MatricesClass 12 Mathematics NCERT Solutions
56 Solutions
Generated by KedovoAI
Solution 1 of 56
Q1EXERCISE 3.1
In the matrix , write:
(i)
The order of the matrix,
(ii)
The number of elements,
(iii)
Write the elements .
Solution
Given:
The matrix
Solution:
(i) The order of the matrix
The matrix A has 3 rows and 4 columns.
Therefore, the order of the matrix A is .
(ii) The number of elements
The number of elements in a matrix of order is .
Here, and .
So, the number of elements is .
(iii) Write the elements
The element refers to the element in the row and column.
- is the element in the 1st row and 3rd column, which is 19.
- is the element in the 2nd row and 1st column, which is 35.
- is the element in the 3rd row and 3rd column, which is -5.
- is the element in the 2nd row and 4th column, which is 12.
- is the element in the 2nd row and 3rd column, which is .
Final Answer:
(i)
The order of the matrix is .
(ii)
The number of elements is 12.
(iii)
The elements are: .
Q2EXERCISE 3.1
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution
Given:
Case 1: A matrix has 24 elements.
Case 2: A matrix has 13 elements.
To Find:
The possible orders for each case.
Solution:
If a matrix is of order , it has elements. To find all possible orders of a matrix with a given number of elements, we need to find all ordered pairs of natural numbers whose product is that number.
Case 1: 24 elements
We need to find all ordered pairs of natural numbers such that .
The pairs are:
(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4).
So, the possible orders are:
.
Case 2: 13 elements
We need to find all ordered pairs of natural numbers such that .
Since 13 is a prime number, its only factors are 1 and 13.
The pairs are:
(1, 13), (13, 1).
So, the possible orders are:
.
Final Answer:
For 24 elements, the possible orders are .
For 13 elements, the possible orders are .
Q3EXERCISE 3.1
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution
Given:
Case 1: A matrix has 18 elements.
Case 2: A matrix has 5 elements.
To Find:
The possible orders for each case.
Solution:
If a matrix is of order , it has elements. We need to find all ordered pairs of natural numbers whose product is the given number of elements.
Case 1: 18 elements
We need to find all ordered pairs of natural numbers such that .
The pairs are:
(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3).
So, the possible orders are:
.
Case 2: 5 elements
We need to find all ordered pairs of natural numbers such that .
Since 5 is a prime number, its only factors are 1 and 5.
The pairs are:
(1, 5), (5, 1).
So, the possible orders are:
.
Final Answer:
For 18 elements, the possible orders are .
For 5 elements, the possible orders are .
Q4EXERCISE 3.1
Construct a matrix, , whose elements are given by:
(i)
(ii)
(iii)
Solution
To Construct: A matrix
Solution:
(i)
So, the matrix is .
(ii)
So, the matrix is .
(iii)
So, the matrix is .
Final Answer:
(i)
(ii)
(iii)
Q5EXERCISE 3.1
Construct a matrix, whose elements are given by:
(i)
(ii)
Solution
To Construct: A matrix
Solution:
(i)
So, the matrix is .
(ii)
So, the matrix is .
Final Answer:
(i)
(ii)
Q6EXERCISE 3.1
Find the values of and from the following equations:
(i)
(ii)
(iii)
Solution
Given: Three matrix equations.
To Find: The values of and .
Principle:
If two matrices are equal, their corresponding elements must be equal.
Solution:
(i)
By comparing the corresponding elements, we get:
(ii)
By comparing the corresponding elements, we get:
From (1), we have . Substituting this into (2):
Factoring the quadratic equation:
So, or .
If , then .
If , then .
Thus, the solutions are () or ().
(iii)
By comparing the corresponding elements, we get:
Substitute (2) into (1):
Substitute into (3):
Substitute into (2):
Final Answer:
(i)
(ii)
or
(iii)
Q7EXERCISE 3.1
Find the value of and from the equation:
Solution
Given: The matrix equation .
To Find: The values of and .
Solution:
Since the two matrices are equal, their corresponding elements must be equal. This gives us a system of four linear equations:
First, solve for and using equations (1) and (2).
From (2), .
Substitute this into (1):
Now find using :
Next, solve for using equation (3) and the value of .
Finally, solve for using equation (4) and the value of .
