Step 5: Calculate

|\mathrm{A}| \mathrm{I}

. I is the identity matrix of order 2,

\mathrm{I} = \left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]

.

|\mathrm{A}| \mathrm{I} = 0 \times \left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

Verification: From steps 3, 4, and 5, we see that:

\mathrm{A}(\operatorname{adj} \mathrm{A}) = (\operatorname{adj} \mathrm{A}) \mathrm{A} = |\mathrm{A}| \mathrm{I} = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

Hence Verified.

Q4EXERCISE 4.4

$\begin{array}{lll}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{rrr}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}\right]

.

To Verify:

\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}

.

Solution: Step 1: Calculate the determinant

|\mathrm{A}|

. Expanding along the second column (

C_2

):

|\mathrm{A}| = -(-1) \left|\begin{array}{cc}3 & -2 \ 1 & 3\end{array}\right| + 0 - 0 = 1((3)(3) - (-2)(1)) = 1(9+2) = 11

Step 2: Find the adjoint of A, adj A. We need the cofactor matrix.

A_{11} = \left|\begin{array}{cc}0 & -2 \ 0 & 3\end{array}\right| = 0

A_{12} = -\left|\begin{array}{cc}3 & -2 \ 1 & 3\end{array}\right| = -(9+2) = -11

A_{13} = \left|\begin{array}{cc}3 & 0 \ 1 & 0\end{array}\right| = 0

A_{21} = -\left|\begin{array}{cc}-1 & 2 \ 0 & 3\end{array}\right| = -(-3-0) = 3

A_{22} = \left|\begin{array}{cc}1 & 2 \ 1 & 3\end{array}\right| = 3-2 = 1

A_{23} = -\left|\begin{array}{cc}1 & -1 \ 1 & 0\end{array}\right| = -(0-(-1)) = -1

A_{31} = \left|\begin{array}{cc}-1 & 2 \ 0 & -2\end{array}\right| = 2-0 = 2

A_{32} = -\left|\begin{array}{cc}1 & 2 \ 3 & -2\end{array}\right| = -(-2-6) = 8

A_{33} = \left|\begin{array}{cc}1 & -1 \ 3 & 0\end{array}\right| = 0-(-3) = 3

Cofactor matrix:

C = \left[\begin{array}{rrr}0 & -11 & 0 \ 3 & 1 & -1 \ 2 & 8 & 3\end{array}\right]

. Adjoint matrix:

\text{adj A} = C^T = \left[\begin{array}{rrr}0 & 3 & 2 \ -11 & 1 & 8 \ 0 & -1 & 3\end{array}\right]

.

Step 3: Calculate

\mathrm{A}(\operatorname{adj} \mathrm{A})

.

\mathrm{A}(\operatorname{adj} \mathrm{A}) = \left[\begin{array}{rrr}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}\right] \left[\begin{array}{rrr}0 & 3 & 2 \ -11 & 1 & 8 \ 0 & -1 & 3\end{array}\right]

= \left[\begin{array}{ccc}0+11+0 & 3-1-2 & 2-8+6 \ 0+0+0 & 9+0+2 & 6+0-6 \ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right] = \left[\begin{array}{ccc}11 & 0 & 0 \ 0 & 11 & 0 \ 0 & 0 & 11\end{array}\right]

Step 4: Calculate

(\operatorname{adj} \mathrm{A}) \mathrm{A}

.

(\operatorname{adj} \mathrm{A}) \mathrm{A} = \left[\begin{array}{rrr}0 & 3 & 2 \ -11 & 1 & 8 \ 0 & -1 & 3\end{array}\right] \left[\begin{array}{rrr}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}\right]

= \left[\begin{array}{ccc}0+9+2 & 0+0+0 & 0-6+6 \ -11+3+8 & 11+0+0 & -22-2+24 \ 0-3+3 & 0+0+0 & 0+2+9\end{array}\right] = \left[\begin{array}{ccc}11 & 0 & 0 \ 0 & 11 & 0 \ 0 & 0 & 11\end{array}\right]

Step 5: Calculate

|\mathrm{A}| \mathrm{I}

. I is the identity matrix of order 3.

|\mathrm{A}| \mathrm{I} = 11 \times \left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}11 & 0 & 0 \ 0 & 11 & 0 \ 0 & 0 & 11\end{array}\right]

Verification: From steps 3, 4, and 5, we see that:

\mathrm{A}(\operatorname{adj} \mathrm{A}) = (\operatorname{adj} \mathrm{A}) \mathrm{A} = |\mathrm{A}| \mathrm{I} = \left[\begin{array}{ccc}11 & 0 & 0 \ 0 & 11 & 0 \ 0 & 0 & 11\end{array}\right]

Hence Verified.

Q5EXERCISE 4.4

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 5. $\begin{array}{cc}2 & -2 \ 4 & 3\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{cc}2 & -2 \ 4 & 3\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Formula:

A^{-1} = \frac{1}{|A|} \text{adj A}

. The inverse exists if and only if

|A| \neq 0

.

Solution: Step 1: Calculate the determinant

|A|

.

|A| = (2)(3) - (-2)(4) = 6 - (-8) = 6 + 8 = 14

Since

|A| = 14 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A. For a 2x2 matrix

\left[\begin{array}{ll}a & b \ c & d\end{array}\right]

, adj A =

\left[\begin{array}{cc}d & -b \ -c & a\end{array}\right]

.

\text{adj A} = \left[\begin{array}{cc}3 & 2 \ -4 & 2\end{array}\right]

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{14} \left[\begin{array}{cc}3 & 2 \ -4 & 2\end{array}\right]

Final Answer:

A^{-1} = \frac{1}{14} \left[\begin{array}{cc}3 & 2 \ -4 & 2\end{array}\right]

.

Q6EXERCISE 4.4

$\begin{array}{rr}-1 & 5 \ -3 & 2\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{rr}-1 & 5 \ -3 & 2\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Formula:

A^{-1} = \frac{1}{|A|} \text{adj A}

. The inverse exists if and only if

|A| \neq 0

.

Solution: Step 1: Calculate the determinant

|A|

.

|A| = (-1)(2) - (5)(-3) = -2 - (-15) = -2 + 15 = 13

Since

|A| = 13 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A.

\text{adj A} = \left[\begin{array}{cc}2 & -5 \ 3 & -1\end{array}\right]

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{13} \left[\begin{array}{cc}2 & -5 \ 3 & -1\end{array}\right]

Final Answer:

A^{-1} = \frac{1}{13} \left[\begin{array}{cc}2 & -5 \ 3 & -1\end{array}\right]

.

Q7EXERCISE 4.4

$\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Formula:

A^{-1} = \frac{1}{|A|} \text{adj A}

. The inverse exists if and only if

|A| \neq 0

.

Solution: Step 1: Calculate the determinant

|A|

. This is an upper triangular matrix, so its determinant is the product of the diagonal elements.

|A| = 1 \times 2 \times 5 = 10

Alternatively, expanding along the first column:

|A| = 1 \left|\begin{array}{ll}2 & 4 \ 0 & 5\end{array}\right| - 0 + 0 = 1(10-0) = 10

Since

|A| = 10 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A. We need the cofactor matrix.

A_{11} = \left|\begin{array}{ll}2 & 4 \ 0 & 5\end{array}\right| = 10

A_{12} = -\left|\begin{array}{ll}0 & 4 \ 0 & 5\end{array}\right| = 0

A_{13} = \left|\begin{array}{ll}0 & 2 \ 0 & 0\end{array}\right| = 0

A_{21} = -\left|\begin{array}{ll}2 & 3 \ 0 & 5\end{array}\right| = -10

A_{22} = \left|\begin{array}{ll}1 & 3 \ 0 & 5\end{array}\right| = 5

A_{23} = -\left|\begin{array}{ll}1 & 2 \ 0 & 0\end{array}\right| = 0

A_{31} = \left|\begin{array}{ll}2 & 3 \ 2 & 4\end{array}\right| = 8-6 = 2

A_{32} = -\left|\begin{array}{ll}1 & 3 \ 0 & 4\end{array}\right| = -4

A_{33} = \left|\begin{array}{ll}1 & 2 \ 0 & 2\end{array}\right| = 2

Cofactor matrix:

C = \left[\begin{array}{ccc}10 & 0 & 0 \ -10 & 5 & 0 \ 2 & -4 & 2\end{array}\right]

. Adjoint matrix:

\text{adj A} = C^T = \left[\begin{array}{ccc}10 & -10 & 2 \ 0 & 5 & -4 \ 0 & 0 & 2\end{array}\right]

.

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{10} \left[\begin{array}{ccc}10 & -10 & 2 \ 0 & 5 & -4 \ 0 & 0 & 2\end{array}\right]

Final Answer:

A^{-1} = \frac{1}{10} \left[\begin{array}{ccc}10 & -10 & 2 \ 0 & 5 & -4 \ 0 & 0 & 2\end{array}\right]

or

\left[\begin{array}{ccc}1 & -1 & 1/5 \ 0 & 1/2 & -2/5 \ 0 & 0 & 1/5\end{array}\right]

.

Q8EXERCISE 4.4

$\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Formula:

A^{-1} = \frac{1}{|A|} \text{adj A}

.

Solution: Step 1: Calculate the determinant

|A|

. This is a lower triangular matrix, so its determinant is the product of the diagonal elements.

|A| = 1 \times 3 \times (-1) = -3

Alternatively, expanding along the first row:

|A| = 1 \left|\begin{array}{cc}3 & 0 \ 2 & -1\end{array}\right| - 0 + 0 = 1(-3-0) = -3

Since

|A| = -3 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A.

A_{11} = \left|\begin{array}{cc}3 & 0 \ 2 & -1\end{array}\right| = -3

A_{12} = -\left|\begin{array}{cc}3 & 0 \ 5 & -1\end{array}\right| = -(-3) = 3

A_{13} = \left|\begin{array}{ll}3 & 3 \ 5 & 2\end{array}\right| = 6-15 = -9

A_{21} = -\left|\begin{array}{cc}0 & 0 \ 2 & -1\end{array}\right| = 0

A_{22} = \left|\begin{array}{cc}1 & 0 \ 5 & -1\end{array}\right| = -1

A_{23} = -\left|\begin{array}{ll}1 & 0 \ 5 & 2\end{array}\right| = -2

A_{31} = \left|\begin{array}{ll}0 & 0 \ 3 & 0\end{array}\right| = 0

A_{32} = -\left|\begin{array}{ll}1 & 0 \ 3 & 0\end{array}\right| = 0

A_{33} = \left|\begin{array}{ll}1 & 0 \ 3 & 3\end{array}\right| = 3

Cofactor matrix:

C = \left[\begin{array}{rrr}-3 & 3 & -9 \ 0 & -1 & -2 \ 0 & 0 & 3\end{array}\right]

. Adjoint matrix:

\text{adj A} = C^T = \left[\begin{array}{rrr}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}\right]

.

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{-3} \left[\begin{array}{rrr}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \ -1 & 1/3 & 0 \ 3 & 2/3 & -1\end{array}\right]

Final Answer:

A^{-1} = -\frac{1}{3} \left[\begin{array}{rrr}-3 & 0 & 0 \ 3 & -1 & 0 \ -9 & -2 & 3\end{array}\right]

or

\left[\begin{array}{ccc}1 & 0 & 0 \ -1 & 1/3 & 0 \ 3 & 2/3 & -1\end{array}\right]

.

Q9EXERCISE 4.4

$\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Solution: Step 1: Calculate the determinant

|A|

. Expanding along the first row:

|A| = 2 \left|\begin{array}{cc}-1 & 0 \ 2 & 1\end{array}\right| - 1 \left|\begin{array}{cc}4 & 0 \ -7 & 1\end{array}\right| + 3 \left|\begin{array}{cc}4 & -1 \ -7 & 2\end{array}\right|

|A| = 2(-1-0) - 1(4-0) + 3(8-7)

|A| = 2(-1) - 1(4) + 3(1) = -2 - 4 + 3 = -3

Since

|A| = -3 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A.

A_{11} = \left|\begin{array}{cc}-1 & 0 \ 2 & 1\end{array}\right| = -1

A_{12} = -\left|\begin{array}{cc}4 & 0 \ -7 & 1\end{array}\right| = -4

A_{13} = \left|\begin{array}{cc}4 & -1 \ -7 & 2\end{array}\right| = 8-7 = 1

A_{21} = -\left|\begin{array}{ll}1 & 3 \ 2 & 1\end{array}\right| = -(1-6) = 5

A_{22} = \left|\begin{array}{cc}2 & 3 \ -7 & 1\end{array}\right| = 2 - (-21) = 23

A_{23} = -\left|\begin{array}{cc}2 & 1 \ -7 & 2\end{array}\right| = -(4 - (-7)) = -11

A_{31} = \left|\begin{array}{cc}1 & 3 \ -1 & 0\end{array}\right| = 0 - (-3) = 3

A_{32} = -\left|\begin{array}{ll}2 & 3 \ 4 & 0\end{array}\right| = -(0-12) = 12

A_{33} = \left|\begin{array}{cc}2 & 1 \ 4 & -1\end{array}\right| = -2-4 = -6

Cofactor matrix:

C = \left[\begin{array}{rrr}-1 & -4 & 1 \ 5 & 23 & -11 \ 3 & 12 & -6\end{array}\right]

. Adjoint matrix:

\text{adj A} = C^T = \left[\begin{array}{rrr}-1 & 5 & 3 \ -4 & 23 & 12 \ 1 & -11 & -6\end{array}\right]

.

