Matrices Chapter 3: NCERT Solutions for Class 12 Maths 

NCERT Solution for Class 12 Maths Chapter Matrices 3 compiled by the best educators around the world is available below. It looks out at all the aspects of the chapter and provides a detailed explanation of each question. It is designed by looking at the recent queries of the students today. NCERT Solutions along with solutions of miscellaneous exercises are written with great detail and explained step by step so that students understand them in detail and with full depth to perform well in their board exams.

Apart from the 12 boards, these solutions offer several perks to the students who are looking for upcoming entrance exams such as JEE, VITEEE, BITSAT, etc. Go through the available NCERT answers of Matrices Chapter 3 where its concepts are clearly explained in the simplest way to avoid any confusion.

Section No. & Topic Name of Matrices Chapter 3

• 3 Matrices: In this chapter, students will be introduced to the fundamentals of matrix and matrix algebra. Association of matrices with different fields is the main highlight of the section.
• 3.1 Introduction: 3.2.1 Order of a matrix: The first section deals with given illustrations to understand how the elements are arranged to form a matrix followed by how its order can be defined.
• 3.2 Matrix: In this section, Matrices are introduced to strengthen the basics of the students and brief the students about the concepts.
• 3.3 Types of Matrices:3.3.1 Equality of matrices: In this section, students learn about the different types of matrices including column matrix, row matrix, square matrix, diagonal matrix, scalar matrix, identity matrix and zero matrix. Equality of matrices is another significant concept in the section provided by proper illustrations.
• 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: This mainly focuses on helping students learn certain operations on matrices including addition, multiplication of a matrix.
• 3.5 Transpose of a Matrix: 3.5.1 This section deals with the properties of transpose of the matrices, the illustrations provided prove the properties of the transpose of a matrix.
• 3.6 Symmetric and Skew Symmetric Matrices: In this section, students will learn the definitions of symmetric and skew symmetric matrices with related examples and theorems.
• 3.7 Elementary Operation (Transformation) of a Matrix: After going through the section, students can comprehend the transformations on a matrix. There are six operations, i.e., transformations on a matrix. Among those three of them are because of columns and the rest three of rows, these are the elementary operations or transformations.
• 3.8 Invertible Matrices: 3.8.1 Inverse of a Matrix by elementary operations.
• Here, students will be briefed with the various concepts and principles to understand the core of the chapters in an easy way. The students will be taught on how to apply elementary operations on the various elements of a matrix to get an inverse matrix.

 Matrices class 12 ncert solutions

Exercise 3.1

Question1. In the matrix A =, write:

(i) The order of the matrix.

(ii) The number of elements.

(iii) Write the elements

Solution :
(i) There are 3 horizontal lines (rows) and 4 vertical lines (columns) in the given matrix A.

Therefore, Order of the matrix is 3 x 4.

(ii) The number of elements in the matrix A is 3 x 4 = 12.

(iii) a13 Element in first row and third column = 19

a21 Element in second row and first column = 35

a33 Element in third row and third column = -5

a24 Element in second row and fourth column = 12

a23 Element in second row and third column = 5/2

Question2. If a matrix has 24 elements, what are possible orders it can order? What, if it has 13 elements?

Solution :
We know that if a matrix is of the order m × 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 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 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 × 13 and 13 × 1.

Question3. If a matrix has 18 elements, what are the possible orders it can have? What if it has 5 elements?

Solution :
We know that if a matrix is of the order m × n, it has mn elements. 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 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 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 × 5 and 5 × 1.

Question4. Construct a 2 x 2 matrix A = [aij ] whose elements are given by:

(i)

(ii) aij = i/j

(iii)

Solution :
(i)

(ii)

(iii)

Question 5. Construct a 3 x 4 matrix, whose elements are given by:

(i)

(ii) aij = 2i – j

Solution :
(i)

(ii)

Question6. Find the values of x,y and z from the following equations:

(i)

(ii)

(iii)

Solution :
(i)Given:

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

x = 1, y = 4 = 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 and 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, y = 2

When x – y = -2 and x + y = 6, we get x = 2, y = 4

∴ x = 4, y = 2, and z = 0, or x = 2, y = 4 and z = 0

(iii)

Question7. Find the values of a,b, c and d from the equation .