Final Answer:
The values are .
Q8EXERCISE 3.1
is a square matrix, if
(A)
(C)
(D)
None of these
Solution
Definition:
A square matrix is a matrix in which the number of rows is equal to the number of columns.
Given:
A matrix where is the number of rows and is the number of columns.
Condition for a square matrix:
For A to be a square matrix, the number of rows must equal the number of columns. Therefore, .
Conclusion:
Comparing this with the given options:
(A) represents a vertical rectangular matrix.
(C) is the condition for a square matrix.
Final Answer: (C)
Q9EXERCISE 3.1
Which of the given values of and make the following pair of matrices equal
(A)
(B)
Not possible to find
(C)
(D)
Solution
Given: The two matrices are equal:
To Find: The values of and .
Solution:
For the matrices to be equal, their corresponding elements must be equal. We set up the equations:
Let's solve these equations.
From (1):
From (4):
We have found two different values for from two different equations ( and ). Since cannot have two different values simultaneously, there is a contradiction. Therefore, it is not possible to find values of and that make the matrices equal.
Let's check the equations for for completeness:
From (2):
From (3):
The value for is consistent (), but the value for is not.
Conclusion:
Because we get conflicting values for , it is not possible to find values for and that make the matrices equal.
Final Answer: (B) Not possible to find
Q10EXERCISE 3.1
The number of all possible matrices of order with each entry 0 or 1 is:
(A)
27
(B)
18
(C)
81
(D)
512
Solution
Given:
Order of the matrix = .
Possible entries for each element = 0 or 1.
To Find:
The total number of possible matrices.
Solution:
A matrix of order has elements.
Each of these 9 elements can be filled in two possible ways (either with 0 or with 1).
Using the fundamental principle of counting, the total number of ways to fill all 9 positions is the product of the number of choices for each position.
Total number of possible matrices = (Number of choices for the 1st element) (Number of choices for the 2nd element) ... (Number of choices for the 9th element)
Total number of possible matrices =
Total number of possible matrices =
Calculating :
Conclusion:
There are 512 possible matrices of order with each entry being either 0 or 1.
Final Answer: (D) 512
Q1EXERCISE 3.2
Let Find each of the following:
(i)
(ii)
(iii)
(iv)
AB
(v)
BA
Solution
Given:
Solution:
(i) A + B
(ii) A - B
(iii) 3A - C
First, find 3A:
Now, subtract C:
(iv) AB
(v) BA
Final Answer:
(i)
(ii)
(iii)
(iv)
(v)
Q2EXERCISE 3.2
Compute the following:
(i)
(ii)
(iii)
(iv)
Solution
To Compute: The sum of the given matrices.
Solution:
(i)
(ii)
Using the algebraic identities and , we get:
(iii)
(iv)
Using the trigonometric identity , we get:
Final Answer:
(i)
(ii)
(iii)
(iv)
Q3EXERCISE 3.2
Compute the indicated products.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution
To Compute: The product of the given matrices.
Solution:
(i)
(ii)
The first matrix is and the second is . The product will be a matrix.
(iii)
The first matrix is and the second is . The product will be a matrix.
(iv)
(v)
The first matrix is and the second is . The product will be a matrix.
(vi)
The first matrix is and the second is . The product will be a matrix.
Final Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Q4EXERCISE 3.2
If and , then compute and . Also, verify that .
Solution
Given:
To Compute and Verify:
- Compute
- Compute
- Verify that
Solution:
1. Compute (A + B)
2. Compute (B - C)
3. Verify A + (B - C) = (A + B) - C
LHS: A + (B - C)
RHS: (A + B) - C
Since LHS = RHS, the associative property of addition for matrices is verified.
Final Answer:
Since LHS = and RHS = , it is verified that .
Q5EXERCISE 3.2
If and , then compute .
Solution
Given:
and
To Compute:
Solution:
First, calculate :
Next, calculate :
Now, compute :
This is a zero matrix of order .
Final Answer:
Q6EXERCISE 3.2
Simplify
Solution
To Simplify: The given matrix expression.
Solution:
First, perform the scalar multiplications:
Now, add the two resulting matrices:
Using the trigonometric identity , we simplify the elements:
This is the identity matrix of order 2, denoted by I.