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{-3} \left[\begin{array}{rrr}-1 & 5 & 3 \ -4 & 23 & 12 \ 1 & -11 & -6\end{array}\right]

Final Answer:

A^{-1} = -\frac{1}{3} \left[\begin{array}{rrr}-1 & 5 & 3 \ -4 & 23 & 12 \ 1 & -11 & -6\end{array}\right]

.

Q10EXERCISE 4.4

$\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Solution: Step 1: Calculate the determinant

|A|

. Expanding along the first column:

|A| = 1 \left|\begin{array}{cc}2 & -3 \ -2 & 4\end{array}\right| - 0 + 3 \left|\begin{array}{cc}-1 & 2 \ 2 & -3\end{array}\right|

|A| = 1(8-6) + 3(3-4) = 1(2) + 3(-1) = 2-3 = -1

Since

|A| = -1 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A.

A_{11} = \left|\begin{array}{cc}2 & -3 \ -2 & 4\end{array}\right| = 8-6 = 2

A_{12} = -\left|\begin{array}{cc}0 & -3 \ 3 & 4\end{array}\right| = -(0 - (-9)) = -9

A_{13} = \left|\begin{array}{cc}0 & 2 \ 3 & -2\end{array}\right| = 0-6 = -6

A_{21} = -\left|\begin{array}{cc}-1 & 2 \ -2 & 4\end{array}\right| = -(-4 - (-4)) = 0

A_{22} = \left|\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right| = 4-6 = -2

A_{23} = -\left|\begin{array}{cc}1 & -1 \ 3 & -2\end{array}\right| = -(-2 - (-3)) = -1

A_{31} = \left|\begin{array}{cc}-1 & 2 \ 2 & -3\end{array}\right| = 3-4 = -1

A_{32} = -\left|\begin{array}{cc}1 & 2 \ 0 & -3\end{array}\right| = -(-3-0) = 3

A_{33} = \left|\begin{array}{cc}1 & -1 \ 0 & 2\end{array}\right| = 2-0 = 2

Cofactor matrix:

C = \left[\begin{array}{rrr}2 & -9 & -6 \ 0 & -2 & -1 \ -1 & 3 & 2\end{array}\right]

. Adjoint matrix:

\text{adj A} = C^T = \left[\begin{array}{rrr}2 & 0 & -1 \ -9 & -2 & 3 \ -6 & -1 & 2\end{array}\right]

.

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{-1} \left[\begin{array}{rrr}2 & 0 & -1 \ -9 & -2 & 3 \ -6 & -1 & 2\end{array}\right] = \left[\begin{array}{rrr}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}\right]

Final Answer:

A^{-1} = \left[\begin{array}{rrr}-2 & 0 & 1 \ 9 & 2 & -3 \ 6 & 1 & -2\end{array}\right]

.

Q11EXERCISE 4.4

$\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$

Solution

Given: The matrix

\mathrm{A} = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}\right]

.

To Find: The inverse of matrix A,

A^{-1}

.

Solution: Step 1: Calculate the determinant

|A|

. Expanding along the first row:

|A| = 1 \left|\begin{array}{cc}\cos \alpha & \sin \alpha \ \sin \alpha & -\cos \alpha\end{array}\right| - 0 + 0

|A| = 1(-\cos^2 \alpha - \sin^2 \alpha) = -(\cos^2 \alpha + \sin^2 \alpha) = -1

Since

|A| = -1 \neq 0

, the inverse exists.

Step 2: Find the adjoint of A, adj A.

A_{11} = \left|\begin{array}{cc}\cos \alpha & \sin \alpha \ \sin \alpha & -\cos \alpha\end{array}\right| = -\cos^2 \alpha - \sin^2 \alpha = -1

A_{12} = -\left|\begin{array}{cc}0 & \sin \alpha \ 0 & -\cos \alpha\end{array}\right| = 0

A_{13} = \left|\begin{array}{cc}0 & \cos \alpha \ 0 & \sin \alpha\end{array}\right| = 0

A_{21} = -\left|\begin{array}{cc}0 & 0 \ \sin \alpha & -\cos \alpha\end{array}\right| = 0

A_{22} = \left|\begin{array}{cc}1 & 0 \ 0 & -\cos \alpha\end{array}\right| = -\cos \alpha

A_{23} = -\left|\begin{array}{cc}1 & 0 \ 0 & \sin \alpha\end{array}\right| = -\sin \alpha

A_{31} = \left|\begin{array}{cc}0 & 0 \ \cos \alpha & \sin \alpha\end{array}\right| = 0

A_{32} = -\left|\begin{array}{cc}1 & 0 \ 0 & \sin \alpha\end{array}\right| = -\sin \alpha

A_{33} = \left|\begin{array}{cc}1 & 0 \ 0 & \cos \alpha\end{array}\right| = \cos \alpha

Cofactor matrix:

C = \left[\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos \alpha & -\sin \alpha \ 0 & -\sin \alpha & \cos \alpha\end{array}\right]

. Adjoint matrix:

\text{adj A} = C^T = \left[\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos \alpha & -\sin \alpha \ 0 & -\sin \alpha & \cos \alpha\end{array}\right]

.

Step 3: Calculate the inverse

A^{-1}

.

A^{-1} = \frac{1}{|A|} \text{adj A} = \frac{1}{-1} \left[\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos \alpha & -\sin \alpha \ 0 & -\sin \alpha & \cos \alpha\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}\right]

Notice that

A^{-1} = A

. This type of matrix is called an involutory matrix.

Final Answer:

A^{-1} = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}\right]

.

Q12EXERCISE 4.4

Let $\mathrm{A}=\begin{array}{ll}3 & 7 \ 2 & 5\end{array}$ and $\mathrm{B}=\begin{array}{ll}6 & 8 \ 7 & 9\end{array}$ . Verify that $(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$ .

Solution

Given: Matrices

\mathrm{A}=\left[\begin{array}{ll}3 & 7 \ 2 & 5\end{array}\right]

and

\mathrm{B}=\left[\begin{array}{ll}6 & 8 \ 7 & 9\end{array}\right]

.

To Verify:

(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}

.

Solution: LHS Calculation: Step 1: Find the product AB.

\mathrm{AB} = \left[\begin{array}{ll}3 & 7 \ 2 & 5\end{array}\right] \left[\begin{array}{ll}6 & 8 \ 7 & 9\end{array}\right] = \left[\begin{array}{ll}3(6)+7(7) & 3(8)+7(9) \ 2(6)+5(7) & 2(8)+5(9)\end{array}\right]

= \left[\begin{array}{ll}18+49 & 24+63 \ 12+35 & 16+45\end{array}\right] = \left[\begin{array}{ll}67 & 87 \ 47 & 61\end{array}\right]

Step 2: Find the determinant of AB.

|\mathrm{AB}| = (67)(61) - (87)(47) = 4087 - 4089 = -2

Step 3: Find the inverse of AB.

(\mathrm{AB})^{-1} = \frac{1}{|\mathrm{AB}|} \text{adj(AB)} = \frac{1}{-2} \left[\begin{array}{cc}61 & -87 \ -47 & 67\end{array}\right]

RHS Calculation: Step 4: Find the inverse of A.

|A| = (3)(5) - (7)(2) = 15 - 14 = 1

A^{-1} = \frac{1}{1} \left[\begin{array}{cc}5 & -7 \ -2 & 3\end{array}\right] = \left[\begin{array}{cc}5 & -7 \ -2 & 3\end{array}\right]

Step 5: Find the inverse of B.

|B| = (6)(9) - (8)(7) = 54 - 56 = -2

B^{-1} = \frac{1}{-2} \left[\begin{array}{cc}9 & -8 \ -7 & 6\end{array}\right]

Step 6: Calculate the product

B^{-1}A^{-1}

.

B^{-1}A^{-1} = \frac{1}{-2} \left[\begin{array}{cc}9 & -8 \ -7 & 6\end{array}\right] \left[\begin{array}{cc}5 & -7 \ -2 & 3\end{array}\right]

= \frac{1}{-2} \left[\begin{array}{ll}9(5)+(-8)(-2) & 9(-7)+(-8)(3) \ -7(5)+6(-2) & -7(-7)+6(3)\end{array}\right]

= \frac{1}{-2} \left[\begin{array}{ll}45+16 & -63-24 \ -35-12 & 49+18\end{array}\right] = \frac{1}{-2} \left[\begin{array}{cc}61 & -87 \ -47 & 67\end{array}\right]

Verification: Comparing the results from Step 3 and Step 6, we find that: LHS =

(AB)^{-1} = \frac{1}{-2} \left[\begin{array}{cc}61 & -87 \ -47 & 67\end{array}\right]

RHS =

B^{-1}A^{-1} = \frac{1}{-2} \left[\begin{array}{cc}61 & -87 \ -47 & 67\end{array}\right]

Since LHS = RHS, the property is verified.

Hence Verified.

Q13EXERCISE 4.4

If $\mathrm{A}=\begin{array}{cc}3 & 1 \ -1 & 2\end{array}$ , show that $\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=\mathrm{O}$ . Hence find $\mathrm{A}^{-1}$ .

Solution

Given: Matrix

\mathrm{A}=\left[\begin{array}{cc}3 & 1 \ -1 & 2\end{array}\right]

.

Part 1: To Show:

\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=\mathrm{O}

.

Proof: First, calculate

A^2

.

A^2 = A \cdot A = \left[\begin{array}{cc}3 & 1 \ -1 & 2\end{array}\right] \left[\begin{array}{cc}3 & 1 \ -1 & 2\end{array}\right] = \left[\begin{array}{cc}3(3)+1(-1) & 3(1)+1(2) \ -1(3)+2(-1) & -1(1)+2(2)\end{array}\right]

A^2 = \left[\begin{array}{cc}9-1 & 3+2 \ -3-2 & -1+4\end{array}\right] = \left[\begin{array}{cc}8 & 5 \ -5 & 3\end{array}\right]

Now, substitute into the expression

\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}

:

= \left[\begin{array}{cc}8 & 5 \ -5 & 3\end{array}\right] - 5\left[\begin{array}{cc}3 & 1 \ -1 & 2\end{array}\right] + 7\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]

= \left[\begin{array}{cc}8 & 5 \ -5 & 3\end{array}\right] - \left[\begin{array}{cc}15 & 5 \ -5 & 10\end{array}\right] + \left[\begin{array}{ll}7 & 0 \ 0 & 7\end{array}\right]

= \left[\begin{array}{cc}8-15+7 & 5-5+0 \ -5-(-5)+0 & 3-10+7\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right] = O

Hence Proved.

Part 2: To Find:

A^{-1}

using the equation

\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=\mathrm{O}

.

Solution: We start with the proven equation:

A^2 - 5A + 7I = O

To find

A^{-1}

, we first check if A is invertible.

|A| = (3)(2) - (1)(-1) = 6+1=7 \neq 0

. So,

A^{-1}

exists. Post-multiplying the equation by

A^{-1}

:

(A^2 - 5A + 7I)A^{-1} = O \cdot A^{-1}

A^2 A^{-1} - 5A A^{-1} + 7I A^{-1} = O

Using the properties

A^2 A^{-1} = A(AA^{-1}) = AI = A

,

AA^{-1} = I

, and

IA^{-1} = A^{-1}

:

A - 5I + 7A^{-1} = O

7A^{-1} = 5I - A

A^{-1} = \frac{1}{7}(5I - A)

Now, substitute the matrices for I and A:

A^{-1} = \frac{1}{7} \left( 5\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right] - \left[\begin{array}{cc}3 & 1 \ -1 & 2\end{array}\right] \right)

A^{-1} = \frac{1}{7} \left( \left[\begin{array}{ll}5 & 0 \ 0 & 5\end{array}\right] - \left[\begin{array}{cc}3 & 1 \ -1 & 2\end{array}\right] \right)

A^{-1} = \frac{1}{7} \left[\begin{array}{cc}5-3 & 0-1 \ 0-(-1) & 5-2\end{array}\right] = \frac{1}{7} \left[\begin{array}{cc}2 & -1 \ 1 & 3\end{array}\right]

Final Answer:

A^{-1} = \frac{1}{7} \left[\begin{array}{cc}2 & -1 \ 1 & 3\end{array}\right]

.

Q14EXERCISE 4.4

For the matrix $\mathrm{A}=\begin{array}{ll}3 & 2 \ 1 & 1\end{array}$ , find the numbers $a$ and $b$ such that $\mathrm{A}^{2}+a \mathrm{~A}+b \mathrm{I}=\mathrm{O}$ .

Solution

Given: Matrix

\mathrm{A}=\left[\begin{array}{ll}3 & 2 \ 1 & 1\end{array}\right]

and the equation

\mathrm{A}^{2}+a \mathrm{~A}+b \mathrm{I}=\mathrm{O}

.

To Find: The values of

a

and

b

.

Solution: Step 1: Calculate

A^2

.

A^2 = A \cdot A = \left[\begin{array}{ll}3 & 2 \ 1 & 1\end{array}\right] \left[\begin{array}{ll}3 & 2 \ 1 & 1\end{array}\right] = \left[\begin{array}{ll}3(3)+2(1) & 3(2)+2(1) \ 1(3)+1(1) & 1(2)+1(1)\end{array}\right]

A^2 = \left[\begin{array}{ll}9+2 & 6+2 \ 3+1 & 2+1\end{array}\right] = \left[\begin{array}{cc}11 & 8 \ 4 & 3\end{array}\right]

Step 2: Substitute

A^2

, A, and I into the given equation.

