Class 12
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Mathematics
Determinants
NCERT Solutions
NCERT Solutions
Determinants
61 Solutions
Exercise:
All Exercises
EXERCISE 4.1
EXERCISE 4.2
EXERCISE 4.3
EXERCISE 4.4
EXERCISE 4.5
Miscellaneous Exercises on Chapter 4
Q1
EXERCISE 4.1
Evaluate the determinants in Exercises 1 and 2.
∣
2
4
−
5
−
1
∣
\left|\begin{array}{rr}2 & 4 \ -5 & -1\end{array}\right|
2
4
−
5
−
1
Q2
EXERCISE 4.1
(i)
∣
cos
θ
−
sin
θ
sin
θ
cos
θ
∣
\left|\begin{array}{cc}\cos \theta & -\sin \theta \ \sin \theta & \cos \theta\end{array}\right|
cos
θ
−
sin
θ
sin
θ
cos
θ
(ii)
∣
x
2
−
x
+
1
x
−
1
x
+
1
x
+
1
∣
\left|\begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1\end{array}\right|
x
2
−
x
+
1
x
−
1
x
+
1
x
+
1
Q3
EXERCISE 4.1
If
A
=
[
1
2
4
2
]
\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
∣
.
Q4
EXERCISE 4.1
If
A
=
[
1
0
1
0
1
2
0
0
4
]
\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
∣
.
Q5
EXERCISE 4.1
Evaluate the determinants
(i)
∣
3
−
1
−
2
0
0
−
1
3
−
5
0
∣
\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
(ii)
∣
3
−
4
5
1
1
−
2
2
3
1
∣
\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
(iii)
∣
0
1
2
−
1
0
−
3
−
2
3
0
∣
\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
(iv)
∣
2
−
1
−
2
0
2
−
1
3
−
5
0
∣
\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
Q6
EXERCISE 4.1
If
A
=
[
1
1
−
2
2
1
−
3
5
4
−
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
∣
.
Q7
EXERCISE 4.1
Find values of
x
x
x
, if
(i)
∣
2
4
5
1
∣
=
∣
2
x
4
6
x
∣
\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
=
2
x
4
6
x
(ii)
∣
2
3
4
5
∣
=
∣
x
3
2
x
5
∣
\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
2
x
5
Q8
EXERCISE 4.1
If
∣
x
2
18
x
∣
=
∣
6
2
18
6
∣
\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
x
x
x
is equal to
(A) 6
(B)
±
6
\pm 6
±
6
(C) -6
(D) 0
Q1
EXERCISE 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
)
(1,0),(6,0),(4,3)
(
1
,
0
)
,
(
6
,
0
)
,
(
4
,
3
)
(ii)
(
2
,
7
)
,
(
1
,
1
)
,
(
10
,
8
)
(2,7),(1,1),(10,8)
(
2
,
7
)
,
(
1
,
1
)
,
(
10
,
8
)
(iii)
(
−
2
,
−
3
)
,
(
3
,
2
)
,
(
−
1
,
−
8
)
(-2,-3),(3,2),(-1,-8)
(
−
2
,
−
3
)
,
(
3
,
2
)
,
(
−
1
,
−
8
)
Q2
EXERCISE 4.2
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.
Q3
EXERCISE 4.2
Find values of
k
k
k
if area of triangle is 4 sq . units and vertices are
(i)
(
k
,
0
)
,
(
4
,
0
)
,
(
0
,
2
)
(k, 0),(4,0),(0,2)
(
k
,
0
)
,
(
4
,
0
)
,
(
0
,
2
)
(ii)
(
−
2
,
0
)
,
(
0
,
4
)
,
(
0
,
k
)
(-2,0),(0,4),(0, k)
(
−
2
,
0
)
,
(
0
,
4
)
,
(
0
,
k
)
Q4
EXERCISE 4.2
(i)
Find equation of line joining
(
1
,
2
)
(1,2)
(
1
,
2
)
and
(
3
,
6
)
(3,6)
(
3
,
6
)
using determinants.