Final Answer:
Q7EXERCISE 3.2
Find X and Y, if
(i)
and
(ii)
and
Solution
To Find: The matrices X and Y.
Solution:
(i)
Let the given equations be:
Add equation (1) and (2) to eliminate Y:
Substitute X back into equation (1) to find Y:
(ii)
Let the given equations be:
To eliminate Y, multiply equation (1) by 2 and equation (2) by 3:
Subtract equation (3) from (4):
Substitute X back into equation (1) to find Y:
Final Answer:
(i)
,
(ii)
,
Q8EXERCISE 3.2
Find X, if and
Solution
Given:
To Find: The matrix X.
Solution:
We have the equation .
To find X, first isolate :
Substitute the given matrix Y into the equation:
Perform the matrix subtraction:
Now, solve for X by multiplying by :
Final Answer:
Q9EXERCISE 3.2
Find and , if
Solution
Given: The matrix equation
To Find: The values of and .
Solution:
First, perform the scalar multiplication and addition on the left-hand side (LHS) of the equation:
LHS =
LHS =
LHS =
Now, set the resulting LHS matrix equal to the right-hand side (RHS) matrix:
For the matrices to be equal, their corresponding elements must be equal. This gives us two equations:
Solve equation (1) for :
Solve equation (2) for :
Final Answer:
and .
Q10EXERCISE 3.2
Solve the equation for and , if
Solution
Given: The matrix equation
To Find: The values of and .
Solution:
First, perform the scalar multiplications on both sides of the equation:
Next, perform the matrix addition on the left-hand side:
For the matrices to be equal, their corresponding elements must be equal. This gives us four equations:
Now, solve each equation:
From (1):
From (2):
From (3):
From (4):
Final Answer:
.
Q11EXERCISE 3.2
If , find the values of and .
Solution
Given: The matrix equation
To Find: The values of and .
Solution:
First, perform the scalar multiplications on the left-hand side (LHS):
LHS =
Next, perform the matrix addition on the LHS:
LHS =
Now, set the LHS equal to the right-hand side (RHS):
By the equality of matrices, we can equate the corresponding elements to get a system of linear equations:
Add equation (1) and (2) to eliminate :
Substitute the value of into equation (2) to find :
Final Answer:
and .
Q12EXERCISE 3.2
Given , find the values of and .
Solution
Given: The matrix equation
To Find: The values of and .
Solution:
First, simplify both sides of the equation.
LHS =
RHS =
Now, equate the LHS and RHS:
By the equality of matrices, we equate the corresponding elements:
Solve each equation:
From (1): .
From (4): .
Substitute into (2):
.
Substitute into (3):
.
Final Answer:
.
Q13EXERCISE 3.2
If , show that .
Solution
Given:
To Show:
Proof:
First, we write out the matrices for , , and .
Now, we compute the product on the LHS: .
LHS =
Performing matrix multiplication:
Row 1:
Element (1,1):
Element (1,2):
Element (1,3):
Row 2:
Element (2,1):
Element (2,2):
Element (2,3):
Row 3:
Element (3,1):
Element (3,2):
Element (3,3):
So, the product matrix is:
Now, we use the trigonometric sum identities:
Substituting these identities into the product matrix:
This is exactly the matrix .
Therefore, LHS = RHS.
Hence Proved.
Q14EXERCISE 3.2
Show that
(i)
(ii)
Solution
To Show: That matrix multiplication is not commutative for the given pairs of matrices.
Solution:
(i)
Let and . We need to show .
Calculate LHS = AB:
Calculate RHS = BA:
Since , we have shown that .
(ii)
Let and . We need to show .
Calculate LHS = CD:
Calculate RHS = DC:
Since , we have shown that .
Hence Shown.
Q15EXERCISE 3.2
Find , if
Solution
Given:
To Find: The value of the matrix expression .
Solution:
First, we need to calculate .
Next, calculate .
Next, write down . Here, I is the identity matrix of order 3.
Now, substitute these matrices into the expression :
Final Answer:
Q16EXERCISE 3.2
If , prove that
Solution
Given:
To Prove:
Proof:
First, calculate .
Next, calculate .
Now, substitute the calculated matrices into the expression :
Expression =
Hence Proved.
Q17EXERCISE 3.2
If and , find so that
Solution
Given:
,
The equation .