A^2 + aA + bI = O

\left[\begin{array}{cc}11 & 8 \ 4 & 3\end{array}\right] + a\left[\begin{array}{ll}3 & 2 \ 1 & 1\end{array}\right] + b\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

\left[\begin{array}{cc}11 & 8 \ 4 & 3\end{array}\right] + \left[\begin{array}{cc}3a & 2a \ a & a\end{array}\right] + \left[\begin{array}{ll}b & 0 \ 0 & b\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

Step 3: Combine the matrices on the LHS.

\left[\begin{array}{cc}11+3a+b & 8+2a \ 4+a & 3+a+b\end{array}\right] = \left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right]

Step 4: Equate the corresponding elements to form a system of equations.

$11 + 3a + b = 0$
$8 + 2a = 0$
$4 + a = 0$
$3 + a + b = 0$

From equation (3), we get:

a = -4

Substitute

a = -4

into equation (2) to check for consistency:

8 + 2(-4) = 8 - 8 = 0

. This is consistent.

Now, substitute

a = -4

into equation (4) to find

b

:

3 + (-4) + b = 0

-1 + b = 0

b = 1

Substitute

a = -4

and

b = 1

into equation (1) to check for consistency:

11 + 3(-4) + 1 = 11 - 12 + 1 = 0

. This is also consistent.

So, the values are

a = -4

and

b = 1

.

Final Answer:

a = -4, b = 1

.

Q15EXERCISE 4.4

For the matrix $\mathrm{A}=\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}$ Show that $\mathrm{A}^{3}-6 \mathrm{~A}^{2}+5 \mathrm{~A}+11 \mathrm{I}=\mathrm{O}$ . Hence, find $\mathrm{A}^{-1}$ .

Solution

Given: Matrix

\mathrm{A}=\left[\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}\right]

.

Part 1: To Show:

\mathrm{A}^{3}-6 \mathrm{~A}^{2}+5 \mathrm{~A}+11 \mathrm{I}=\mathrm{O}

.

Proof: Step 1: Calculate

A^2

.

A^2 = \left[\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}\right] \left[\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}\right] = \left[\begin{array}{ccc}1+1+2 & 1+2-1 & 1-3+3 \ 1+2-6 & 1+4+3 & 1-6-9 \ 2-1+6 & 2-2-3 & 2+3+9\end{array}\right] = \left[\begin{array}{ccc}4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14\end{array}\right]

Step 2: Calculate

A^3 = A^2 \cdot A

.

A^3 = \left[\begin{array}{ccc}4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14\end{array}\right] \left[\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}\right] = \left[\begin{array}{ccc}4+2+2 & 4+4-1 & 4-6+3 \ -3+8-28 & -3+16+14 & -3-24-42 \ 7-3+28 & 7-6-14 & 7+9+42\end{array}\right] = \left[\begin{array}{ccc}8 & 7 & 1 \ -23 & 27 & -69 \ 32 & -13 & 58\end{array}\right]

Step 3: Substitute into the expression

\mathrm{A}^{3}-6 \mathrm{~A}^{2}+5 \mathrm{~A}+11 \mathrm{I}

.

A^3 = \left[\begin{array}{ccc}8 & 7 & 1 \ -23 & 27 & -69 \ 32 & -13 & 58\end{array}\right]

-6A^2 = -6 \left[\begin{array}{ccc}4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14\end{array}\right] = \left[\begin{array}{ccc}-24 & -12 & -6 \ 18 & -48 & 84 \ -42 & 18 & -84\end{array}\right]

5A = 5 \left[\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}\right] = \left[\begin{array}{ccc}5 & 5 & 5 \ 5 & 10 & -15 \ 10 & -5 & 15\end{array}\right]

11I = 11 \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}11 & 0 & 0 \ 0 & 11 & 0 \ 0 & 0 & 11\end{array}\right]

Summing these matrices:

\left[\begin{array}{ccc}8-24+5+11 & 7-12+5+0 & 1-6+5+0 \ -23+18+5+0 & 27-48+10+11 & -69+84-15+0 \ 32-42+10+0 & -13+18-5+0 & 58-84+15+11\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0\end{array}\right] = O

Hence Proved.

Part 2: To Find:

A^{-1}

.

Solution: Start with the equation:

A^3 - 6A^2 + 5A + 11I = O

. Post-multiply by

A^{-1}

(we assume it exists, and will verify

|A| \neq 0

later).

A^2 - 6A + 5I + 11A^{-1} = O

11A^{-1} = -A^2 + 6A - 5I

A^{-1} = \frac{1}{11}(-A^2 + 6A - 5I)

Substitute the matrices:

A^{-1} = \frac{1}{11} \left( -\left[\begin{array}{ccc}4 & 2 & 1 \ -3 & 8 & -14 \ 7 & -3 & 14\end{array}\right] + 6\left[\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}\right] - 5\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] \right)

A^{-1} = \frac{1}{11} \left( \left[\begin{array}{ccc}-4 & -2 & -1 \ 3 & -8 & 14 \ -7 & 3 & -14\end{array}\right] + \left[\begin{array}{ccc}6 & 6 & 6 \ 6 & 12 & -18 \ 12 & -6 & 18\end{array}\right] - \left[\begin{array}{ccc}5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5\end{array}\right] \right)

A^{-1} = \frac{1}{11} \left[\begin{array}{ccc}-4+6-5 & -2+6-0 & -1+6-0 \ 3+6-0 & -8+12-5 & 14-18-0 \ -7+12-0 & 3-6-0 & -14+18-5\end{array}\right] = \frac{1}{11} \left[\begin{array}{ccc}-3 & 4 & 5 \ 9 & -1 & -4 \ 5 & -3 & -1\end{array}\right]

Let's check

|A|

to confirm the inverse exists.

|A| = 1(6-3) - 1(3 - (-6)) + 1(-1-4) = 3 - 9 - 5 = -11 \neq 0

.

Final Answer:

A^{-1} = \frac{1}{11} \left[\begin{array}{ccc}-3 & 4 & 5 \ 9 & -1 & -4 \ 5 & -3 & -1\end{array}\right]

.

Q16EXERCISE 4.4

If $\mathrm{A}=\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}$ Verify that $\mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I}=\mathrm{O}$ and hence find $\mathrm{A}^{-1}$

Solution

Given: Matrix

\mathrm{A}=\left[\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}\right]

.

Part 1: To Verify:

\mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I}=\mathrm{O}

.

Verification: Step 1: Calculate

A^2

.

A^2 = \left[\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}\right] \left[\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}\right] = \left[\begin{array}{ccc}4+1+1 & -2-2-1 & 2+1+2 \ -2-2-1 & 1+4+1 & -1-2-2 \ 2+1+2 & -1-2-2 & 1+1+4\end{array}\right] = \left[\begin{array}{ccc}6 & -5 & 5 \ -5 & 6 & -5 \ 5 & -5 & 6\end{array}\right]

Step 2: Calculate

A^3 = A^2 \cdot A

.

A^3 = \left[\begin{array}{ccc}6 & -5 & 5 \ -5 & 6 & -5 \ 5 & -5 & 6\end{array}\right] \left[\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}\right] = \left[\begin{array}{ccc}12+5+5 & -6-10-5 & 6+5+10 \ -10-6-5 & 5+12+5 & -5-6-10 \ 10+5+6 & -5-10-6 & 5+5+12\end{array}\right] = \left[\begin{array}{ccc}22 & -21 & 21 \ -21 & 22 & -21 \ 21 & -21 & 22\end{array}\right]

Step 3: Substitute into the expression

\mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I}

.

A^3 = \left[\begin{array}{ccc}22 & -21 & 21 \ -21 & 22 & -21 \ 21 & -21 & 22\end{array}\right]

-6A^2 = -6 \left[\begin{array}{ccc}6 & -5 & 5 \ -5 & 6 & -5 \ 5 & -5 & 6\end{array}\right] = \left[\begin{array}{ccc}-36 & 30 & -30 \ 30 & -36 & 30 \ -30 & 30 & -36\end{array}\right]

9A = 9 \left[\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}\right] = \left[\begin{array}{ccc}18 & -9 & 9 \ -9 & 18 & -9 \ 9 & -9 & 18\end{array}\right]

-4I = -4 \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}-4 & 0 & 0 \ 0 & -4 & 0 \ 0 & 0 & -4\end{array}\right]

Summing these matrices:

\left[\begin{array}{ccc}22-36+18-4 & -21+30-9+0 & 21-30+9+0 \ -21+30-9+0 & 22-36+18-4 & -21+30-9+0 \ 21-30+9+0 & -21+30-9+0 & 22-36+18-4\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0\end{array}\right] = O

Hence Verified.

Part 2: To Find:

A^{-1}

.

Solution: Start with the equation:

A^3 - 6A^2 + 9A - 4I = O

. Post-multiply by

A^{-1}

:

A^2 - 6A + 9I - 4A^{-1} = O

4A^{-1} = A^2 - 6A + 9I

A^{-1} = \frac{1}{4}(A^2 - 6A + 9I)

Substitute the matrices:

A^{-1} = \frac{1}{4} \left( \left[\begin{array}{ccc}6 & -5 & 5 \ -5 & 6 & -5 \ 5 & -5 & 6\end{array}\right] - 6\left[\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}\right] + 9\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] \right)

A^{-1} = \frac{1}{4} \left( \left[\begin{array}{ccc}6 & -5 & 5 \ -5 & 6 & -5 \ 5 & -5 & 6\end{array}\right] - \left[\begin{array}{ccc}12 & -6 & 6 \ -6 & 12 & -6 \ 6 & -6 & 12\end{array}\right] + \left[\begin{array}{ccc}9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9\end{array}\right] \right)

A^{-1} = \frac{1}{4} \left[\begin{array}{ccc}6-12+9 & -5+6+0 & 5-6+0 \ -5+6+0 & 6-12+9 & -5+6+0 \ 5-6+0 & -5+6+0 & 6-12+9\end{array}\right] = \frac{1}{4} \left[\begin{array}{ccc}3 & 1 & -1 \ 1 & 3 & 1 \ -1 & 1 & 3\end{array}\right]

Final Answer:

A^{-1} = \frac{1}{4} \left[\begin{array}{ccc}3 & 1 & -1 \ 1 & 3 & 1 \ -1 & 1 & 3\end{array}\right]

.

Q17EXERCISE 4.4

Let A be a nonsingular square matrix of order $3 \times 3$ . Then $|\operatorname{adj} \mathrm{A}|$ is equal to

(A)

|\mathrm{A}|

(B)

|\mathrm{A}|^{2}

(C)

|\mathrm{A}|^{3}

(D)

3|\mathrm{~A}|

Solution

Given: A is a nonsingular square matrix of order

n=3

.

To Find: The value of

|\operatorname{adj} \mathrm{A}|

.

Formula: For any square matrix A of order

n

, we have the property:

|\operatorname{adj} \mathrm{A}| = |\mathrm{A}|^{n-1}

Derivation of the formula: We know that

\mathrm{A}(\operatorname{adj} \mathrm{A}) = |\mathrm{A}| \mathrm{I}

. Taking the determinant on both sides:

|\mathrm{A}(\operatorname{adj} \mathrm{A})| = ||\mathrm{A}| \mathrm{I}|

Using the property

|XY| = |X||Y|

:

|\mathrm{A}| |\operatorname{adj} \mathrm{A}| = |\mathrm{A}|^n |\mathrm{I}|

Since

|I|=1

:

|\mathrm{A}| |\operatorname{adj} \mathrm{A}| = |\mathrm{A}|^n

Since A is nonsingular,

|A| \neq 0

, so we can divide by

|A|

:

|\operatorname{adj} \mathrm{A}| = |\mathrm{A}|^{n-1}

Solution: In this problem, the order of the matrix is

n=3

. Substituting

n=3

into the formula:

|\operatorname{adj} \mathrm{A}| = |\mathrm{A}|^{3-1} = |\mathrm{A}|^2

This corresponds to option (B).

Final Answer: The correct option is (B).

Q18EXERCISE 4.4

If A is an invertible matrix of order 2, then $\operatorname{det}\left(\mathrm{A}^{-1}\right)$ is equal to

(A)

\operatorname{det}(\mathrm{A})

(B)

\frac{1}{\operatorname{det}(\mathrm{A})}

(C)

1

(D)

0

Solution

Given: A is an invertible matrix of order 2.

To Find: The value of

\operatorname{det}(A^{-1})

.

Property: For an invertible matrix A, we have the definition:

A A^{-1} = I

where I is the identity matrix.

Solution: Taking the determinant on both sides of the equation

A A^{-1} = I

:

\operatorname{det}(A A^{-1}) = \operatorname{det}(I)

Using the property

\operatorname{det}(XY) = \operatorname{det}(X) \operatorname{det}(Y)

:

\operatorname{det}(A) \operatorname{det}(A^{-1}) = \operatorname{det}(I)

We know that the determinant of the identity matrix is 1, so

\operatorname{det}(I) = 1

.

\operatorname{det}(A) \operatorname{det}(A^{-1}) = 1

Since A is invertible, we know that

\operatorname{det}(A) \neq 0

. Therefore, we can divide both sides by

\operatorname{det}(A)

:

\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}

This corresponds to option (B).

Final Answer: The correct option is (B).

Q1EXERCISE 4.5

Examine the consistency of the system of equations in Exercises 1 to 6. $x+2 y=2$ $2 x+3 y=3$

Solution

Given: The system of linear equations:

x+2 y=2

2 x+3 y=3

To Examine: The consistency of the system.