(ii)
Find equation of line joining
(
3
,
1
)
(3,1)
(
3
,
1
)
and
(
9
,
3
)
(9,3)
(
9
,
3
)
using determinants.
Q5
EXERCISE 4.2
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
k
k
k
is
(A) 12
(B) -2
(C)
−
12
,
−
2
-12,-2
−
12
,
−
2
(D)
12
,
−
2
12,-2
12
,
−
2
Q1
EXERCISE 4.3
Write Minors and Cofactors of the elements of following determinants:
(i)
∣
2
−
4
0
3
∣
\left|\begin{array}{rr}2 & -4 \ 0 & 3\end{array}\right|
2
−
4
0
3
(ii)
∣
a
c
b
d
∣
\left|\begin{array}{ll}a & c \ b & d\end{array}\right|
a
c
b
d
Q2
EXERCISE 4.3
Write Minors and Cofactors of the elements of following determinants:
(i)
∣
1
0
0
0
1
0
0
0
1
∣
\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
(ii)
∣
1
0
4
3
5
−
1
0
1
2
∣
\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
Q3
EXERCISE 4.3
Using Cofactors of elements of second row, evaluate
Δ
=
∣
5
3
8
2
0
1
1
2
3
∣
\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
.
Q4
EXERCISE 4.3
Using Cofactors of elements of third column, evaluate
Δ
=
∣
1
x
y
z
1
y
z
x
1
z
x
y
∣
\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
z
x
1
z
x
y
.
Q5
EXERCISE 4.3
If
Δ
=
∣
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
∣
\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|
Δ
=
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
and
A
i
j
\mathrm{A}_{i j}
A
ij
is Cofactors of
a
i
j
a_{i j}
a
ij
, then value of
Δ
\Delta
Δ
is given by
(A)
a
11
A
31
+
a
12
A
32
+
a
13
A
33
a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33}
a
11
A
31
+
a
12
A
32
+
a
13
A
33
(B)
a
11
A
11
+
a
12
A
21
+
a
13
A
31
a_{11} \mathrm{~A}_{11}+a_{12} \mathrm{~A}_{21}+a_{13} \mathrm{~A}_{31}
a
11
A
11
+
a
12
A
21
+
a
13
A
31
(C)
a
21
A
11
+
a
22
A
12
+
a
23
A
13
a_{21} \mathrm{~A}_{11}+a_{22} \mathrm{~A}_{12}+a_{23} \mathrm{~A}_{13}
a
21
A
11
+
a
22
A
12
+
a
23
A
13
(D)
a
11
A
11
+
a
21
A
21
+
a
31
A
31
a_{11} \mathrm{~A}_{11}+a_{21} \mathrm{~A}_{21}+a_{31} \mathrm{~A}_{31}
a
11
A
11
+
a
21
A
21
+
a
31
A
31
Q1
EXERCISE 4.4
Find adjoint of each of the matrices in Exercises 1 and 2.
1
2
3
4
\begin{array}{ll}1 & 2 \ 3 & 4\end{array}
1
2
3
4
Q2
EXERCISE 4.4
1
−
1
2
2
3
5
−
2
0
1
\begin{array}{ccc}1 & -1 & 2 \ 2 & 3 & 5 \ -2 & 0 & 1\end{array}
1
−
1
2
2
3
5
−
2
0
1
Q3
EXERCISE 4.4
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
(
adj
A
)
=
(
adj
A
)
A
=
∣
A
∣
I
in Exercises 3 and 4 3.
2
3
−
4
−
6
\begin{array}{cc}2 & 3 \ -4 & -6\end{array}
2
3
−
4
−
6
Q4
EXERCISE 4.4
1
−
1
2
3
0
−
2
1
0
3
\begin{array}{lll}1 & -1 & 2 \ 3 & 0 & -2 \ 1 & 0 & 3\end{array}
1
−
1
2
3
0
−
2
1
0
3
Q5
EXERCISE 4.4
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 5.
