Here is a step by step NCERT Solutions for Class 12 Maths Chapter 3 Matrices prepared by experts. This solutions covers all questions of Maths Chapter 3 Matrices Class 12 as per CBSE Board guidelines from the latest NCERT book for class 12 maths. Following topics and sub-topics in Class 12 Maths Chapter 3 Matrices are covered.
3.1 Introduction
3.2 Matrix
3.2.1 Order of a matrix
3.3 Types of Matrices
3.3.1 Equality of matrices
3.4 Operations on Matrices
3.4.1 Addition of matrices
3.4.2 Multiplication of a matrix by a scalar
3.4.3 Properties of matrix addition
3.4.4 Properties of scalar multiplication of a matrix
3.4.5 Multiplication of matrices
3.4.6 Properties of multiplication of matrices
3.5. Transpose of a Matrix
3.5.1 Properties of transpose of the matrices
3.6 Symmetric and Skew Symmetric Matrices
3.7 Elementary Operation (Transformation) of a Matrix
3.8 Invertible Matrices
3.8.1 Inverse of a matrix by elementary operations.

Matrices NCERT Solutions – Class 12 Maths

NCERT Solutions for Class 12 Maths Chapter 3

Exercise 3.1 : Solutions of Questions on Page Number : 64
Q1 :In the matrix, write:
(i) The order of the matrix (ii) The number of elements,
(iii) Write the elements a13, a21, a33, a24, a23
Answer :
(i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 ×4.
(ii) Since the order of the matrix is 3 ×4, there are 3 ×4 = 12 elements in it.
(iii) a13= 19, a21= 35, a33= – 5, a24= 12, a23=


Q2 :If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?
Answer :
We know that if a matrix is of the order m x n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and
(6, 4)
Hence, the possible orders of a matrix having 24 elements are:
1 x 24, 24 x 1, 2 x 12, 12 x 2, 3 x 8, 8 x 3, 4 x 6, and 6 x 4
(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 x 13 and 13 x 1.


Q3 :If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Answer :
We know that if a matrix is of the order m x n, it has mnelements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are:
1 x 18, 18 x 1, 2 x 9, 9 x 2, 3 x 6, and 6 x 3
(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 x 5 and 5 x 1.


Q4 :Construct a 2 ×2 matrix,, whose elements are given by:
(i)
(ii)
(iii)
Answer :
In general, a 2 ×2 matrix is given by
(i)
Therefore, the required matrix is
(ii)
Therefore, the required matrix is
(iii)
Therefore, the required matrix is


Q5 :Construct a 3 ×4 matrix, whose elements are given by
(i) (ii)
Answer :
In general, a 3 ×4 matrix is given by
(i)
Therefore, the required matrix is
(ii)
Therefore, the required matrix is


Q6 :Find the value of x, y, and z from the following equation:
(i) (ii)
(iii)
Answer :
(i)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x= 1, y = 4, and z = 3
(ii)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x+ y = 6, xy = 8, 5 + z = 5
Now, 5 + z = 5 ⇒ z = 0
We know that:
(x – y)2= (x + y)2 – 4xy
⇒ (x – y)2= 36 – 32 = 4
⇒ x – y= ±2
Now, when x – y = 2 and x +y = 6, we get x= 4 and y= 2
When x – y = – 2 and x + y = 6, we get x = 2 and y = 4
∴x= 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0
(iii)
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x+ y + z= 9 … (1)
x+ z = 5 … (2)
y+ z = 7 … (3)
From (1) and (2), we have:
y + 5 = 9
⇒ y = 4
Then, from (3), we have:
4+ z = 7
⇒ z = 3
∴ x +z = 5
⇒ x = 2
∴ x = 2, y= 4, and z= 3


Q7 :Find the value of a, b, c, and d from the equation:
Answer : As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
a – b= – 1 … (1)
2a – b= 0 … (2)
2a+ c = 5 … (3)
3c+ d = 13 … (4)
From (2), we have:
b= 2a
Then, from (1), we have:
a – 2a= – 1
⇒ a= 1
⇒ b= 2
Now, from (3), we have:
2×1 + c= 5
⇒ c= 3
From (4) we have:
3×3 + d= 13
⇒ 9 + d = 13 ⇒ d = 4
∴a= 1, b = 2, c = 3, and d = 4


Q8 :is a square matrix, if
(A) m < n (B) m> n
(C) m= n
(D) None of these
Answer :
The correct answer is C.
It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.
Therefore, is a square matrix, if m= n.