Concept: A system of equations

AX=B

is consistent if

|A| \neq 0

. If

|A| = 0

, we must check

(\text{adj A})B

. If

(\text{adj A})B \neq O

, the system is inconsistent. If

(\text{adj A})B = O

, the system may be consistent (infinitely many solutions) or inconsistent (no solution).

Solution: The system can be written in matrix form

AX=B

, where:

A = \left[\begin{array}{ll}1 & 2 \ 2 & 3\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{l}2 \ 3\end{array}\right]

First, calculate the determinant of the coefficient matrix A:

|A| = \left|\begin{array}{ll}1 & 2 \ 2 & 3\end{array}\right| = (1)(3) - (2)(2) = 3 - 4 = -1

Since

|A| = -1 \neq 0

, the matrix A is non-singular. Therefore, the system of equations has a unique solution and is consistent.

Final Answer: The system of equations is consistent.

Q2EXERCISE 4.5

$2 x-y=5$ $x+y=4$

Solution

Given: The system of linear equations:

2x - y = 5

x + y = 4

To Examine: The consistency of the system.

Solution: The system can be written in matrix form

AX=B

, where:

A = \left[\begin{array}{cc}2 & -1 \ 1 & 1\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{l}5 \ 4\end{array}\right]

Calculate the determinant of the coefficient matrix A:

|A| = \left|\begin{array}{cc}2 & -1 \ 1 & 1\end{array}\right| = (2)(1) - (-1)(1) = 2 - (-1) = 3

Since

|A| = 3 \neq 0

, the matrix A is non-singular. Therefore, the system of equations has a unique solution and is consistent.

Final Answer: The system of equations is consistent.

Q3EXERCISE 4.5

$x+3 y=5$ $2 x+6 y=8$

Solution

Given: The system of linear equations:

x+3y=5

2x+6y=8

To Examine: The consistency of the system.

Solution: The system can be written in matrix form

AX=B

, where:

A = \left[\begin{array}{ll}1 & 3 \ 2 & 6\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{l}5 \ 8\end{array}\right]

Calculate the determinant of the coefficient matrix A:

|A| = \left|\begin{array}{ll}1 & 3 \ 2 & 6\end{array}\right| = (1)(6) - (3)(2) = 6 - 6 = 0

Since

|A| = 0

, the matrix A is singular. We need to calculate

(\text{adj A})B

. First, find adj A:

\text{adj A} = \left[\begin{array}{cc}6 & -3 \ -2 & 1\end{array}\right]

Now, calculate

(\text{adj A})B

:

(\text{adj A})B = \left[\begin{array}{cc}6 & -3 \ -2 & 1\end{array}\right] \left[\begin{array}{l}5 \ 8\end{array}\right] = \left[\begin{array}{l}6(5)+(-3)(8) \ -2(5)+1(8)\end{array}\right] = \left[\begin{array}{l}30-24 \ -10+8\end{array}\right] = \left[\begin{array}{c}6 \ -2\end{array}\right]

Since

(\text{adj A})B \neq O

(the zero matrix), the system of equations has no solution.

Final Answer: The system of equations is inconsistent.

Q4EXERCISE 4.5

$x+y+z=1$ $2 x+3 y+2 z=2$ $a x+a y+2 a z=4$

Solution

Given: The system of linear equations:

x+y+z=1

2x+3y+2z=2

ax+ay+2az=4

To Examine: The consistency of the system.

Solution: The system can be written in matrix form

AX=B

, where:

A = \left[\begin{array}{ccc}1 & 1 & 1 \ 2 & 3 & 2 \ a & a & 2a\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{l}1 \ 2 \ 4\end{array}\right]

Calculate the determinant of the coefficient matrix A:

|A| = 1 \left|\begin{array}{cc}3 & 2 \ a & 2a\end{array}\right| - 1 \left|\begin{array}{cc}2 & 2 \ a & 2a\end{array}\right| + 1 \left|\begin{array}{cc}2 & 3 \ a & a\end{array}\right|

|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)

|A| = 4a - 2a - a = a

Case 1: If

a \neq 0

. Then

|A| \neq 0

. The system has a unique solution and is consistent.

Case 2: If

a = 0

. Then

|A| = 0

. The system becomes:

x+y+z=1

2x+3y+2z=2

0=4

The last equation

0=4

is a contradiction, which means the system has no solution. Therefore, if

a=0

, the system is inconsistent.

Assuming

a

is a non-zero constant from the context of the problem setup.

Final Answer: The system is consistent if

a \neq 0

and inconsistent if

a = 0

.

Q5EXERCISE 4.5

$3 x-y-2 z=2$ $2 y-z=-1$ $3 x-5 y=3$

Solution

Given: The system of linear equations:

3x - y - 2z = 2

0x + 2y - z = -1

3x - 5y + 0z = 3

To Examine: The consistency of the system.

Solution: The system can be written in matrix form

AX=B

, where:

A = \left[\begin{array}{ccc}3 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{c}2 \ -1 \ 3\end{array}\right]

Calculate the determinant of the coefficient matrix A:

|A| = 3 \left|\begin{array}{cc}2 & -1 \ -5 & 0\end{array}\right| - (-1) \left|\begin{array}{cc}0 & -1 \ 3 & 0\end{array}\right| + (-2) \left|\begin{array}{cc}0 & 2 \ 3 & -5\end{array}\right|

|A| = 3(0 - 5) + 1(0 - (-3)) - 2(0 - 6)

|A| = 3(-5) + 1(3) - 2(-6) = -15 + 3 + 12 = 0

Since

|A| = 0

, we need to calculate

(\text{adj A})B

. First, find adj A:

A_{11} = \left|\begin{array}{cc}2 & -1 \ -5 & 0\end{array}\right| = -5

A_{12} = -\left|\begin{array}{cc}0 & -1 \ 3 & 0\end{array}\right| = -3

A_{13} = \left|\begin{array}{cc}0 & 2 \ 3 & -5\end{array}\right| = -6

A_{21} = -\left|\begin{array}{cc}-1 & -2 \ -5 & 0\end{array}\right| = -(0 - 10) = 10

A_{22} = \left|\begin{array}{cc}3 & -2 \ 3 & 0\end{array}\right| = 0 - (-6) = 6

A_{23} = -\left|\begin{array}{cc}3 & -1 \ 3 & -5\end{array}\right| = -(-15 - (-3)) = -(-12) = 12

A_{31} = \left|\begin{array}{cc}-1 & -2 \ 2 & -1\end{array}\right| = 1 - (-4) = 5

A_{32} = -\left|\begin{array}{cc}3 & -2 \ 0 & -1\end{array}\right| = -(-3 - 0) = 3

A_{33} = \left|\begin{array}{cc}3 & -1 \ 0 & 2\end{array}\right| = 6 - 0 = 6

\text{adj A} = \left[\begin{array}{ccc}-5 & 10 & 5 \ -3 & 6 & 3 \ -6 & 12 & 6\end{array}\right]

Now, calculate

(\text{adj A})B

:

(\text{adj A})B = \left[\begin{array}{ccc}-5 & 10 & 5 \ -3 & 6 & 3 \ -6 & 12 & 6\end{array}\right] \left[\begin{array}{c}2 \ -1 \ 3\end{array}\right] = \left[\begin{array}{c}-10-10+15 \ -6-6+9 \ -12-12+18\end{array}\right] = \left[\begin{array}{c}-5 \ -3 \ -6\end{array}\right]

Since

|A| = 0

and

(\text{adj A})B \neq O

, the system of equations has no solution.

Final Answer: The system of equations is inconsistent.

Q6EXERCISE 4.5

$5 x-y+4 z=5$ $2 x+3 y+5 z=2$ $5 x-2 y+6 z=-1$

Solution

Given: The system of linear equations:

5x-y+4z=5

2x+3y+5z=2

5x-2y+6z=-1

To Examine: The consistency of the system.

Solution: The system can be written in matrix form

AX=B

, where:

A = \left[\begin{array}{ccc}5 & -1 & 4 \ 2 & 3 & 5 \ 5 & -2 & 6\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{c}5 \ 2 \ -1\end{array}\right]

Calculate the determinant of the coefficient matrix A:

|A| = 5 \left|\begin{array}{cc}3 & 5 \ -2 & 6\end{array}\right| - (-1) \left|\begin{array}{ll}2 & 5 \ 5 & 6\end{array}\right| + 4 \left|\begin{array}{cc}2 & 3 \ 5 & -2\end{array}\right|

|A| = 5(18 - (-10)) + 1(12 - 25) + 4(-4 - 15)

|A| = 5(28) + 1(-13) + 4(-19) = 140 - 13 - 76 = 51

Since

|A| = 51 \neq 0

, the matrix A is non-singular. Therefore, the system of equations has a unique solution and is consistent.

Final Answer: The system of equations is consistent.

Q7EXERCISE 4.5

Solve system of linear equations, using matrix method, in Exercises 7 to 14. 7. $5 x+2 y=4$ $7 x+3 y=5$

Solution

Given: The system of linear equations:

5x+2y=4

7x+3y=5

To Find: The solution for

x

and

y

using the matrix method.

Method: The solution is given by

X = A^{-1}B

, where

A^{-1} = \frac{1}{|A|}\text{adj A}

.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ll}5 & 2 \ 7 & 3\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{l}4 \ 5\end{array}\right]

Step 1: Find

|A|

.

|A| = (5)(3) - (2)(7) = 15 - 14 = 1

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

\text{adj A} = \left[\begin{array}{cc}3 & -2 \ -7 & 5\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{|A|}\text{adj A} = \frac{1}{1} \left[\begin{array}{cc}3 & -2 \ -7 & 5\end{array}\right] = \left[\begin{array}{cc}3 & -2 \ -7 & 5\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \left[\begin{array}{l}x \ y\end{array}\right] = \left[\begin{array}{cc}3 & -2 \ -7 & 5\end{array}\right] \left[\begin{array}{l}4 \ 5\end{array}\right]

= \left[\begin{array}{l}3(4) + (-2)(5) \ -7(4) + 5(5)\end{array}\right] = \left[\begin{array}{c}12-10 \ -28+25\end{array}\right] = \left[\begin{array}{c}2 \ -3\end{array}\right]

Therefore,

x=2

and

y=-3

.

Final Answer:

x=2, y=-3

.

Q8EXERCISE 4.5

$2 x-y=-2$ $3 x+4 y=3$

Solution

Given: The system of linear equations:

2x-y=-2

3x+4y=3

To Find: The solution for

x

and

y

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{cc}2 & -1 \ 3 & 4\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{c}-2 \ 3\end{array}\right]

Step 1: Find

|A|

.

|A| = (2)(4) - (-1)(3) = 8 - (-3) = 11

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

\text{adj A} = \left[\begin{array}{cc}4 & 1 \ -3 & 2\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{|A|}\text{adj A} = \frac{1}{11} \left[\begin{array}{cc}4 & 1 \ -3 & 2\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \left[\begin{array}{l}x \ y\end{array}\right] = \frac{1}{11} \left[\begin{array}{cc}4 & 1 \ -3 & 2\end{array}\right] \left[\begin{array}{c}-2 \ 3\end{array}\right]

= \frac{1}{11} \left[\begin{array}{l}4(-2) + 1(3) \ -3(-2) + 2(3)\end{array}\right] = \frac{1}{11} \left[\begin{array}{c}-8+3 \ 6+6\end{array}\right] = \frac{1}{11} \left[\begin{array}{c}-5 \ 12\end{array}\right] = \left[\begin{array}{c}-5/11 \ 12/11\end{array}\right]

Therefore,

x = -5/11

and

y = 12/11

.

Final Answer:

x = -5/11, y = 12/11

.

Q9EXERCISE 4.5

$4 x-3 y=3$ $3 x-5 y=7$

Solution

Given: The system of linear equations:

4x-3y=3

3x-5y=7

To Find: The solution for

x

and

y

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ll}4 & -3 \ 3 & -5\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{l}3 \ 7\end{array}\right]

Step 1: Find

|A|

.

|A| = (4)(-5) - (-3)(3) = -20 - (-9) = -20 + 9 = -11

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

\text{adj A} = \left[\begin{array}{ll}-5 & 3 \ -3 & 4\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{|A|}\text{adj A} = \frac{1}{-11} \left[\begin{array}{ll}-5 & 3 \ -3 & 4\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \left[\begin{array}{l}x \ y\end{array}\right] = -\frac{1}{11} \left[\begin{array}{ll}-5 & 3 \ -3 & 4\end{array}\right] \left[\begin{array}{l}3 \ 7\end{array}\right]

= -\frac{1}{11} \left[\begin{array}{l}-5(3) + 3(7) \ -3(3) + 4(7)\end{array}\right] = -\frac{1}{11} \left[\begin{array}{c}-15+21 \ -9+28\end{array}\right] = -\frac{1}{11} \left[\begin{array}{c}6 \ 19\end{array}\right] = \left[\begin{array}{c}-6/11 \ -19/11\end{array}\right]

Therefore,

x = -6/11

and

y = -19/11

.

Final Answer:

x = -6/11, y = -19/11

.