Solution :

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

Question8. A = [aij]m x n is a square matrix if:

(A) m < n (B) m > n (C) m = n (D) None of these

Solution :
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. A = [aij]m x n

option (C) is correct.

Question9. Which of the given values of x and y make the following pairs of matrices equal:

(B) Not possible to find 

Solution :

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 x and y for which the given matrices are equal.

Therefore, option (B) is correct.

Question10. 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

Solution :
The correct answer is D.

The given matrix of the order 3 × 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

Question1. Let A = . Find each of the following:

(i) A + B

(ii) A – B

(iii) 3A – C

(iv) AB

(v) BA

Solution :
(i)

(ii)

(iii)

(iv)

(v)

Question2. Compute the following:

(i)

(ii)

(iii)

(iv)

Solution :

(ii)

(iii)

(iv)

Question3. Compute the indicated products:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution :
(i)

(ii)

(iii) =

(iv)=

(v)=

(vi)=

Question4. If then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

Solution :

L.H.S. = R.H.S. Proved.

Question5. If then compute 3A – 5B.

Solution :

Question6. Simplify:

Solution :

Question7. Find X and Y, if:

(i)

(ii)

Solution :
(i)

(ii)

Question8. Find X if Y =

Solution :

Question9. Find x and y if

Solution :

Question10. Solve the equation for x,y,z and t if

Solution :

Question11. Iffind the values of x and y

Solution :

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

Question12. Given: find the values of x ,y,z and w

Solution :

Question13. If show that F(x) F(y) = F(x + y)

Solution :

Question14. Show that:

(i)

(ii)

Solution :
(i)

(ii)

Question15. Find A2 – 5A + 6I if A = .

Solution:

Question16. If A = prove that A3 – 6A2 + 7A + 2I = 0.

Solution :
L.H.S. = A3 – 6A2 + 7A + 2I

Exercise 3.3

Question1. Find the transpose of each of the following matrices:

(i)

(ii)

(iii)

Solution :
(i) Let A =

Transpose of A = A’ or AT = [ 51/2 -1]

(ii)

Transpose of A = A’ or AT=

(iii)

Transpose of A = A’ or AT=

Question2. If then verify that:

(i) (A + B)’ = A’ + B’

(ii) (A – B)’ = A’ – B’

Solution :

Question3. If then verify that:

(i) (A + B)’ = A’ + B’

(ii) (A – B)’ = A’ – B’

Solution :

 

Question4. If then find (A + 2B)’.

Solution :

Question5. For the matrices A and B, verify that (AB)’ = B’A’, where:

(i)

(ii)

Solution :

Question6. (i) If A = then verify that A’A = I.

(ii) If A = then verify that A’A = I.

Solution :

Question7. (i) Show that the matrix A = is a symmetric matrix.

(ii) Show that the matrix A = is a skew symmetric matrix.

Solution :
(i) Given: A =

Changing rows of matrix A as the columns of new matrix A’ = = A

∴ A’ = A

Therefore, by definitions of symmetric matrix, A is a symmetric matrix.

(ii) Given: A =

A’ = = – = – A

∴ A’ = – A

Therefore, by definition matrix A is a skew-symmetric matrix

Question8. For a matrix A = verify that:

(i) (A + A’) is a symmetric matrix.

(ii) (A – A’) is a skew symmetric matrix.

Solution :

Question9. Find 1/2 (A + A’) and 1/2(A – A’) when A =

Solution :

Question10. Express the following matrices as the sum of a symmetric and skew symmetric matrix:

(i)

(ii)

(iii)

(iv)

Solution :

(iii)

(iv)

Choose the correct answer in Exercises 11 and 12.