Q9 : Which of the given values of xand y make the following pair of matrices equal
(A)
(B) Not possible to find
(C)
(D)
Answer :
The correct answer is B.
It is given that
Equatingthe corresponding elements, we get:
We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.
Hence, it is not possible to find the values of xand y for which the given matrices are equal.


Q10 :The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:
(A) 27
(B) 18
(C) 81
(D) 512
Answer :
The correct answer is D.
The given matrix of the order 3 x 3 has 9 elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is 29= 512


Exercise 3.2 : Solutions of Questions on Page Number : 80


Q1 :Let
Find each of the following
(i) (ii) (iii)(iv)
(v)

Answer :
(i)
(ii)
(iii)
(iv) Matrix Ahas 2 columns. This number is equal to the number of rows in matrix B. Therefore, ABis defined as:
(v) MatrixB has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BAis defined as:


Q2 :Compute the following:
(i) (ii)
(iii)
(v)
Answer :
(i)
(ii)
(iii)
(iv)


Q3 :Compute the indicated products
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer :
(i)
(ii)
(iii)
(iv)
(v)
(vi)


Q4 :If, and, then compute and. Also, verify that
Answer :


Q5 :If and then compute.
Answer :


Q6 :Simplify
Answer :


Q7 :Find X and Y, if
(i) and
(ii) and
Answer :
(i)
Adding equations (1) and (2), we get:
(ii)
Multiplying equation (3) with (2), we get: Multiplying equation (4) with (3), we get:
From (5) and (6), we have: Now,


Q8 :Find X, if and
Answer :


Q9 :Find x and y, if
Answer :
Comparing the corresponding elements of these two matrices, we have:
∴x= 3 and y= 3


Q10 :Solve the equation for x, y, zand t if
Answer :
Comparing the corresponding elements of these two matrices, we get:


Q11 :If, find values of xand y.
Answer :
Comparing the corresponding elements of these two matrices, we get:
2x – y= 10 and 3x+ y = 5
Adding these two equations, we have:
5x= 15
⇒ x= 3
Now, 3x+ y = 5
⇒ y= 5 – 3x
⇒ y= 5 – 9 = – 4
∴x= 3 and y= – 4


Q12 :Given, find the values of x, y, zand
w.
Answer :
Comparing the corresponding elements of these two matrices, we get:


Q13 :If, show that.
Answer :


Q14 :Show that
(i)
(ii)
Answer :
(i)
(ii)


Q15 :Find if
Answer :
We have A2= A × A


Q16 :If, prove that
Answer :


Q17 :If and, find k so that Answer :
Comparing the corresponding elements, we have:
Thus, the value of kis 1.


Q18 :If and I is the identity matrix of order 2, show that
Answer :


Q19 :A trust fund has Rs 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 Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a) Rs 1,800 (b) Rs 2,000
Answer :
(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 – x).
It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year.
Therefore, in order to obtain an annual total interest of Rs 1800, we have:
Thus, in order to obtain an annual total interest of Rs 1800, the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond.
(b) Let Rs xbe invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 – x).
Therefore, in order to obtain an annual total interest of Rs 2000, we have:
Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.


Q20 :The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Answer :
The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.
Theselling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40.
The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:
Thus, the bookshop will receive Rs 20160 from the sale of all these books.


Q21 :Assume X, Y, Z, W and P are matrices of order, and respectively. The restriction on n, k and p so that will be defined are:
A. k = 3, p = n
B. k is arbitrary, p = 2
C. p is arbitrary, k = 3
D. k = 2, p = 3
Answer :
Matrices P and Y are of the orders p × k and 3 × k respectively.
Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k.
Matrices W and Y are of the orders n × 3 and 3 × k respectively.
Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order n × k.
Matrices PY and WY can be added only when their orders are the same.
However, PY is of the order p × k and WY is of the order n × k. Therefore, we must have p = n.
Thus, k = 3 and p = n are the restrictions on n, k, and p so that will be defined.


Q22 :Assume X, Y, Z, Wand Pare matrices of order, and respectively. If n= p, then the order of the matrix is
A p ×2 B 2 × n C n ×3 D p × n
Answer :
The correct answer is B.
Matrix X is of the order 2 x n.
Therefore, matrix 7X is also of the same order.
Matrix Z is of the order 2 x p, i.e., 2 x n [Since n= p]
Therefore, matrix 5Z is also of the same order.
Now, both the matrices 7X and 5Z are of the order 2 x n.
Thus, matrix 7X- 5Z is well-defined and is of the order 2 x n.