Q10EXERCISE 4.5

$5 x+2 y=3$ $3 x+2 y=5$

Solution

Given: The system of linear equations:

5x+2y=3

3x+2y=5

To Find: The solution for

x

and

y

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ll}5 & 2 \ 3 & 2\end{array}\right], X = \left[\begin{array}{l}x \ y\end{array}\right], B = \left[\begin{array}{l}3 \ 5\end{array}\right]

Step 1: Find

|A|

.

|A| = (5)(2) - (2)(3) = 10 - 6 = 4

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

\text{adj A} = \left[\begin{array}{cc}2 & -2 \ -3 & 5\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{|A|}\text{adj A} = \frac{1}{4} \left[\begin{array}{cc}2 & -2 \ -3 & 5\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \left[\begin{array}{l}x \ y\end{array}\right] = \frac{1}{4} \left[\begin{array}{cc}2 & -2 \ -3 & 5\end{array}\right] \left[\begin{array}{l}3 \ 5\end{array}\right]

= \frac{1}{4} \left[\begin{array}{l}2(3) + (-2)(5) \ -3(3) + 5(5)\end{array}\right] = \frac{1}{4} \left[\begin{array}{c}6-10 \ -9+25\end{array}\right] = \frac{1}{4} \left[\begin{array}{c}-4 \ 16\end{array}\right] = \left[\begin{array}{c}-1 \ 4\end{array}\right]

Therefore,

x=-1

and

y=4

.

Final Answer:

x=-1, y=4

.

Q11EXERCISE 4.5

$2 x+y+z=1$ $x-2 y-z=\frac{3}{2}$ $3 y-5 z=9$

Solution

Given: The system of linear equations:

2x+y+z=1

x-2y-z=3/2

0x+3y-5z=9

To avoid fractions, we can multiply the second equation by 2:

2x+y+z=1

2x-4y-2z=3

3y-5z=9

To Find: The solution for

x, y, z

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ccc}2 & 1 & 1 \ 2 & -4 & -2 \ 0 & 3 & -5\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{l}1 \ 3 \ 9\end{array}\right]

Step 1: Find

|A|

.

|A| = 2 \left|\begin{array}{cc}-4 & -2 \ 3 & -5\end{array}\right| - 1 \left|\begin{array}{cc}2 & -2 \ 0 & -5\end{array}\right| + 1 \left|\begin{array}{cc}2 & -4 \ 0 & 3\end{array}\right|

|A| = 2(20 - (-6)) - 1(-10 - 0) + 1(6 - 0)

|A| = 2(26) - 1(-10) + 1(6) = 52 + 10 + 6 = 68

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

A_{11} = \left|\begin{array}{cc}-4 & -2 \ 3 & -5\end{array}\right| = 20+6=26

A_{12} = -\left|\begin{array}{cc}2 & -2 \ 0 & -5\end{array}\right| = -(-10)=10

A_{13} = \left|\begin{array}{cc}2 & -4 \ 0 & 3\end{array}\right| = 6

A_{21} = -\left|\begin{array}{cc}1 & 1 \ 3 & -5\end{array}\right| = -(-5-3)=8

A_{22} = \left|\begin{array}{cc}2 & 1 \ 0 & -5\end{array}\right| = -10

A_{23} = -\left|\begin{array}{cc}2 & 1 \ 0 & 3\end{array}\right| = -6

A_{31} = \left|\begin{array}{cc}1 & 1 \ -4 & -2\end{array}\right| = -2-(-4)=2

A_{32} = -\left|\begin{array}{cc}2 & 1 \ 2 & -2\end{array}\right| = -(-4-2)=6

A_{33} = \left|\begin{array}{cc}2 & 1 \ 2 & -4\end{array}\right| = -8-2=-10

\text{adj A} = \left[\begin{array}{ccc}26 & 8 & 2 \ 10 & -10 & 6 \ 6 & -6 & -10\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{68} \left[\begin{array}{ccc}26 & 8 & 2 \ 10 & -10 & 6 \ 6 & -6 & -10\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \frac{1}{68} \left[\begin{array}{ccc}26 & 8 & 2 \ 10 & -10 & 6 \ 6 & -6 & -10\end{array}\right] \left[\begin{array}{l}1 \ 3 \ 9\end{array}\right] = \frac{1}{68} \left[\begin{array}{c}26(1)+8(3)+2(9) \ 10(1)-10(3)+6(9) \ 6(1)-6(3)-10(9)\end{array}\right]

= \frac{1}{68} \left[\begin{array}{c}26+24+18 \ 10-30+54 \ 6-18-90\end{array}\right] = \frac{1}{68} \left[\begin{array}{c}68 \ 34 \ -102\end{array}\right] = \left[\begin{array}{c}1 \ 1/2 \ -3/2\end{array}\right]

Final Answer:

x=1, y=1/2, z=-3/2

.

Q12EXERCISE 4.5

$x-y+z=4$ $2 x+y-3 z=0$ $x+y+z=2$

Solution

Given: The system of linear equations:

x-y+z=4

2x+y-3z=0

x+y+z=2

To Find: The solution for

x, y, z

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ccc}1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{l}4 \ 0 \ 2\end{array}\right]

Step 1: Find

|A|

.

|A| = 1 \left|\begin{array}{cc}1 & -3 \ 1 & 1\end{array}\right| - (-1) \left|\begin{array}{cc}2 & -3 \ 1 & 1\end{array}\right| + 1 \left|\begin{array}{ll}2 & 1 \ 1 & 1\end{array}\right|

|A| = 1(1 - (-3)) + 1(2 - (-3)) + 1(2 - 1)

|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

A_{11} = \left|\begin{array}{cc}1 & -3 \ 1 & 1\end{array}\right| = 4

A_{12} = -\left|\begin{array}{cc}2 & -3 \ 1 & 1\end{array}\right| = -5

A_{13} = \left|\begin{array}{ll}2 & 1 \ 1 & 1\end{array}\right| = 1

A_{21} = -\left|\begin{array}{cc}-1 & 1 \ 1 & 1\end{array}\right| = -(-1-1)=2

A_{22} = \left|\begin{array}{ll}1 & 1 \ 1 & 1\end{array}\right| = 0

A_{23} = -\left|\begin{array}{cc}1 & -1 \ 1 & 1\end{array}\right| = -(1-(-1))=-2

A_{31} = \left|\begin{array}{cc}-1 & 1 \ 1 & -3\end{array}\right| = 3-1=2

A_{32} = -\left|\begin{array}{cc}1 & 1 \ 2 & -3\end{array}\right| = -(-3-2)=5

A_{33} = \left|\begin{array}{cc}1 & -1 \ 2 & 1\end{array}\right| = 1-(-2)=3

\text{adj A} = \left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{10} \left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \frac{1}{10} \left[\begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3\end{array}\right] \left[\begin{array}{l}4 \ 0 \ 2\end{array}\right] = \frac{1}{10} \left[\begin{array}{c}4(4)+2(0)+2(2) \ -5(4)+0(0)+5(2) \ 1(4)-2(0)+3(2)\end{array}\right]

= \frac{1}{10} \left[\begin{array}{c}16+0+4 \ -20+0+10 \ 4+0+6\end{array}\right] = \frac{1}{10} \left[\begin{array}{c}20 \ -10 \ 10\end{array}\right] = \left[\begin{array}{c}2 \ -1 \ 1\end{array}\right]

Final Answer:

x=2, y=-1, z=1

.

Q13EXERCISE 4.5

$2 x+3 y+3 z=5$ $x-2 y+z=-4$ $3 x-y-2 z=3$

Solution

Given: The system of linear equations:

2x+3y+3z=5

x-2y+z=-4

3x-y-2z=3

To Find: The solution for

x, y, z

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ccc}2 & 3 & 3 \ 1 & -2 & 1 \ 3 & -1 & -2\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{c}5 \ -4 \ 3\end{array}\right]

Step 1: Find

|A|

.

|A| = 2 \left|\begin{array}{cc}-2 & 1 \ -1 & -2\end{array}\right| - 3 \left|\begin{array}{cc}1 & 1 \ 3 & -2\end{array}\right| + 3 \left|\begin{array}{cc}1 & -2 \ 3 & -1\end{array}\right|

|A| = 2(4 - (-1)) - 3(-2 - 3) + 3(-1 - (-6))

|A| = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

A_{11} = \left|\begin{array}{cc}-2 & 1 \ -1 & -2\end{array}\right| = 5

A_{12} = -\left|\begin{array}{cc}1 & 1 \ 3 & -2\end{array}\right| = -(-5)=5

A_{13} = \left|\begin{array}{cc}1 & -2 \ 3 & -1\end{array}\right| = 5

A_{21} = -\left|\begin{array}{cc}3 & 3 \ -1 & -2\end{array}\right| = -(-6-(-3))=3

A_{22} = \left|\begin{array}{cc}2 & 3 \ 3 & -2\end{array}\right| = -4-9=-13

A_{23} = -\left|\begin{array}{cc}2 & 3 \ 3 & -1\end{array}\right| = -(-2-9)=11

A_{31} = \left|\begin{array}{cc}3 & 3 \ -2 & 1\end{array}\right| = 3-(-6)=9

A_{32} = -\left|\begin{array}{ll}2 & 3 \ 1 & 1\end{array}\right| = -(2-3)=1

A_{33} = \left|\begin{array}{cc}2 & 3 \ 1 & -2\end{array}\right| = -4-3=-7

\text{adj A} = \left[\begin{array}{ccc}5 & 3 & 9 \ 5 & -13 & 1 \ 5 & 11 & -7\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{40} \left[\begin{array}{ccc}5 & 3 & 9 \ 5 & -13 & 1 \ 5 & 11 & -7\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \frac{1}{40} \left[\begin{array}{ccc}5 & 3 & 9 \ 5 & -13 & 1 \ 5 & 11 & -7\end{array}\right] \left[\begin{array}{c}5 \ -4 \ 3\end{array}\right] = \frac{1}{40} \left[\begin{array}{c}5(5)+3(-4)+9(3) \ 5(5)-13(-4)+1(3) \ 5(5)+11(-4)-7(3)\end{array}\right]

= \frac{1}{40} \left[\begin{array}{c}25-12+27 \ 25+52+3 \ 25-44-21\end{array}\right] = \frac{1}{40} \left[\begin{array}{c}40 \ 80 \ -40\end{array}\right] = \left[\begin{array}{c}1 \ 2 \ -1\end{array}\right]

Final Answer:

x=1, y=2, z=-1

.

Q14EXERCISE 4.5

$x-y+2 z=7$ $3 x+4 y-5 z=-5$ $2 x-y+3 z=12$

Solution

Given: The system of linear equations:

x-y+2z=7

3x+4y-5z=-5

2x-y+3z=12

To Find: The solution for

x, y, z

using the matrix method.

Solution: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{ccc}1 & -1 & 2 \ 3 & 4 & -5 \ 2 & -1 & 3\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{c}7 \ -5 \ 12\end{array}\right]

Step 1: Find

|A|

.

|A| = 1 \left|\begin{array}{cc}4 & -5 \ -1 & 3\end{array}\right| - (-1) \left|\begin{array}{cc}3 & -5 \ 2 & 3\end{array}\right| + 2 \left|\begin{array}{cc}3 & 4 \ 2 & -1\end{array}\right|

|A| = 1(12 - 5) + 1(9 - (-10)) + 2(-3 - 8)

|A| = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

A_{11} = \left|\begin{array}{cc}4 & -5 \ -1 & 3\end{array}\right| = 7

A_{12} = -\left|\begin{array}{cc}3 & -5 \ 2 & 3\end{array}\right| = -19

A_{13} = \left|\begin{array}{cc}3 & 4 \ 2 & -1\end{array}\right| = -11

A_{21} = -\left|\begin{array}{cc}-1 & 2 \ -1 & 3\end{array}\right| = -(-3-(-2))=1

A_{22} = \left|\begin{array}{ll}1 & 2 \ 2 & 3\end{array}\right| = 3-4=-1

A_{23} = -\left|\begin{array}{cc}1 & -1 \ 2 & -1\end{array}\right| = -(-1-(-2))=-1

A_{31} = \left|\begin{array}{cc}-1 & 2 \ 4 & -5\end{array}\right| = 5-8=-3

A_{32} = -\left|\begin{array}{cc}1 & 2 \ 3 & -5\end{array}\right| = -(-5-6)=11

A_{33} = \left|\begin{array}{cc}1 & -1 \ 3 & 4\end{array}\right| = 4-(-3)=7

\text{adj A} = \left[\begin{array}{ccc}7 & 1 & -3 \ -19 & -1 & 11 \ -11 & -1 & 7\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{4} \left[\begin{array}{ccc}7 & 1 & -3 \ -19 & -1 & 11 \ -11 & -1 & 7\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \frac{1}{4} \left[\begin{array}{ccc}7 & 1 & -3 \ -19 & -1 & 11 \ -11 & -1 & 7\end{array}\right] \left[\begin{array}{c}7 \ -5 \ 12\end{array}\right] = \frac{1}{4} \left[\begin{array}{c}7(7)+1(-5)-3(12) \ -19(7)-1(-5)+11(12) \ -11(7)-1(-5)+7(12)\end{array}\right]

= \frac{1}{4} \left[\begin{array}{c}49-5-36 \ -133+5+132 \ -77+5+84\end{array}\right] = \frac{1}{4} \left[\begin{array}{c}8 \ 4 \ 12\end{array}\right] = \left[\begin{array}{c}2 \ 1 \ 3\end{array}\right]

Final Answer:

x=2, y=1, z=3

.

Q15EXERCISE 4.5

If $\mathrm{A}=\left[\begin{array}{rrr}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}\right]$ , find $\mathrm{A}^{-1}$ . Using $\mathrm{A}^{-1}$ solve the system of equations $2 x-3 y+5 z=11$ $3 x+2 y-4 z=-5$ $x+y-2 z=-3$

Solution

Part 1: Find $A^{-1}$

Given: The matrix

\mathrm{A} = \left[\begin{array}{rrr}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}\right]

.