To Find: The value of .
Solution:
First, calculate .
Now, calculate the right-hand side (RHS) of the given equation:
Now, equate with :
By the equality of matrices, we can equate the corresponding elements:
Solving each equation for :
From (1): .
From (2): .
From (3): .
From (4): .
All equations give the same value .
Final Answer:
.
Q18EXERCISE 3.2
If and I is the identity matrix of order 2, show that
Solution
Given:
, I is the identity matrix of order 2.
To Show:
Proof:
Let . Then .
Calculate LHS = I + A:
LHS =
Calculate RHS = (I - A) Rotation Matrix:
First, find I - A:
We use the half-angle formulas for and in terms of :
So the rotation matrix is .
Now, calculate the product for the RHS:
RHS =
Let's multiply the matrices:
Element (1,1):
Element (1,2):
Element (2,1):
Element (2,2):
So, RHS =
Comparing LHS and RHS:
LHS = and RHS = .
Since LHS = RHS, the identity is proved.
Hence Shown.
Q19EXERCISE 3.2
A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a)
₹1800
(b)
₹2000
Solution
Given:
Total investment = ₹ 30,000
Interest rate of first bond = 5% = 0.05
Interest rate of second bond = 7% = 0.07
To Find: How to divide the investment to get specific total interests.
Let:
Let the amount invested in the first bond be ₹ .
Then the amount invested in the second bond will be ₹ .
We can represent the investments as a row matrix A and the interest rates as a column matrix B.
Investment matrix
Interest rate matrix
The total annual interest is given by the product AB.
Total Interest =
Total Interest =
Total Interest =
Total Interest =
Solution:
(a) Annual total interest of ₹1800
We set the total interest equal to 1800.
By equality of matrices:
So, investment in the first bond = ₹ = ₹ 15,000.
Investment in the second bond = ₹ = ₹ = ₹ 15,000.
(b) Annual total interest of ₹2000
We set the total interest equal to 2000.
By equality of matrices:
So, investment in the first bond = ₹ = ₹ 5,000.
Investment in the second bond = ₹ = ₹ = ₹ 25,000.
Final Answer:
(a) To obtain an annual interest of ₹1800, the trust must invest ₹ 15,000 in the first bond and ₹ 15,000 in the second bond.
(b) To obtain an annual interest of ₹2000, the trust must invest ₹ 5,000 in the first bond and ₹ 25,000 in the second bond.
Q20EXERCISE 3.2
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution
Given:
Number of chemistry books = 10 dozen =
Number of physics books = 8 dozen =
Number of economics books = 10 dozen =
Selling price of a chemistry book = ₹ 80
Selling price of a physics book = ₹ 60
Selling price of an economics book = ₹ 40
To Find: The total amount received from selling all the books, using matrix algebra.
Solution:
Let's represent the number of books as a row matrix Q (for quantity).
Let's represent the selling prices as a column matrix P (for price).
The total amount received is the product of these two matrices, QP.
Total Amount =
Total Amount =
Total Amount =
Total Amount =
So, the total amount the bookshop will receive is ₹ 20,160.
Final Answer:
The total amount the bookshop will receive from selling all the books is ₹ 20,160.
Q21EXERCISE 3.2
Assume X, Y, Z, W and P are matrices of order and , respectively. Choose the correct answer in Exercises 21 and 22. 21. The restriction on and so that PY + WY will be defined are:
(A)
(B)
is arbitrary,
(C)
is arbitrary,
(D)
Solution
Given:
Order of matrix P is .
Order of matrix Y is .
Order of matrix W is .
To Find: The restrictions on and for the expression PY + WY to be defined.
Solution:
For the sum PY + WY to be defined, both matrix products PY and WY must be defined, and the resulting matrices must have the same order.
Step 1: Condition for PY to be defined
The product of two matrices is defined only if the number of columns of the first matrix is equal to the number of rows of the second matrix.
For PY:
Order of P is .
Order of Y is .
Number of columns of P = .
Number of rows of Y = .
For PY to be defined, we must have .
If , the order of the resulting matrix PY is , which is .
Step 2: Condition for WY to be defined
For WY:
Order of W is .
Order of Y is .
Number of columns of W = .
Number of rows of Y = .