2
−
2
4
3
\begin{array}{cc}2 & -2 \ 4 & 3\end{array}
2
−
2
4
3
Q6
EXERCISE 4.4
−
1
5
−
3
2
\begin{array}{rr}-1 & 5 \ -3 & 2\end{array}
−
1
5
−
3
2
Q7
EXERCISE 4.4
1
2
3
0
2
4
0
0
5
\begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5\end{array}
1
2
3
0
2
4
0
0
5
Q8
EXERCISE 4.4
1
0
0
3
3
0
5
2
−
1
\begin{array}{ccc}1 & 0 & 0 \ 3 & 3 & 0 \ 5 & 2 & -1\end{array}
1
0
0
3
3
0
5
2
−
1
Q9
EXERCISE 4.4
2
1
3
4
−
1
0
−
7
2
1
\begin{array}{ccc}2 & 1 & 3 \ 4 & -1 & 0 \ -7 & 2 & 1\end{array}
2
1
3
4
−
1
0
−
7
2
1
Q10
EXERCISE 4.4
1
−
1
2
0
2
−
3
3
−
2
4
\begin{array}{ccc}1 & -1 & 2 \ 0 & 2 & -3 \ 3 & -2 & 4\end{array}
1
−
1
2
0
2
−
3
3
−
2
4
Q11
EXERCISE 4.4
[
1
0
0
0
cos
α
sin
α
0
sin
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
cos
α
sin
α
0
sin
α
−
cos
α
]
Q12
EXERCISE 4.4
Let
A
=
3
7
2
5
\mathrm{A}=\begin{array}{ll}3 & 7 \ 2 & 5\end{array}
A
=
3
7
2
5
and
B
=
6
8
7
9
\mathrm{B}=\begin{array}{ll}6 & 8 \ 7 & 9\end{array}
B
=
6
8
7
9
. Verify that
(
A
B
)
−
1
=
B
−
1
A
−
1
(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}
(
AB
)
−
1
=
B
−
1
A
−
1
.
Q13
EXERCISE 4.4
If
A
=
3
1
−
1
2
\mathrm{A}=\begin{array}{cc}3 & 1 \ -1 & 2\end{array}
A
=
3
1
−
1
2
, show that
A
2
−
5
A
+
7
I
=
O
\mathrm{A}^{2}-5 \mathrm{~A}+7 \mathrm{I}=\mathrm{O}
A
2
−
5
A
+
7
I
=
O
. Hence find
A
−
1
\mathrm{A}^{-1}
A
−
1
.
Q14
EXERCISE 4.4
For the matrix
A
=
3
2
1
1
\mathrm{A}=\begin{array}{ll}3 & 2 \ 1 & 1\end{array}
A
=
3
2
1
1
, find the numbers
a
a
a
and
b
b
b
such that
A
2
+
a
A
+
b
I
=
O
\mathrm{A}^{2}+a \mathrm{~A}+b \mathrm{I}=\mathrm{O}
A
2
+
a
A
+
b
I
=
O
.
Q15
EXERCISE 4.4
For the matrix
A
=
1
1
1
1
2
−
3
2
−
1
3
\mathrm{A}=\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3\end{array}
A
=
1
1
1
1
2
−
3
2
−
1
3
Show that
A
3
−
6
A
2
+
5
A
+
11
I
=
O
\mathrm{A}^{3}-6 \mathrm{~A}^{2}+5 \mathrm{~A}+11 \mathrm{I}=\mathrm{O}
A
3
−
6
A
2
+
5
A
+
11
I
=
O
. Hence, find
A
−
1
\mathrm{A}^{-1}
A
−
1
.