Exercise 3.3 : Solutions of Questions on Page Number : 88


Q1 :Find the transpose of each of the following matrices:
(i) (ii) (iii)
Answer :


Q2 :If and, then verify that
(i)
(ii)
Answer :
(i)
(ii)


Q3 :If and, then verify that
(i)

(ii)
Answer :
(i) It is known that
Therefore, we have:
(ii)


Q4 :If and, then find
Answer :
We know that


Q5 :For the matrices A and B, verify that (AB)’ = where
(i)
(ii)
Answer :
(i)
(ii)


Q6 :If (i) , then verify that
(ii) , then verify that
Answer :
(i)
(ii)


Q7 :(i) Show that the matrix is a symmetric matrix
(ii) Show that the matrix is a skew symmetric matrix
Answer :
(i) We have:
Hence, A is a symmetric matrix.
(ii) We have:
Hence, A is a skew-symmetric matrix.


Q8 :For the matrix, verify that
(i) is a symmetric matrix
(ii) is a skew symmetric matrix
Answer :
(i)
Hence, is a symmetric matrix.
(ii)
Hence, is a skew-symmetric matrix.


Q9 :Find and, when
Answer :
The given matrix is


Q10 :Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i) 
(ii)
(iii)
(iv)
Answer :
(i)
Thus, is a symmetric matrix.

Thus,   is a skew-symmetric matrix.
Representing A as the sum of P and Q:

(ii)
Thus,  is a symmetric matrix.

Thus,  is a skew-symmetric matrix.
Representing A as the sum of P and Q:

(iii)

Thus,   is a symmetric matrix.

Thus,  is a skew-symmetric matrix.
Representing A as the sum of P and Q:

(iv)
Thus,  is a symmetric matrix.

Thus,  is a skew-symmetric matrix.
Representing A as the sum of P and Q:


Q11 :If A, Bare symmetric matrices of same order, then AB- BA is a
A. Skew symmetric matrix B. Symmetric matrix
C. Zero matrix D. Identity matrix
Answer :
The correct answer is A.
Aand B are symmetric matrices, therefore, we have:
Thus, (AB – BA) is a skew-symmetric matrix.


Q12 :
If , then, if the value of α is
A. B.
C. π D.
Answer :
The correct answer is B.

Comparing the corresponding elements of the two matrices, we have:


Exercise 3.4 : Solutions of Questions on Page Number : 97


Q1 :Find the inverse of each of the matrices, if it exists.

Answer :
We know that A= IA


Q2 :Find the inverse of each of the matrices, if it exists.

Answer :
We know that A= IA


Q3 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA


Q4 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA


Q5 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA


Q6 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA


Q7 :Find the inverse of each of the matrices, if it exists.
Answer :We know that A= AI


Q8 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA


Q9 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA


Q10 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= AI


Q11 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= AI


Q12 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA
Now, in the above equation, we can see all the zeros in the second row of the matrix on the L.H.S.
Therefore, A – 1 does not exist.


Q13 : Find the inverse of each of the matrices, if it exists.
Answer :We know that A= IA


Q14 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA
Applying, we have:
Now, in the above equation, we can see all the zeros in the first row of the matrix on the L.H.S.
Therefore, A – 1does not exist.


Q15 :Find the inverse of each of the matrices, if it exists.
 Answer :
We know that A= IA


Q16 :Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IA
Applying R R2 + 3R1and R3  → R3 – 2R1, we have:


Q17 : Find the inverse of each of the matrices, if it exists.
Answer :
We know that A= IAAppling,We have


Q18 :Matrices A and Bwill be inverse of each other only if
A. AB= BA
C. AB= 0, BA = I
B. AB= BA = 0
D. AB= BA = I
Answer :
Answer: D
We know that if Ais a square matrix of order m, and if there exists another square matrix Bof the same order m, such that AB= BA = I, then B is said to be the inverse of A. In this case, it is clear that Ais the inverse of B.
Thus, matrices Aand B will be inverses of each other only if AB= BA = I.