Since the number of columns of W is equal to the number of rows of Y, the product WY is always defined.
The order of the resulting matrix WY is . Since we found , the order of WY is .
Step 3: Condition for PY + WY to be defined
The sum of two matrices is defined only if they have the same order.
Order of PY is .
Order of WY is .
For the sum to be defined, we must have: Order of PY = Order of WY.
This implies that .
Conclusion:
The restrictions are and .
This corresponds to option (A).
Final Answer: (A)
Q22EXERCISE 3.2
If , then the order of the matrix is:
(A)
(B)
(C)
(D)
Solution
Given:
Order of matrix X is .
Order of matrix Z is .
The condition .
To Find: The order of the matrix .
Solution:
For the expression to be defined, the matrices X and Z must have the same order.
Order of X is .
Order of Z is .
We are given that .
Substituting into the order of Z, we find that the order of Z is .
Since both X and Z have the same order (), the difference is defined.
Scalar multiplication does not change the order of a matrix. So, the order of is and the order of is .
The result of adding or subtracting matrices of the same order is a matrix of that same order.
Therefore, the order of the matrix is .
Since , the order can also be written as . Looking at the options, is available.
Conclusion:
The order of is .
This corresponds to option (B).
Final Answer: (B)
Q1EXERCISE 3.3
Find the transpose of each of the following matrices:
(i)
(ii)
(iii)
Solution
Definition:
The transpose of a matrix is obtained by interchanging its rows and columns. The transpose of a matrix A is denoted by A' or A^T.
Solution:
(i)
Let . This is a column matrix.
Its transpose will be a row matrix.
(ii)
Let . This is a matrix.
To find the transpose, interchange the rows and columns.
(iii)
Let . This is a matrix.
To find the transpose, interchange the rows and columns.
Final Answer:
(i)
(ii)
(iii)
Q2EXERCISE 3.3
If and , then verify that
(i)
,
(ii)
Solution
Given:
,
Solution:
First, let's find the transposes of A and B.
(i) Verify
LHS:
First, find A + B:
Now, find the transpose of (A+B):
RHS:
Since LHS = RHS, the property is verified.
(ii) Verify
LHS:
First, find A - B:
Now, find the transpose of (A-B):
RHS:
Since LHS = RHS, the property is verified.
Final Answer:
(i)
LHS = RHS = . Hence verified.
(ii)
LHS = RHS = . Hence verified.
Q3EXERCISE 3.3
If and , then verify that
(i)
(ii)
Solution
Given:
,
Solution:
First, we need to find matrices A and B'.
From A', we can find A by taking its transpose:
From B, we find B':
(i) Verify
LHS:
First, find A + B:
Now, find the transpose of (A+B):
RHS:
Since LHS = RHS, the property is verified.
(ii) Verify
LHS:
First, find A - B:
Now, find the transpose of (A-B):
RHS:
Since LHS = RHS, the property is verified.
Final Answer:
(i)
LHS = RHS = . Hence verified.
(ii)
LHS = RHS = . Hence verified.
Q4EXERCISE 3.3
If and , then find
Solution
Given:
,
To Find:
Solution:
We can use the property of transpose: . Also, .
So, .
First, we need to find B'.
Now, we can compute .
Alternative Method:
First find A, then A + 2B, and then its transpose.
Both methods give the same result.
Final Answer:
Q5EXERCISE 3.3
For the matrices A and B, verify that , where
(i)
(ii)
Solution
To Verify: The reversal law for transpose, .
Solution:
(i)
Given: ,
LHS:
First, find AB. A is and B is , so AB will be .
Now, find the transpose of AB:
RHS:
First, find A' and B'.
Now, find the product B'A'. B' is and A' is , so B'A' will be .
Since LHS = RHS, the property is verified.
(ii)
Given:
LHS:
First, find AB. A is and B is , so AB will be .
Now, find the transpose of AB:
RHS:
First, find A' and B'.
Now, find the product B'A'. B' is and A' is , so B'A' will be .
Since LHS = RHS, the property is verified.
Final Answer:
(i)
LHS = RHS = . Hence verified.
(ii)
LHS = RHS = . Hence verified.
Q6EXERCISE 3.3
If (i) , then verify that (ii) If , then verify that
Solution
To Verify: , where I is the identity matrix of order 2.