Solution: Step 1: Find

|A|

.

|A| = 2 \left|\begin{array}{cc}2 & -4 \ 1 & -2\end{array}\right| - (-3) \left|\begin{array}{cc}3 & -4 \ 1 & -2\end{array}\right| + 5 \left|\begin{array}{ll}3 & 2 \ 1 & 1\end{array}\right|

|A| = 2(-4 - (-4)) + 3(-6 - (-4)) + 5(3 - 2)

|A| = 2(0) + 3(-2) + 5(1) = 0 - 6 + 5 = -1

Since

|A| \neq 0

,

A^{-1}

exists.

Step 2: Find adj A.

A_{11} = \left|\begin{array}{cc}2 & -4 \ 1 & -2\end{array}\right| = 0

A_{12} = -\left|\begin{array}{cc}3 & -4 \ 1 & -2\end{array}\right| = -(-2)=2

A_{13} = \left|\begin{array}{ll}3 & 2 \ 1 & 1\end{array}\right| = 1

A_{21} = -\left|\begin{array}{cc}-3 & 5 \ 1 & -2\end{array}\right| = -(6-5)=-1

A_{22} = \left|\begin{array}{cc}2 & 5 \ 1 & -2\end{array}\right| = -4-5=-9

A_{23} = -\left|\begin{array}{cc}2 & -3 \ 1 & 1\end{array}\right| = -(2-(-3))=-5

A_{31} = \left|\begin{array}{cc}-3 & 5 \ 2 & -4\end{array}\right| = 12-10=2

A_{32} = -\left|\begin{array}{cc}2 & 5 \ 3 & -4\end{array}\right| = -(-8-15)=23

A_{33} = \left|\begin{array}{cc}2 & -3 \ 3 & 2\end{array}\right| = 4-(-9)=13

\text{adj A} = \left[\begin{array}{ccc}0 & -1 & 2 \ 2 & -9 & 23 \ 1 & -5 & 13\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{|A|}\text{adj A} = \frac{1}{-1} \left[\begin{array}{ccc}0 & -1 & 2 \ 2 & -9 & 23 \ 1 & -5 & 13\end{array}\right] = \left[\begin{array}{ccc}0 & 1 & -2 \ -2 & 9 & -23 \ -1 & 5 & -13\end{array}\right]

Part 2: Solve the system of equations

Given:

2 x-3 y+5 z=11

3 x+2 y-4 z=-5

x+y-2 z=-3

Solution: The system can be written as

AX=B

, where A is the same matrix as in Part 1.

A = \left[\begin{array}{rrr}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{c}11 \ -5 \ -3\end{array}\right]

The solution is

X = A^{-1}B

. Using the

A^{-1}

found in Part 1:

X = \left[\begin{array}{ccc}0 & 1 & -2 \ -2 & 9 & -23 \ -1 & 5 & -13\end{array}\right] \left[\begin{array}{c}11 \ -5 \ -3\end{array}\right]

= \left[\begin{array}{c}0(11)+1(-5)-2(-3) \ -2(11)+9(-5)-23(-3) \ -1(11)+5(-5)-13(-3)\end{array}\right] = \left[\begin{array}{c}0-5+6 \ -22-45+69 \ -11-25+39\end{array}\right] = \left[\begin{array}{l}1 \ 2 \ 3\end{array}\right]

Final Answer:

A^{-1} = \left[\begin{array}{ccc}0 & 1 & -2 \ -2 & 9 & -23 \ -1 & 5 & -13\end{array}\right]

and the solution to the system is

x=1, y=2, z=3

.

Q16EXERCISE 4.5

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg by matrix method.

Solution

Given:

Cost of 4 kg onion + 3 kg wheat + 2 kg rice = ₹ 60
Cost of 2 kg onion + 4 kg wheat + 6 kg rice = ₹ 90
Cost of 6 kg onion + 2 kg wheat + 3 kg rice = ₹ 70

To Find: The cost per kg of onion, wheat, and rice.

Let:

Cost of 1 kg onion = $x$
Cost of 1 kg wheat = $y$
Cost of 1 kg rice = $z$

Equations: From the given information, we can form the following system of linear equations:

$4x + 3y + 2z = 60$
$2x + 4y + 6z = 90$
$6x + 2y + 3z = 70$

Solution using Matrix Method: The system in matrix form is

AX=B

, where:

A = \left[\begin{array}{lll}4 & 3 & 2 \ 2 & 4 & 6 \ 6 & 2 & 3\end{array}\right], X = \left[\begin{array}{l}x \ y \ z\end{array}\right], B = \left[\begin{array}{l}60 \ 90 \ 70\end{array}\right]

Step 1: Find

|A|

.

|A| = 4(12-12) - 3(6-36) + 2(4-24) = 4(0) - 3(-30) + 2(-20) = 90 - 40 = 50

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

A_{11} = (12-12)=0

A_{12} = -(6-36)=30

A_{13} = (4-24)=-20

A_{21} = -(9-4)=-5

A_{22} = (12-12)=0

A_{23} = -(8-18)=10

A_{31} = (18-8)=10

A_{32} = -(24-4)=-20

A_{33} = (16-6)=10

\text{adj A} = \left[\begin{array}{ccc}0 & -5 & 10 \ 30 & 0 & -20 \ -20 & 10 & 10\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{50} \left[\begin{array}{ccc}0 & -5 & 10 \ 30 & 0 & -20 \ -20 & 10 & 10\end{array}\right]

Step 4: Find

X = A^{-1}B

.

X = \frac{1}{50} \left[\begin{array}{ccc}0 & -5 & 10 \ 30 & 0 & -20 \ -20 & 10 & 10\end{array}\right] \left[\begin{array}{l}60 \ 90 \ 70\end{array}\right] = \frac{1}{50} \left[\begin{array}{c}0(60)-5(90)+10(70) \ 30(60)+0(90)-20(70) \ -20(60)+10(90)+10(70)\end{array}\right]

= \frac{1}{50} \left[\begin{array}{c}0-450+700 \ 1800+0-1400 \ -1200+900+700\end{array}\right] = \frac{1}{50} \left[\begin{array}{c}250 \ 400 \ 400\end{array}\right] = \left[\begin{array}{l}5 \ 8 \ 8\end{array}\right]

So,

x=5, y=8, z=8

.

Final Answer: The cost of onion is ₹ 5 per kg, the cost of wheat is ₹ 8 per kg, and the cost of rice is ₹ 8 per kg.

Q1Miscellaneous Exercises on Chapter 4

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \ -\sin \theta & -x & 1 \ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$ .

Solution

To Prove: The determinant is independent of

\theta

.

Proof: Let the given determinant be

\Delta

.

\Delta = \left|\begin{array}{ccc}x & \sin \theta & \cos \theta \ -\sin \theta & -x & 1 \ \cos \theta & 1 & x\end{array}\right|

We evaluate the determinant by expanding along the first row (

R_1

).

\Delta = x \left|\begin{array}{cc}-x & 1 \ 1 & x\end{array}\right| - \sin \theta \left|\begin{array}{cc}-\sin \theta & 1 \ \cos \theta & x\end{array}\right| + \cos \theta \left|\begin{array}{cc}-\sin \theta & -x \ \cos \theta & 1\end{array}\right|

\Delta = x(-x^2 - 1) - \sin \theta(-x \sin \theta - \cos \theta) + \cos \theta(-\sin \theta - (-x \cos \theta))

\Delta = -x^3 - x + x \sin^2 \theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x \cos^2 \theta

Combine the terms with

x \sin^2 \theta

and

x \cos^2 \theta

, and notice that

\sin \theta \cos \theta - \sin \theta \cos \theta = 0

.

\Delta = -x^3 - x + x(\sin^2 \theta + \cos^2 \theta)

Using the trigonometric identity

\sin^2 \theta + \cos^2 \theta = 1

:

\Delta = -x^3 - x + x(1)

\Delta = -x^3 - x + x

\Delta = -x^3

Since the final value of the determinant,

-x^3

, does not contain the term

\theta

, it is independent of

\theta

.

Hence Proved.

Q2Miscellaneous Exercises on Chapter 4

Evaluate $\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \ -\sin \beta & \cos \beta & 0 \ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$ .

Solution

To Find: The value of the given determinant.

Solution: Let the determinant be

\Delta

.

\Delta = \left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \ -\sin \beta & \cos \beta & 0 \ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|

We expand along the third column (

C_3

) for simplicity.

\Delta = -\sin \alpha \left|\begin{array}{cc}-\sin \beta & \cos \beta \ \sin \alpha \cos \beta & \sin \alpha \sin \beta\end{array}\right| - 0 + \cos \alpha \left|\begin{array}{cc}\cos \alpha \cos \beta & \cos \alpha \sin \beta \ -\sin \beta & \cos \beta\end{array}\right|

Evaluate the two smaller determinants: First part:

-\sin \alpha [(-\sin \beta)(\sin \alpha \sin \beta) - (\cos \beta)(\sin \alpha \cos \beta)]

= -\sin \alpha [-\sin \alpha \sin^2 \beta - \sin \alpha \cos^2 \beta]

= -\sin \alpha [-\sin \alpha (\sin^2 \beta + \cos^2 \beta)]

= -\sin \alpha [-\sin \alpha (1)] = \sin^2 \alpha

Second part:

\cos \alpha [(\cos \alpha \cos \beta)(\cos \beta) - (\cos \alpha \sin \beta)(-\sin \beta)]

= \cos \alpha [\cos \alpha \cos^2 \beta + \cos \alpha \sin^2 \beta]

= \cos \alpha [\cos \alpha (\cos^2 \beta + \sin^2 \beta)]

= \cos \alpha [\cos \alpha (1)] = \cos^2 \alpha

Now, add the results of the two parts:

\Delta = \sin^2 \alpha + \cos^2 \alpha

Using the identity

\sin^2 \alpha + \cos^2 \alpha = 1

:

\Delta = 1

Final Answer: The value of the determinant is 1.

Q3Miscellaneous Exercises on Chapter 4

If $\mathrm{A}^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1\end{array}\right]$ , find $(\mathrm{AB})^{-1}$

Solution

Given:

A^{-1} = \left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}\right]

B = \left[\begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1\end{array}\right]

To Find:

(AB)^{-1}

.

Property: We know that

(AB)^{-1} = B^{-1} A^{-1}

. We are already given

A^{-1}

, so we only need to find

B^{-1}

.

Solution: Step 1: Find the determinant of B.

|B| = 1\left|\begin{array}{cc}3 & 0 \ -2 & 1\end{array}\right| - 2\left|\begin{array}{cc}-1 & 0 \ 0 & 1\end{array}\right| + (-2)\left|\begin{array}{cc}-1 & 3 \ 0 & -2\end{array}\right|

|B| = 1(3-0) - 2(-1-0) - 2(2-0) = 3 + 2 - 4 = 1

Since

|B|=1 \neq 0

,

B^{-1}

exists.

Step 2: Find the adjoint of B.

B_{11} = \left|\begin{array}{cc}3 & 0 \ -2 & 1\end{array}\right| = 3

B_{12} = -\left|\begin{array}{cc}-1 & 0 \ 0 & 1\end{array}\right| = 1

B_{13} = \left|\begin{array}{cc}-1 & 3 \ 0 & -2\end{array}\right| = 2

B_{21} = -\left|\begin{array}{cc}2 & -2 \ -2 & 1\end{array}\right| = -(2-4) = 2

B_{22} = \left|\begin{array}{cc}1 & -2 \ 0 & 1\end{array}\right| = 1

B_{23} = -\left|\begin{array}{cc}1 & 2 \ 0 & -2\end{array}\right| = -(-2) = 2

B_{31} = \left|\begin{array}{cc}2 & -2 \ 3 & 0\end{array}\right| = 0 - (-6) = 6

B_{32} = -\left|\begin{array}{cc}1 & -2 \ -1 & 0\end{array}\right| = -(0-2) = 2

B_{33} = \left|\begin{array}{cc}1 & 2 \ -1 & 3\end{array}\right| = 3 - (-2) = 5

\text{adj B} = \left[\begin{array}{lll}3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 2 & 5\end{array}\right]

Step 3: Find

B^{-1}

.

B^{-1} = \frac{1}{|B|} \text{adj B} = \frac{1}{1} \left[\begin{array}{lll}3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 2 & 5\end{array}\right] = \left[\begin{array}{lll}3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 2 & 5\end{array}\right]

Step 4: Calculate

(AB)^{-1} = B^{-1}A^{-1}

.

(AB)^{-1} = \left[\begin{array}{lll}3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 2 & 5\end{array}\right] \left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}\right]

= \left[\begin{array}{ccc}3(3)+2(-15)+6(5) & 3(-1)+2(6)+6(-2) & 3(1)+2(-5)+6(2) \ 1(3)+1(-15)+2(5) & 1(-1)+1(6)+2(-2) & 1(1)+1(-5)+2(2) \ 2(3)+2(-15)+5(5) & 2(-1)+2(6)+5(-2) & 2(1)+2(-5)+5(2)\end{array}\right]

= \left[\begin{array}{ccc}9-30+30 & -3+12-12 & 3-10+12 \ 3-15+10 & -1+6-4 & 1-5+4 \ 6-30+25 & -2+12-10 & 2-10+10\end{array}\right] = \left[\begin{array}{ccc}9 & -3 & 5 \ -2 & 1 & 0 \ 1 & 0 & 2\end{array}\right]

Final Answer:

(AB)^{-1} = \left[\begin{array}{ccc}9 & -3 & 5 \ -2 & 1 & 0 \ 1 & 0 & 2\end{array}\right]

.