Q16
EXERCISE 4.4
If
A
=
2
−
1
1
−
1
2
−
1
1
−
1
2
\mathrm{A}=\begin{array}{ccc}2 & -1 & 1 \ -1 & 2 & -1 \ 1 & -1 & 2\end{array}
A
=
2
−
1
1
−
1
2
−
1
1
−
1
2
Verify that
A
3
−
6
A
2
+
9
A
−
4
I
=
O
\mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I}=\mathrm{O}
A
3
−
6
A
2
+
9
A
−
4
I
=
O
and hence find
A
−
1
\mathrm{A}^{-1}
A
−
1
Q17
EXERCISE 4.4
Let A be a nonsingular square matrix of order
3
×
3
3 \times 3
3
×
3
. Then
∣
adj
A
∣
|\operatorname{adj} \mathrm{A}|
∣
adj
A
∣
is equal to
(A)
∣
A
∣
|\mathrm{A}|
∣
A
∣
(B)
∣
A
∣
2
|\mathrm{A}|^{2}
∣
A
∣
2
(C)
∣
A
∣
3
|\mathrm{A}|^{3}
∣
A
∣
3
(D)
3
∣
A
∣
3|\mathrm{~A}|
3∣
A
∣
Q18
EXERCISE 4.4
If A is an invertible matrix of order 2, then
det
(
A
−
1
)
\operatorname{det}\left(\mathrm{A}^{-1}\right)
det
(
A
−
1
)
is equal to
(A)
det
(
A
)
\operatorname{det}(\mathrm{A})
det
(
A
)
(B)
1
det
(
A
)
\frac{1}{\operatorname{det}(\mathrm{A})}
det
(
A
)
1
(C) 1
(D) 0
Q1
EXERCISE 4.5
Examine the consistency of the system of equations in Exercises 1 to 6.
x
+
2
y
=
2
x+2 y=2
x
+
2
y
=
2
2
x
+
3
y
=
3
2 x+3 y=3
2
x
+
3
y
=
3
Q2
EXERCISE 4.5
2
x
−
y
=
5
2 x-y=5
2
x
−
y
=
5
x
+
y
=
4
x+y=4
x
+
y
=
4
Q3
EXERCISE 4.5
x
+
3
y
=
5
x+3 y=5
x
+
3
y
=
5
2
x
+
6
y
=
8
2 x+6 y=8
2
x
+
6
y
=
8
Q4
EXERCISE 4.5
x
+
y
+
z
=
1
x+y+z=1
x
+
y
+
z
=
1
2
x
+
3
y
+
2
z
=
2
2 x+3 y+2 z=2
2
x
+
3
y
+
2
z
=
2
a
x
+
a
y
+
2
a
z
=
4
a x+a y+2 a z=4
a
x
+
a
y
+
2
a
z
=
4
Q5
EXERCISE 4.5
3
x
−
y
−
2
z
=
2
3 x-y-2 z=2
3
x
−
y
−
2
z
=
2
2
y
−
z
=
−
1
2 y-z=-1
2
y
−
z
=
−
1
3
x
−
5
y
=
3
3 x-5 y=3
3
x
−
5
y
=
3
Q6
EXERCISE 4.5
5
x
−
y
+
4
z
=
5
5 x-y+4 z=5
5
x
−
y
+
4
z
=
5
2
x
+
3
y
+
5
z
=
2
2 x+3 y+5 z=2
2
x
+
3
y
+
5
z
=
2
5
x
−
2
y
+
6
z
=
−
1
5 x-2 y+6 z=-1
5
x
−
2
y
+
6
z
=
−
1
Q7
EXERCISE 4.5
Solve system of linear equations, using matrix method, in Exercises 7 to 14. 7.