Solution:
(i)
Given:
First, find the transpose of A:
Now, compute the product A'A:
Using the trigonometric identity :
Hence, is verified.
(ii)
Given:
First, find the transpose of A:
Now, compute the product A'A:
Using the trigonometric identity :
Hence, is verified.
Final Answer:
In both cases, it is verified that .
Q7EXERCISE 3.3
(i) Show that the matrix A=left[ right] is a symmetric matrix. (ii) Show that the matrix A=left[ right] is a…
(i)
Show that the matrix is a symmetric matrix.
(ii)
Show that the matrix is a skew symmetric matrix.
Solution
Definitions:
A square matrix A is symmetric if .
A square matrix A is skew-symmetric if .
Solution:
(i) Show that A is a symmetric matrix.
Given:
To show that A is symmetric, we need to find its transpose A' and check if A' = A.
Comparing A' with A, we see that they are identical.
Since , the matrix A is a symmetric matrix.
Hence Shown.
(ii) Show that A is a skew symmetric matrix.
Given:
To show that A is skew-symmetric, we need to find its transpose A' and check if A' = -A.
First, find A':
Next, find -A:
Comparing A' with -A, we see that they are identical.
Since , the matrix A is a skew-symmetric matrix.
Hence Shown.
Q8EXERCISE 3.3
For the matrix , verify that
(i)
is a symmetric matrix
(ii)
is a skew symmetric matrix
Solution
Given:
Solution:
First, find the transpose of A, A'.
(i) Verify that is a symmetric matrix
Let . We need to show that P is symmetric, i.e., .
First, calculate P:
Now, find the transpose of P:
Since , the matrix is symmetric.
Hence Verified.
(ii) Verify that is a skew symmetric matrix
Let . We need to show that Q is skew-symmetric, i.e., .
First, calculate Q:
Now, find the transpose of Q:
Next, find -Q:
Since , the matrix is skew-symmetric.
Hence Verified.
Q9EXERCISE 3.3
Find and , when
Solution
Given:
To Find: and .
Solution:
First, find the transpose of A, A'.
Now, calculate :
Then, calculate :
Next, calculate :
Then, calculate :
Note: The given matrix A is a skew-symmetric matrix, so .
Therefore, , and .
This confirms our results.
Final Answer:
Q10EXERCISE 3.3
Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i)
(ii)
(iii)
(iv)
Solution
Theorem: Any square matrix A can be expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where and .
Solution:
(i)
(ii)
This matrix is already symmetric, since .
(iii)
(iv)
Final Answer:
(i)
(ii)
(iii)
(iv)
Q11EXERCISE 3.3
Choose the correct answer in the Exercises 11 and 12. 11. If A, B are symmetric matrices of same order, then AB - BA is a
(A)
Skew symmetric matrix
(B)
Symmetric matrix
(C)
Zero matrix
(D)
Identity matrix
Solution
Given:
A and B are symmetric matrices of the same order.
This means and .
To Determine: The nature of the matrix .
Solution:
Let .
To check if C is symmetric or skew-symmetric, we find its transpose, C'.
Using the property , we get:
Using the reversal law for transpose, , we get:
Now, substitute the given conditions and :
We can factor out -1 to compare with C:
Since , we have:
The condition is the definition of a skew-symmetric matrix.
Therefore, is a skew-symmetric matrix.
Final Answer: (A) Skew symmetric matrix
Q12EXERCISE 3.3
If , and , then the value of is
(A)
(B)
(C)
(D)
Solution
Given:
, where I is the identity matrix of order 2.
To Find: The value of .
Solution:
First, find the transpose of A, A'.
Now, use the given equation .
Perform the matrix addition on the left side:
By the equality of matrices, we can equate the corresponding elements.
Solve for :
From the given options, we need to find which value of satisfies this equation.
(A)
(B)
(C)
(D)
The correct value is .
Final Answer: (B)
Q1EXERCISE 3.4
Matrices A and B will be inverse of each other only if
(A)
(B)
(C)
(D)
Solution
Definition of Inverse Matrix:
If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that (where I is the identity matrix of order m), then B is called the inverse of A (denoted by ) and A is the inverse of B.
Analysis of Options:
(A) : This is the condition for commutativity of multiplication, not for being inverses.