Q4Miscellaneous Exercises on Chapter 4

Let $\mathrm{A}=\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}$ . Verify that

(i)

$[\operatorname{adj} \mathrm{A}]^{-1}=\operatorname{adj}\left(\mathrm{A}^{-1}\right)$

(ii)

$\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}$

Solution

Given: Matrix

\mathrm{A}=\left[\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}\right]

.

First, we need to find

|A|

, adj A, and

A^{-1}

.

Calculate $|A|$ :

|A| = 1(15-1) - (-2)(-10-1) + 1(-2-3) = 14 + 2(-11) - 5 = 14 - 22 - 5 = -13

Since

|A| \neq 0

, A is invertible.

Calculate adj A:

A_{11} = 14, A_{12} = -(-11) = 11, A_{13} = -5

A_{21} = -(-10-1) = 11, A_{22} = 5-1 = 4, A_{23} = -(1-(-2)) = -3

A_{31} = -2-3 = -5, A_{32} = -(1-(-2)) = -3, A_{33} = 3-4 = -1

\text{adj A} = \left[\begin{array}{ccc}14 & 11 & -5 \ 11 & 4 & -3 \ -5 & -3 & -1\end{array}\right]

Calculate $A^{-1}$ :

A^{-1} = \frac{1}{|A|} \text{adj A} = -\frac{1}{13} \left[\begin{array}{ccc}14 & 11 & -5 \ 11 & 4 & -3 \ -5 & -3 & -1\end{array}\right]

(i) To Verify:

[\operatorname{adj} \mathrm{A}]^{-1}=\operatorname{adj}\left(\mathrm{A}^{-1}\right)

LHS Calculation:

[\operatorname{adj} \mathrm{A}]^{-1}

Let

C = \text{adj A}

. We need to find

C^{-1}

.

|C| = |\text{adj A}| = |A|^{n-1} = (-13)^{3-1} = (-13)^2 = 169

.

C^{-1} = \frac{1}{|C|} \text{adj C}

. We know the property

\text{adj(adj A)} = |A|^{n-2} A

. For

n=3

, this is

|A|A

.

\text{adj C} = \text{adj(adj A)} = -13 A = -13 \left[\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}\right]

.

[\operatorname{adj} \mathrm{A}]^{-1} = \frac{1}{169} \left( -13 \left[\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}\right] \right) = -\frac{1}{13} \left[\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}\right]

RHS Calculation:

\operatorname{adj}\left(\mathrm{A}^{-1}\right)

Let

B = A^{-1} = kA'

, where

k = -1/13

and

A' = \text{adj A}

. We use the property

\text{adj}(kA) = k^{n-1} \text{adj}(A)

. So,

\text{adj}(A^{-1}) = \text{adj}(-\frac{1}{13} \text{adj A}) = (-\frac{1}{13})^{3-1} \text{adj(adj A)}

.

\text{adj(adj A)} = |A|A = -13A

.

\text{adj}(A^{-1}) = \frac{1}{169}(-13A) = -\frac{1}{13}A = -\frac{1}{13} \left[\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}\right]

Since LHS = RHS, the property is verified.

(ii) To Verify:

\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}

LHS Calculation:

(A^{-1})^{-1}

Let

B = A^{-1}

. We need to find

B^{-1}

.

B^{-1} = \frac{1}{|B|} \text{adj B}

.

|B| = |A^{-1}| = \frac{1}{|A|} = \frac{1}{-13}

.

\text{adj B} = \text{adj}(A^{-1}) = -\frac{1}{13}A

(from part i).

B^{-1} = \frac{1}{1/(-13)} \left( -\frac{1}{13}A \right) = (-13) \left( -\frac{1}{13}A \right) = A

RHS: A. Since LHS = RHS, the property is verified.

Hence Verified.

Q5Miscellaneous Exercises on Chapter 4

Evaluate $\left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}\right|$

Solution

To Find: The value of the given determinant.

Solution: Let

\Delta = \left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}\right|

.

Apply the column operation

C_1 \rightarrow C_1 + C_2 + C_3

.

\Delta = \left|\begin{array}{ccc}x+y+(x+y) & y & x+y \ y+(x+y)+x & x+y & x \ (x+y)+x+y & x & y\end{array}\right| = \left|\begin{array}{ccc}2(x+y) & y & x+y \ 2(x+y) & x+y & x \ 2(x+y) & x & y\end{array}\right|

Take

2(x+y)

common from the first column (

C_1

).

\Delta = 2(x+y) \left|\begin{array}{ccc}1 & y & x+y \ 1 & x+y & x \ 1 & x & y\end{array}\right|

Now apply row operations

R_2 \rightarrow R_2 - R_1

and

R_3 \rightarrow R_3 - R_1

.

\Delta = 2(x+y) \left|\begin{array}{ccc}1 & y & x+y \ 0 & x & -y \ 0 & x-y & -x\end{array}\right|

Expand along the first column (

C_1

).

\Delta = 2(x+y) \left[ 1 \left|\begin{array}{cc}x & -y \ x-y & -x\end{array}\right| - 0 + 0 \right]

\Delta = 2(x+y) [x(-x) - (-y)(x-y)]

\Delta = 2(x+y) [-x^2 - (-xy+y^2)]

\Delta = 2(x+y) [-x^2 + xy - y^2]

\Delta = -2(x+y)(x^2 - xy + y^2)

Using the identity

a^3+b^3 = (a+b)(a^2-ab+b^2)

:

\Delta = -2(x^3 + y^3)

Final Answer:

-2(x^3 + y^3)

.

Q6Miscellaneous Exercises on Chapter 4

Evaluate $\left|\begin{array}{ccc}1 & x & y \ 1 & x+y & y \ 1 & x & x+y\end{array}\right|$

Solution

To Find: The value of the given determinant.

Solution: Let

\Delta = \left|\begin{array}{ccc}1 & x & y \ 1 & x+y & y \ 1 & x & x+y\end{array}\right|

.

Apply row operations

R_2 \rightarrow R_2 - R_1

and

R_3 \rightarrow R_3 - R_1

.

\Delta = \left|\begin{array}{ccc}1 & x & y \ 1-1 & (x+y)-x & y-y \ 1-1 & x-x & (x+y)-y\end{array}\right|

\Delta = \left|\begin{array}{ccc}1 & x & y \ 0 & y & 0 \ 0 & 0 & x\end{array}\right|

This is an upper triangular matrix. The determinant of a triangular matrix is the product of its diagonal elements.

\Delta = 1 \times y \times x = xy

Alternatively, we can expand along the first column (

C_1

).

\Delta = 1 \left|\begin{array}{ll}y & 0 \ 0 & x\end{array}\right| - 0 + 0

\Delta = 1(yx - 0) = xy

Final Answer:

xy

.

Q7Miscellaneous Exercises on Chapter 4

Solve the system of equations $& \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 \n& \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \n& \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2 \end{aligned}$$$

Solution

Given: The system of equations:

\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4

\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1

\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2

To Find: The values of

x, y, z

.

Solution: Let

u = \frac{1}{x}, v = \frac{1}{y}, w = \frac{1}{z}

. The system becomes a linear system in

u, v, w

:

2u + 3v + 10w = 4

4u - 6v + 5w = 1

6u + 9v - 20w = 2

We solve this using the matrix method. The system is

AU=B

, where:

A = \left[\begin{array}{ccc}2 & 3 & 10 \ 4 & -6 & 5 \ 6 & 9 & -20\end{array}\right], U = \left[\begin{array}{c}u \ v \ w\end{array}\right], B = \left[\begin{array}{l}4 \ 1 \ 2\end{array}\right]

Step 1: Find

|A|

.

|A| = 2(120-45) - 3(-80-30) + 10(36 - (-36))

|A| = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200

Since

|A| \neq 0

, a unique solution exists.

Step 2: Find adj A.

A_{11} = 75, A_{12} = -(-110)=110, A_{13} = 72

A_{21} = -(-60-90)=150, A_{22} = -40-60=-100, A_{23} = -(18-18)=0

A_{31} = 15-(-60)=75, A_{32} = -(10-40)=30, A_{33} = -12-12=-24

\text{adj A} = \left[\begin{array}{ccc}75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{1200} \left[\begin{array}{ccc}75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24\end{array}\right]

Step 4: Find

U = A^{-1}B

.

U = \frac{1}{1200} \left[\begin{array}{ccc}75 & 150 & 75 \ 110 & -100 & 30 \ 72 & 0 & -24\end{array}\right] \left[\begin{array}{l}4 \ 1 \ 2\end{array}\right] = \frac{1}{1200} \left[\begin{array}{c}300+150+150 \ 440-100+60 \ 288+0-48\end{array}\right]

= \frac{1}{1200} \left[\begin{array}{l}600 \ 400 \ 240\end{array}\right] = \left[\begin{array}{l}600/1200 \ 400/1200 \ 240/1200\end{array}\right] = \left[\begin{array}{l}1/2 \ 1/3 \ 1/5\end{array}\right]

So,

u = 1/2, v = 1/3, w = 1/5

. Now, find

x, y, z

.

x = 1/u = 2

y = 1/v = 3

z = 1/w = 5

Final Answer:

x=2, y=3, z=5

.

Q8Miscellaneous Exercises on Chapter 4

If $x, y, z$ are nonzero real numbers, then the inverse of matrix $\mathrm{A}=\left[\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}\right]$ is

(A)

\left[\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}\right]

(B)

x y z\left[\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}\right]

(C)

\frac{1}{x y z}\left[\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}\right]

(D)

\frac{1}{x y z}\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right]

Solution

Given: The matrix

\mathrm{A}=\left[\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}\right]

, where

x,y,z

are non-zero.

To Find: The inverse of matrix A,

A^{-1}

.

Solution: We use the formula

A^{-1} = \frac{1}{|A|} \text{adj A}

.

Step 1: Find

|A|

. Since A is a diagonal matrix, its determinant is the product of its diagonal elements.

|A| = xyz

Since

x, y, z

are non-zero,

|A| \neq 0

and the inverse exists.

Step 2: Find adj A. We find the cofactors:

A_{11} = yz, A_{12} = 0, A_{13} = 0

A_{21} = 0, A_{22} = xz, A_{23} = 0

A_{31} = 0, A_{32} = 0, A_{33} = xy

The cofactor matrix is

\left[\begin{array}{ccc}yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy\end{array}\right]

. The adjoint is the transpose of the cofactor matrix, which is the same in this case because it is a diagonal matrix.

\text{adj A} = \left[\begin{array}{ccc}yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy\end{array}\right]

Step 3: Find

A^{-1}

.

A^{-1} = \frac{1}{xyz} \left[\begin{array}{ccc}yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy\end{array}\right]

Now, multiply the scalar

1/(xyz)

into the matrix:

A^{-1} = \left[\begin{array}{ccc}\frac{yz}{xyz} & 0 & 0 \ 0 & \frac{xz}{xyz} & 0 \ 0 & 0 & \frac{xy}{xyz}\end{array}\right] = \left[\begin{array}{ccc}\frac{1}{x} & 0 & 0 \ 0 & \frac{1}{y} & 0 \ 0 & 0 & \frac{1}{z}\end{array}\right] = \left[\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}\right]

This matches option (A).

Final Answer: The correct option is (A).

Q9Miscellaneous Exercises on Chapter 4

Let $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \ -\sin \theta & 1 & \sin \theta \ -1 & -\sin \theta & 1\end{array}\right]$ , where $0 \leq \theta \leq 2 \pi$ . Then

(A)

\operatorname{Det}(\mathrm{A})=0

(B)

\operatorname{Det}(\mathrm{A}) \in(2, \infty)

(C)

\operatorname{Det}(\mathrm{A}) \in(2,4)

(D)

\operatorname{Det}(\mathrm{A}) \in[2,4]

Solution

Given: The matrix

\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \ -\sin \theta & 1 & \sin \theta \ -1 & -\sin \theta & 1\end{array}\right]

, with

0 \leq \theta \leq 2 \pi

.

To Find: The range of values for Det(A).

Solution: First, we calculate the determinant of A, Det(A).

\operatorname{Det}(A) = 1 \left|\begin{array}{cc}1 & \sin \theta \ -\sin \theta & 1\end{array}\right| - \sin \theta \left|\begin{array}{cc}-\sin \theta & \sin \theta \ -1 & 1\end{array}\right| + 1 \left|\begin{array}{cc}-\sin \theta & 1 \ -1 & -\sin \theta\end{array}\right|

\operatorname{Det}(A) = 1(1 - (-\sin^2 \theta)) - \sin \theta(-\sin \theta - (-\sin \theta)) + 1(\sin^2 \theta - (-1))

\operatorname{Det}(A) = (1 + \sin^2 \theta) - \sin \theta(-\sin \theta + \sin \theta) + (\sin^2 \theta + 1)

\operatorname{Det}(A) = (1 + \sin^2 \theta) - \sin \theta(0) + (\sin^2 \theta + 1)

\operatorname{Det}(A) = 1 + \sin^2 \theta + \sin^2 \theta + 1

\operatorname{Det}(A) = 2 + 2\sin^2 \theta

Now, we need to find the range of this expression for

0 \leq \theta \leq 2 \pi

. We know the range of the sine function:

-1 \leq \sin \theta \leq 1

Squaring this inequality, we get:

0 \leq \sin^2 \theta \leq 1

Now, we build the expression for Det(A): Multiply by 2:

0 \leq 2\sin^2 \theta \leq 2

Add 2 to all parts of the inequality:

2+0 \leq 2+2\sin^2 \theta \leq 2+2

2 \leq \operatorname{Det}(A) \leq 4

So, the value of the determinant lies in the closed interval [2, 4]. This corresponds to option (D).