Exercise Miscellaneous : Solutions of Questions on Page Number : 100


Q1 :Let, show that, where I is the identity matrix of order 2 and n ∈ N
Answer :It is given that

We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
Therefore, the result is true for n = 1.
Let the result be true for n= k.
That is,
Now, we prove that the result is true for n= k + 1.
Consider
From (1), we have:
Therefore, the result is true for n= k + 1.
Thus, by the principle of mathematical induction, we have:


Q2 :If, prove that
Answer :
It is given that
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
Therefore, the result is true for n= 1.
Let the result be true for n= k.
That is
Now, we prove that the result is true for n= k + 1.
Therefore, the result is true for n= k + 1.
Thus by the principle of mathematical induction, we have:


Q3 :If, then prove where n is any positive integer
Answer :
It is given that
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
Therefore, the result is true for n= 1.
Let the result be true for n= k.
That is,
Now, we prove that the result is true for n= k + 1.
Therefore, the result is true for n= k + 1.
Thus, by the principle of mathematical induction, we have:


Q4 :If Aand B are symmetric matrices, prove that AB- BAis a skew symmetric matrix.
Answer :
It is given that Aand B are symmetric matrices. Therefore, we have:
Thus, (AB – BA) is a skew-symmetric matrix.


Q5 :Show that the matrix is symmetric or skew symmetric according as Ais symmetric or skew symmetric.
Answer :
We suppose that Ais a symmetric matrix, then … (1)
Consider
Thus, if A is a symmetric matrix, thenis a symmetric matrix.
Now, we suppose that Ais a skew-symmetric matrix.
Then,Thus, if A is a skew-symmetric matrix, thenis a skew-symmetric matrix. Hence, if A is a symmetric or skew-symmetric matrix, thenis a symmetric or skew-symmetric matrix accordingly.


Q6 :Find the values of x, y, zif the matrix satisfy the equation
Answer :
Now,
On comparing the corresponding elements, we have:


Q7 :For what values of?
Answer :
We have:
∴4 + 4x = 0
⇒ x= – 1
Thus, the required value of xis – 1


Q8 :If, show that
Answer :


Q9 :Find x, if
Answer :
We have:


Q10 :A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:
Market
Products
I
10000
2000
18000
II
6000
20000
8000
(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 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 Rs 2.00, Rs 1.00 and 50 paise respectively. Find the gross profit.
Answer :(a) The unit sale prices of x, y, and z are respectively given as Rs 2.50, Rs 1.50, and Rs 1.00.
Consequently, the total revenue in market I can be represented in the form of a matrix as:
The total revenue in market II can be represented in the form of a matrix as:
Therefore, the total revenue in market I isRs 46000 and the same in market II isRs 53000.
(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise.
Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:
Since the total revenue in market I isRs 46000, the gross profit in this marketis (Rs 46000 – Rs 31000) Rs 15000.
The total cost prices of all the products in market II can be represented in the form of a matrix as:
Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs 53000 – Rs 36000) Rs 17000.


Q11 : Find the matrix X so that
Answer :
It is given that:
The matrix given on the R.H.S. of the equation is a 2 ×3 matrix and the one given on the L.H.S. of the equation is a 2 ×3 matrix. Therefore, Xhas to be a 2 ×2 matrix.
Now, let
Therefore, we have:
Equating the corresponding elements of the two matrices, we have:
Thus, a = 1, b= 2, c = – 2, d= 0
Hence, the required matrix Xis


Q12 :If Aand B are square matrices of the same order such that AB= BA, then prove by induction that. Further, prove that for all n ∈ N
Answer :

Aand B are square matrices of the same order such that AB= BA.
For n = 1, we have:Therefore, the result is true for n= 1.
Let the result be true for n= k.Now, we prove that the result is true for n= k + 1.Therefore, the result is true for n= k + 1.
Thus, by the principle of mathematical induction, we haveNow, we prove that for all n ∈ N
For n = 1, we have:Therefore, the result is true for n= 1.
Let the result be true for n= k.Now, we prove that the result is true for n= k + 1.Therefore, the result is true for n= k + 1.
Thus, by the principle of mathematical induction, we have, for all natural numbers.


Q13 :Choose the correct answer in the following questions:
If is such that then
A.B.C.D.Answer :
Answer: C

On comparing the corresponding elements, we have:


Q14 :If the matrix Ais both symmetric and skew symmetric, then
A. Ais a diagonal matrix
B. Ais a zero matrix
C. Ais a square matrix
D. None of these
Answer :
Answer: B

If Ais both symmetric and skew-symmetric matrix, then we should haveTherefore, A is a zero matrix.


Q15 :If Ais square matrix such that then is equal to
A. A  B. I – A  C. I  D. 3A
Answer :
Answer: C