(B) : This means the product is the zero matrix, not the identity matrix.
(C) : This is not the definition of an inverse. The product in both orders must be the identity matrix.
(D) : This is the precise definition of two matrices being inverses of each other.
Conclusion:
Based on the definition, two matrices A and B are inverses of each other if and only if their product, in both orders, is the identity matrix.
Final Answer: (D)
Q1Miscellaneous Exercise on Chapter 3
If A and B are symmetric matrices, prove that AB - BA is a skew symmetric matrix.
Solution
Given:
A and B are symmetric matrices.
This implies and .
To Prove:
AB - BA is a skew-symmetric matrix.
Proof:
Let .
To prove that C is a skew-symmetric matrix, we must show that .
Let's find the transpose of C:
Using the property of transpose of a difference, :
Using the reversal law for the transpose of a product, :
Now, we use the given information that A and B are symmetric, so and .
Substitute these into the expression for C':
To compare this with C, we can factor out a -1:
Since we defined , we have:
This is the definition of a skew-symmetric matrix. Therefore, AB - BA is a skew-symmetric matrix.
Hence Proved.
Q2Miscellaneous Exercise on Chapter 3
Show that the matrix B'AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Solution
To Show:
(i)
If A is symmetric, then B'AB is symmetric.
(ii)
If A is skew-symmetric, then B'AB is skew-symmetric.
Proof:
Let . To determine if C is symmetric or skew-symmetric, we need to find its transpose, C'.
Using the reversal law for transpose, , we get:
Using the property , we have .
So, .
Now we consider the two cases for matrix A.
Case (i): A is symmetric
If A is symmetric, then .
Substitute this into the expression for C':
Since , we have .
This is the definition of a symmetric matrix. Thus, if A is symmetric, B'AB is symmetric.
Case (ii): A is skew-symmetric
If A is skew-symmetric, then .
Substitute this into the expression for C':
Since , we have .
This is the definition of a skew-symmetric matrix. Thus, if A is skew-symmetric, B'AB is skew-symmetric.
Hence Shown.
Q3Miscellaneous Exercise on Chapter 3
Find the values of x, y, z if the matrix satisfy the equation .
Solution
Given:
The equation .
To Find: The values of .
Solution:
First, find the transpose of A, A'.
Now, compute the product A'A.
We are given that , where I is the identity matrix of order 3.
Equating A'A with I:
By the equality of matrices, we equate the corresponding diagonal elements:
Final Answer:
, ,
Q4Miscellaneous Exercise on Chapter 3
For what values of : ?
Solution
Given: The matrix equation
where O is the zero matrix, [0].
To Find: The value of .
Solution:
We perform the matrix multiplication from left to right.
First, multiply the first two matrices. The first is and the second is , so the result is a matrix.
Now, substitute this result back into the original equation:
Perform the final matrix multiplication. The first matrix is and the second is , so the result is a matrix.
By equality of matrices:
Final Answer:
Q5Miscellaneous Exercise on Chapter 3
If , show that .
Solution
Given:
To Show: , where I is the identity matrix and 0 is the zero matrix.
Proof:
First, calculate .
Next, calculate .
Next, write down .
Now, substitute these matrices into the expression :
Hence Shown.
Q6Miscellaneous Exercise on Chapter 3
Find , if
Solution
Given: The matrix equation
where O is the zero matrix, [0].
To Find: The value of .
Solution:
We perform the matrix multiplication from left to right.
First, multiply the first two matrices. The first is and the second is , so the result is a matrix.
Now, substitute this result back into the original equation:
Perform the final matrix multiplication. The first matrix is and the second is , so the result is a matrix.
By equality of matrices:
Final Answer:
Q7Miscellaneous Exercise on Chapter 3
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below: Market | Products --- | --- I | 10,000 | 2,000 | 18,000 II | 6,000 | 20,000 | 8,000
(a)
If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b)
If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Solution
Given:
Annual sales data for three products (x, y, z) in two markets (I, II).
Solution:
Let the annual sales be represented by a matrix A.
(a) Total Revenue in each market
Let the unit sale prices be represented by a column matrix P.
Prices are ₹ 2.50, ₹ 1.50, ₹ 1.00.