5
x
+
2
y
=
4
5 x+2 y=4
5
x
+
2
y
=
4
7
x
+
3
y
=
5
7 x+3 y=5
7
x
+
3
y
=
5
Q8
EXERCISE 4.5
2
x
−
y
=
−
2
2 x-y=-2
2
x
−
y
=
−
2
3
x
+
4
y
=
3
3 x+4 y=3
3
x
+
4
y
=
3
Q9
EXERCISE 4.5
4
x
−
3
y
=
3
4 x-3 y=3
4
x
−
3
y
=
3
3
x
−
5
y
=
7
3 x-5 y=7
3
x
−
5
y
=
7
Q10
EXERCISE 4.5
5
x
+
2
y
=
3
5 x+2 y=3
5
x
+
2
y
=
3
3
x
+
2
y
=
5
3 x+2 y=5
3
x
+
2
y
=
5
Q11
EXERCISE 4.5
2
x
+
y
+
z
=
1
2 x+y+z=1
2
x
+
y
+
z
=
1
x
−
2
y
−
z
=
3
2
x-2 y-z=\frac{3}{2}
x
−
2
y
−
z
=
2
3
3
y
−
5
z
=
9
3 y-5 z=9
3
y
−
5
z
=
9
Q12
EXERCISE 4.5
x
−
y
+
z
=
4
x-y+z=4
x
−
y
+
z
=
4
2
x
+
y
−
3
z
=
0
2 x+y-3 z=0
2
x
+
y
−
3
z
=
0
x
+
y
+
z
=
2
x+y+z=2
x
+
y
+
z
=
2
Q13
EXERCISE 4.5
2
x
+
3
y
+
3
z
=
5
2 x+3 y+3 z=5
2
x
+
3
y
+
3
z
=
5
x
−
2
y
+
z
=
−
4
x-2 y+z=-4
x
−
2
y
+
z
=
−
4
3
x
−
y
−
2
z
=
3
3 x-y-2 z=3
3
x
−
y
−
2
z
=
3
Q14
EXERCISE 4.5
x
−
y
+
2
z
=
7
x-y+2 z=7
x
−
y
+
2
z
=
7
3
x
+
4
y
−
5
z
=
−
5
3 x+4 y-5 z=-5
3
x
+
4
y
−
5
z
=
−
5
2
x
−
y
+
3
z
=
12
2 x-y+3 z=12
2
x
−
y
+
3
z
=
12
Q15
EXERCISE 4.5
If
A
=
[
2
−
3
5
3
2
−
4
1
1
−
2
]
\mathrm{A}=\left[\begin{array}{rrr}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}\right]
A
=
[
2
−
3
5
3
2
−
4
1
1
−
2
]
, find
A
−
1
\mathrm{A}^{-1}
A
−
1
. Using
A
−
1
\mathrm{A}^{-1}
A
−
1
solve the system of equations
2
x
−
3
y
+
5
z
=
11
2 x-3 y+5 z=11
2
x
−
3
y
+
5
z
=
11
3
x
+
2
y
−
4
z
=
−
5
3 x+2 y-4 z=-5
3
x
+
2
y
−
4
z
=
−
5
x
+
y
−
2
z
=
−
3
x+y-2 z=-3
x
+
y
−
2
z
=
−
3
Q16
EXERCISE 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.
Q1
Miscellaneous Exercises on Chapter 4
Prove that the determinant
∣
x
sin
θ
cos
θ
−
sin
θ
−
x
1
cos
θ
1
x
∣
\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \ -\sin \theta & -x & 1 \ \cos \theta & 1 & x\end{array}\right|
x
sin
θ
cos
θ
−
sin
θ
−
x
1
cos
θ
1
x
is independent of
θ
\theta
θ
.
Q2
Miscellaneous Exercises on Chapter 4
Evaluate
∣
cos
α
cos
β
cos
α
sin
β
−
sin
α
−
sin
β
cos
β
0
sin
α
cos
β
sin
α
sin
β
cos
α
∣
\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|
cos
α
cos
β
cos
α
sin
β
−
sin
α
−
sin
β
cos
β
0
sin
α
cos
β
sin
α
sin
β
cos
α
.