Final Answer: The correct option is (D).

DeterminantsClass 12 Mathematics NCERT Solutions

Evaluate the determinants in Exercises 1 and 2. ∣24 −5−1∣\left|\begin{array}{rr}2 & 4 \ -5 & -1\end{array}\right|​2​4 −5​−1​​

Solution

(i) left right (ii) left right

∣cos⁡θ−sin⁡θ sin⁡θcos⁡θ∣\left|\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}\right|​cosθ​−sinθ sinθ​cosθ​​

∣x2−x+1x−1 x+1x+1∣\left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}\right|​x2−x+1​x−1 x+1​x+1​​

Solution

If A=[12 42]\mathrm{A}=\left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}\right]A=[1​2 4​2​], then show that ∣2 A∣=4∣ A∣|2 \mathrm{~A}|=4|\mathrm{~A}|∣2 A∣=4∣ A∣.

Solution

If A=[101 012 004]\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4\end{array}\right]A=[1​0​1 0​1​2 0​0​4​], then show that ∣3 A∣=27∣ A∣|3 \mathrm{~A}|=27|\mathrm{~A}|∣3 A∣=27∣ A∣.

Solution

Evaluate the determinants

∣3−1−2 00−1 3−50∣\left|\begin{array}{rrr}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}\right|​3​−1​−2 0​0​−1 3​−5​0​​

∣3−45 11−2 231∣\left|\begin{array}{rrr}3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1\end{array}\right|​3​−4​5 1​1​−2 2​3​1​​

∣012 −10−3 −230∣\left|\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0\end{array}\right|​0​1​2 −1​0​−3 −2​3​0​​

∣2−1−2 02−1 3−50∣\left|\begin{array}{rrr}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}\right|​2​−1​−2 0​2​−1 3​−5​0​​

Solution

If A=[11−2 21−3 54−9]\mathrm{A}=\left[\begin{array}{lll}1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9\end{array}\right]A=[1​1​−2 2​1​−3 5​4​−9​], find ∣A∣|\mathrm{A}|∣A∣.

Solution

Find values of xxx, if

∣24 51∣=∣2x4 6x∣\left|\begin{array}{ll}2 & 4 \ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \ 6 & x\end{array}\right|​2​4 5​1​​=​2x​4 6​x​​

∣23 45∣=∣x3 2x5∣\left|\begin{array}{ll}2 & 3 \ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \ 2 x & 5\end{array}\right|​2​3 4​5​​=​x​3 2x​5​​

Solution

If ∣x2 18x∣=∣62 186∣\left|\begin{array}{cc}x & 2 \ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \ 18 & 6\end{array}\right|​x​2 18​x​​=​6​2 18​6​​, then xxx is equal to

Solution

Find area of the triangle with vertices at the point given in each of the following:

(1,0),(6,0),(4,3)(1,0),(6,0),(4,3)(1,0),(6,0),(4,3)

(2,7),(1,1),(10,8)(2,7),(1,1),(10,8)(2,7),(1,1),(10,8)

(−2,−3),(3,2),(−1,−8)(-2,-3),(3,2),(-1,-8)(−2,−3),(3,2),(−1,−8)

Solution

Show that points A(a,b+c),B(b,c+a),C(c,a+b)\mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b)A(a,b+c),B(b,c+a),C(c,a+b) are collinear.

Solution

Find values of kkk if area of triangle is 4 sq . units and vertices are

(k,0),(4,0),(0,2)(k, 0),(4,0),(0,2)(k,0),(4,0),(0,2)

(−2,0),(0,4),(0,k)(-2,0),(0,4),(0, k)(−2,0),(0,4),(0,k)

Solution

(i) Find equation of line joining (1,2) and (3,6) using determinants. (ii) Find equation of line joining…

Find equation of line joining (1,2)(1,2)(1,2) and (3,6)(3,6)(3,6) using determinants.

Find equation of line joining (3,1)(3,1)(3,1) and (9,3)(9,3)(9,3) using determinants.

Solution

If area of triangle is 35 sq units with vertices (2,−6),(5,4)(2,-6),(5,4)(2,−6),(5,4) and (k,4)(k, 4)(k,4). Then kkk is

Solution

Write Minors and Cofactors of the elements of following determinants:

∣2−4 03∣\left|\begin{array}{rr}2 & -4 \ 0 & 3\end{array}\right|​2​−4 0​3​​

∣ac bd∣\left|\begin{array}{ll}a & c \ b & d\end{array}\right|​a​c b​d​​

Solution

Write Minors and Cofactors of the elements of following determinants:

∣100 010 001∣\left|\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right|​1​0​0 0​1​0 0​0​1​​

∣104 35−1 012∣\left|\begin{array}{rrr}1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2\end{array}\right|​1​0​4 3​5​−1 0​1​2​​

Solution

Using Cofactors of elements of second row, evaluate Δ=∣538 201 123∣\Delta=\left|\begin{array}{lll}5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3\end{array}\right|Δ=​5​3​8 2​0​1 1​2​3​​.

Solution

Using Cofactors of elements of third column, evaluate Δ=∣1xyz 1yzx 1zxy∣\Delta=\left|\begin{array}{lll}1 & x & y z \ 1 & y & z x \ 1 & z & x y\end{array}\right|Δ=​1​x​yz 1​y​zx 1​z​xy​​.

Solution

Solution

Find adjoint of each of the matrices in Exercises 1 and 2. 12 34\begin{array}{ll}1 & 2 \ 3 & 4\end{array}1​2 3​4​

Solution

1−12 235 −201\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}1​−1​2 2​3​5 −2​0​1​

Solution

Verify A(adj⁡A)=(adj⁡A)A=∣A∣I\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}A(adjA)=(adjA)A=∣A∣I in Exercises 3 and 4 3. 23 −4−6\begin{array}{cc}2 & 3 \ -4 & -6\end{array}2​3 −4​−6​

Solution

1−12 30−2 103\begin{array}{lll}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}1​−1​2 3​0​−2 1​0​3​

Solution

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 5. 2−2 43\begin{array}{cc}2 & -2 \ 4 & 3\end{array}2​−2 4​3​

Solution

−15 −32\begin{array}{rr}-1 & 5 \ -3 & 2\end{array}−1​5 −3​2​

Solution

123 024 005\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}1​2​3 0​2​4 0​0​5​

Solution

100 330 52−1\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}1​0​0 3​3​0 5​2​−1​

Solution

213 4−10 −721\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}2​1​3 4​−1​0 −7​2​1​

Solution

1−12 02−3 3−24\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}1​−1​2 0​2​−3 3​−2​4​

Solution

[100 0cos⁡αsin⁡α 0sin⁡α−cos⁡α]\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}\right][1​0​0 0​cosα​sinα 0​sinα​−cosα​]

Solution

Let A=37 25\mathrm{A}=\begin{array}{ll}3 & 7 \ 2 & 5\end{array}A=3​7 2​5​ and B=68 79\mathrm{B}=\begin{array}{ll}6 & 8 \ 7 & 9\end{array}B=6​8 7​9​. Verify that (AB)−1=B−1 A−1(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}(AB)−1=B−1 A−1.

Solution

If A=31 −12\mathrm{A}=\begin{array}{cc}3 & 1 \ -1 & 2\end{array}A=3​1 −1​2​, show that A2−5 A+7I=O\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=\mathrm{O}A2−5 A+7I=O. Hence find A−1\mathrm{A}^{-1}A−1.

Evaluate the determinants in Exercises 1 and 2. $\left|\begin{array}{rr}2 & 4 \ -5 & -1\end{array}\right|$

$\left|\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}\right|$

$\left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}\right|$

If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \ 4 & 2\end{array}\right]$ , then show that $|2 \mathrm{~A}|=4|\mathrm{~A}|$ .

If $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4\end{array}\right]$ , then show that $|3 \mathrm{~A}|=27|\mathrm{~A}|$ .

$\left|\begin{array}{rrr}3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0\end{array}\right|$

$\left|\begin{array}{rrr}3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1\end{array}\right|$

$\left|\begin{array}{ccc}0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0\end{array}\right|$

$\left|\begin{array}{rrr}2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0\end{array}\right|$

If $\mathrm{A}=\left[\begin{array}{lll}1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9\end{array}\right]$ , find $|\mathrm{A}|$ .

Find values of $x$ , if

$\left|\begin{array}{ll}2 & 4 \ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \ 6 & x\end{array}\right|$

$\left|\begin{array}{ll}2 & 3 \ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \ 2 x & 5\end{array}\right|$

If $\left|\begin{array}{cc}x & 2 \ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \ 18 & 6\end{array}\right|$ , then $x$ is equal to

$(1,0),(6,0),(4,3)$

$(2,7),(1,1),(10,8)$

$(-2,-3),(3,2),(-1,-8)$

Show that points $\mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b)$ are collinear.

Find values of $k$ if area of triangle is 4 sq . units and vertices are

$(k, 0),(4,0),(0,2)$

$(-2,0),(0,4),(0, k)$

Find equation of line joining $(1,2)$ and $(3,6)$ using determinants.

Find equation of line joining $(3,1)$ and $(9,3)$ using determinants.

If area of triangle is 35 sq units with vertices $(2,-6),(5,4)$ and $(k, 4)$ . Then $k$ is

$\left|\begin{array}{rr}2 & -4 \ 0 & 3\end{array}\right|$

$\left|\begin{array}{ll}a & c \ b & d\end{array}\right|$

$\left|\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right|$

$\left|\begin{array}{rrr}1 & 0 & 4 \ 3 & 5 & -1 \ 0 & 1 & 2\end{array}\right|$

Using Cofactors of elements of second row, evaluate $\Delta=\left|\begin{array}{lll}5 & 3 & 8 \ 2 & 0 & 1 \ 1 & 2 & 3\end{array}\right|$ .

Using Cofactors of elements of third column, evaluate $\Delta=\left|\begin{array}{lll}1 & x & y z \ 1 & y & z x \ 1 & z & x y\end{array}\right|$ .

Find adjoint of each of the matrices in Exercises 1 and 2. $\begin{array}{ll}1 & 2 \ 3 & 4\end{array}$

$\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}$

Verify $\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}$ in Exercises 3 and 4 3. $\begin{array}{cc}2 & 3 \ -4 & -6\end{array}$

$\begin{array}{lll}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}$

Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 5. $\begin{array}{cc}2 & -2 \ 4 & 3\end{array}$

$\begin{array}{rr}-1 & 5 \ -3 & 2\end{array}$

$\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}$

$\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}$

$\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}$

$\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}$

$\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos \alpha & \sin \alpha \ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$

Let $\mathrm{A}=\begin{array}{ll}3 & 7 \ 2 & 5\end{array}$ and $\mathrm{B}=\begin{array}{ll}6 & 8 \ 7 & 9\end{array}$ . Verify that $(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$ .

If $\mathrm{A}=\begin{array}{cc}3 & 1 \ -1 & 2\end{array}$ , show that $\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=\mathrm{O}$ . Hence find $\mathrm{A}^{-1}$ .

For the matrix $\mathrm{A}=\begin{array}{ll}3 & 2 \ 1 & 1\end{array}$ , find the numbers $a$ and $b$ such that $\mathrm{A}^{2}+a \mathrm{~A}+b \mathrm{I}=\mathrm{O}$ .

Let A be a nonsingular square matrix of order $3 \times 3$ . Then $|\operatorname{adj} \mathrm{A}|$ is equal to

If A is an invertible matrix of order 2, then $\operatorname{det}\left(\mathrm{A}^{-1}\right)$ is equal to

Examine the consistency of the system of equations in Exercises 1 to 6. $x+2 y=2$ $2 x+3 y=3$

$2 x-y=5$ $x+y=4$

$x+3 y=5$ $2 x+6 y=8$

$x+y+z=1$ $2 x+3 y+2 z=2$ $a x+a y+2 a z=4$

$3 x-y-2 z=2$ $2 y-z=-1$ $3 x-5 y=3$

$5 x-y+4 z=5$ $2 x+3 y+5 z=2$ $5 x-2 y+6 z=-1$

Solve system of linear equations, using matrix method, in Exercises 7 to 14. 7. $5 x+2 y=4$ $7 x+3 y=5$

$2 x-y=-2$ $3 x+4 y=3$

$4 x-3 y=3$ $3 x-5 y=7$

$5 x+2 y=3$ $3 x+2 y=5$

$2 x+y+z=1$ $x-2 y-z=\frac{3}{2}$ $3 y-5 z=9$

$x-y+z=4$ $2 x+y-3 z=0$ $x+y+z=2$

$2 x+3 y+3 z=5$ $x-2 y+z=-4$ $3 x-y-2 z=3$

$x-y+2 z=7$ $3 x+4 y-5 z=-5$ $2 x-y+3 z=12$

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \ -\sin \theta & -x & 1 \ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$ .

Let $\mathrm{A}=\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}$ . Verify that

$[\operatorname{adj} \mathrm{A}]^{-1}=\operatorname{adj}\left(\mathrm{A}^{-1}\right)$

$\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}$

Evaluate $\left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}\right|$

Evaluate $\left|\begin{array}{ccc}1 & x & y \ 1 & x+y & y \ 1 & x & x+y\end{array}\right|$

Solve the system of equations $& \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 \n& \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \n& \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2 \end{aligned}$$$

If $x, y, z$ are nonzero real numbers, then the inverse of matrix $\mathrm{A}=\left[\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}\right]$ is