The total revenue in each market is given by the product AP. The result will be a matrix, where the first row is the revenue for Market I and the second row is for Market II.
Revenue =
Total revenue for Market I is ₹ 46,000.
Total revenue for Market II is ₹ 53,000.
(b) Gross Profit
Let the unit costs be represented by a column matrix C.
Costs are ₹ 2.00, ₹ 1.00, and 50 paise (₹ 0.50).
The total cost for each market is given by the product AC.
Total Cost =
Gross Profit = Total Revenue - Total Cost.
Gross Profit =
Gross profit for Market I is ₹ 15,000.
Gross profit for Market II is ₹ 17,000.
Total gross profit = ₹ 15,000 + ₹ 17,000 = ₹ 32,000.
Final Answer:
(a) The total revenue in Market I is ₹ 46,000 and in Market II is ₹ 53,000.
(b) The gross profit from both markets is ₹ 15,000 (Market I) and ₹ 17,000 (Market II), for a total of ₹ 32,000.
Q8Miscellaneous Exercise on Chapter 3
Find the matrix X so that
Solution
Given: The matrix equation
To Find: The matrix X.
Solution:
Let the given matrices be and .
The equation is .
First, determine the order of matrix X.
Order of A is .
Order of B is .
Let the order of X be .
The product XA is defined if the number of columns of X () is equal to the number of rows of A (2). So, .
The order of the resulting matrix XA is .
We are given that the order of B is .
So, , which implies .
Therefore, the order of matrix X is .
Let .
Now, substitute this into the equation:
Perform the matrix multiplication on the left side:
By equality of matrices, we can set up systems of linear equations by comparing corresponding elements.
From the first row:
From the second row:
Solve for and using (1) and (3):
Subtract (3) from (1): .
Substitute into (3): .
Check with (2): . This is correct.
Solve for and using (4) and (6):
Subtract (6) from (4): .
Substitute into (6): .
Check with (5): . This is correct.
So, .
The matrix X is .
Final Answer:
Q9Miscellaneous Exercise on Chapter 3
If is such that , then
(A)
(B)
(C)
(D)
Solution
Given:
, where I is the identity matrix.
To Find: The relationship between and .
Solution:
First, calculate .
We are given that .
Equating with I:
By equality of matrices, we equate the corresponding elements:
Rearranging this equation to match the format of the options:
This corresponds to option (C).
Final Answer: (C)
Q10Miscellaneous Exercise on Chapter 3
If the matrix A is both symmetric and skew symmetric, then
(A)
A is a diagonal matrix
(B)
A is a zero matrix
(C)
A is a square matrix
(D)
None of these
Solution
Given:
The matrix A is both symmetric and skew-symmetric.
To Find: The nature of matrix A.
Solution:
By the definition of a symmetric matrix, we have:
By the definition of a skew-symmetric matrix, we have:
Since both conditions are true for matrix A, we can equate the expressions for A':
Let . The equation becomes:
This implies that for every element in the matrix, .
Since this must be true for all elements , every element of the matrix A must be zero.
A matrix where all elements are zero is called a zero matrix (or null matrix).
Analysis of Options:
(A) A is a diagonal matrix: A zero matrix is a special case of a diagonal matrix, but this is not the most specific correct answer.
(B) A is a zero matrix: This is the direct conclusion from our derivation.
(C) A is a square matrix: Both symmetric and skew-symmetric matrices must be square, so this is true, but again, not the most specific conclusion.
(D) None of these: Incorrect as (B) is correct.
The most precise description of matrix A is that it is a zero matrix.
Final Answer: (B) A is a zero matrix
Q11Miscellaneous Exercise on Chapter 3
If A is square matrix such that , then is equal to
(A)
A
(B)
I - A
(C)
I
(D)
3 A
Solution
Given:
A is a square matrix such that .
To Find: The value of the expression .
Solution:
We start by expanding . We can use the binomial expansion formula . Since matrix multiplication is not always commutative, we must be careful. However, the identity matrix I commutes with any square matrix A of the same order (i.e., IA = AI = A). So, the standard binomial expansion is valid.
Now, we simplify each term using the given property and properties of the identity matrix I.
- (since )
Substitute these simplified terms back into the expansion:
Now, substitute this result into the original expression:
This corresponds to option (C).
Final Answer: (C) I