Q3
Miscellaneous Exercises on Chapter 4
If
A
−
1
=
[
3
−
1
1
−
15
6
−
5
5
−
2
2
]
\mathrm{A}^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2\end{array}\right]
A
−
1
=
[
3
−
1
1
−
15
6
−
5
5
−
2
2
]
and
B
=
[
1
2
−
2
−
1
3
0
0
−
2
1
]
\mathrm{B}=\left[\begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1\end{array}\right]
B
=
[
1
2
−
2
−
1
3
0
0
−
2
1
]
, find
(
A
B
)
−
1
(\mathrm{AB})^{-1}
(
AB
)
−
1
Q4
Miscellaneous Exercises on Chapter 4
Let
A
=
1
−
2
1
−
2
3
1
1
1
5
\mathrm{A}=\begin{array}{ccc}1 & -2 & 1 \ -2 & 3 & 1 \ 1 & 1 & 5\end{array}
A
=
1
−
2
1
−
2
3
1
1
1
5
. Verify that
(i)
[
adj
A
]
−
1
=
adj
(
A
−
1
)
[\operatorname{adj} \mathrm{A}]^{-1}=\operatorname{adj}\left(\mathrm{A}^{-1}\right)
[
adj
A
]
−
1
=
adj
(
A
−
1
)
(ii)
(
A
−
1
)
−
1
=
A
\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}
(
A
−
1
)
−
1
=
A
Q5
Miscellaneous Exercises on Chapter 4
Evaluate
∣
x
y
x
+
y
y
x
+
y
x
x
+
y
x
y
∣
\left|\begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & y\end{array}\right|
x
y
x
+
y
y
x
+
y
x
x
+
y
x
y
Q6
Miscellaneous Exercises on Chapter 4
Evaluate
∣
1
x
y
1
x
+
y
y
1
x
x
+
y
∣
\left|\begin{array}{ccc}1 & x & y \ 1 & x+y & y \ 1 & x & x+y\end{array}\right|
1
x
y
1
x
+
y
y
1
x
x
+
y
Q7
Miscellaneous 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}$$
Q8
Miscellaneous Exercises on Chapter 4
If
x
,
y
,
z
x, y, z
x
,
y
,
z
are nonzero real numbers, then the inverse of matrix
A
=
[
x
0
0
0
y
0
0
0
z
]
\mathrm{A}=\left[\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}\right]
A
=
[
x
0
0
0
y
0
0
0
z
]
is
(A)
[
x
−
1
0
0
0
y
−
1
0
0
0
z
−
1
]
\left[\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}\right]
[
x
−
1
0
0
0
y
−
1
0
0
0
z
−
1
]
(B)
x
y
z
[
x
−
1
0
0
0
y
−
1
0
0
0
z
−
1
]
x y z\left[\begin{array}{ccc}x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1}\end{array}\right]
x
yz
[
x
−
1
0
0
0
y
−
1
0
0
0
z
−
1
]
(C)
1
x
y
z
[
x
0
0
0
y
0
0
0
z
]
\frac{1}{x y z}\left[\begin{array}{ccc}x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z\end{array}\right]
x
yz
1
[
x
0
0
0
y
0
0
0
z
]
(D)
1
x
y
z
[
1
0
0
0
1
0
0
0
1
]
\frac{1}{x y z}\left[\begin{array}{lll}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right]
x
yz
1
[
1
0
0
0
1
0
0
0
1
]
Q9
Miscellaneous Exercises on Chapter 4
Let
A
=
[
1
sin
θ
1
−
sin
θ
1
sin
θ
−
1
−
sin
θ
1
]
\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \ -\sin \theta & 1 & \sin \theta \ -1 & -\sin \theta & 1\end{array}\right]
A
=
[
1
sin
θ
1
−
sin
θ
1
sin
θ
−
1
−
sin
θ
1
]
, where
0
≤
θ
≤
2
π
0 \leq \theta \leq 2 \pi
0
≤
θ
≤
2
π
. Then
(A)
Det
(
A
)
=
0
\operatorname{Det}(\mathrm{A})=0
Det
(
A
)
=
0
(B)
Det
(
A
)
∈
(
2
,
∞
)
\operatorname{Det}(\mathrm{A}) \in(2, \infty)
Det
(
A
)
∈
(
2
,
∞
)
(C)
Det
(
A
)
∈
(
2
,
4
)
\operatorname{Det}(\mathrm{A}) \in(2,4)
Det
(
A
)
∈
(
2
,
4
)
(D)
Det
(
A
)
∈
[
2
,
4
]
\operatorname{Det}(\mathrm{A}) \in[2,4]
Det
(
A
)
∈
[
2
,
4
]
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