Question11. If A and B are symmetric matrices of same order, 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 ∴ A = A’ and B = B’

Now, (AB – BA)’ = (AB)’ – (BA)’ ∴ (AB – BA)’ = B’A’ – A’B’ [Reversal law]

∴ (AB – BA)’ = BA – AB [From eq. (i)] ∴ (AB – BA)’ = – (AB – BA)

∴ (AB – BA) is a skew matrix.

Therefore, option (A) is correct.

Question12. If A = , then A + A’ = I, if the value of α is:

(A) π/6

(B) π/3

(C) π

(D) 3π/2

Solution :

Therefore, option (B) is correct.

= R.H.S. Proved.

Question17. If find k so that A2 = kA – 2I

Solution :

Question18. If and I is the identity matrix of order 2, show that I + A = (I – A)

Solution :

Question19. A trust fund has ` 30,000 that must be invested in two different types of bond. 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 in two types of bonds, if the trust fund must obtain an annual interest of (a) ` 1800, (b) ` 2000.

Solution :
(a) It is given that Rs.30,000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs. x be invested in bonds of the first type. Thus, Rs. (30,000 − x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = [x 30,000 − x]

Annual interest obtained is Rs. 1800. We know, Interest = PTR/100

Here, the time is one year and thus T = 1.

Hence, the interest obtained after one year can be expressed in matrix representation as-

⇒ 5x + 210000 − 7x = 180000

⇒ – 2x = – 30000

∴ x = 15000

Amount invested in the first bond = x = Rs. 15000

⇒ Amount invested second bond = Rs. (30000 − x) = Rs. (30000 − 15000) = Rs. 15000

∴ The trust has to invest Rs. 15000 each in both the bonds in order to obtain an annual interest of Rs. 1800.

(b) It is given that Rs.30,000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs. x be invested in bonds of the first type. Thus, Rs. (30,000 − x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = [x 30,000 − x]

Annual interest obtained is Rs. 2000.

Hence, the interest obtained after one year can be expressed in matrix representation as-

⇒ 5x + 210000 − 7x = 200000

⇒ −2x = − 10000

∴ x = 5000

Amount invested in the first bond = x = Rs. 5000

⇒ Amount invested second bond = Rs. (30000 − x) = Rs. (30000 − 5000) = Rs. 25000

∴ The trust has to invest Rs.5000 in the first bond and Rs.25000 in the second bond in order to obtain an annual interest of Rs.2000.

Question20. 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 :
The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.

The selling 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.

Question21. Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p, n x 3 and p x k repectively. The restriction on n,k and p so that PY + WY will be define are:

1. k = 3, p = n
2. k is arbitrary, p = 2
3. p is arbitrary, k = 3
4. k = 2, p = 3

Solution :
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 PY + WY will be defined.

Therefore, option (A) is correct.

Question22. Assume X, Y, Z, W and P are matrices of order 2 x n,3 x k,2 x p, n x 3 and p x k repectively. If n = p then order of matrix 7X – 5Z is:

(A) p × 2

(B) 2 × n

(C) n × 3

(D)p × n

Solution :

Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of the order 2 × p, i.e., 2 × 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 × n.

Thus, matrix 7X − 5Z is well-defined and is of the order 2 × n.

The correct answer is B.

Exercise 3.4

Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.

Question 1.

Solution :

Question 2.

Solution :

Question3.

Solution :

Question4.

Solution :

Question5.

Solution :

Question6.

Solution :

Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 7 to 14.

Question7.

Solution :

Question8.

Solution :
Let A =

Question9.

Solution :
Let A =

Question10.

Solution :
Let A =

Question11.

Solution :
Let A =

Question12.

Solution :
Let A =

Question13.

Solution :
Let A =

Question14.

Solution :
Let A =

Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 15 to 17.

Question15.

Solution :
Let A =

Question16.

Solution :
Let A = ,

Question17.

Solution :
Let A = ,

Question18. Matrices A and B will be inverse of each other only if:

(A) AB = BA

(B) AB = BA = 0 

(C) AB = 0, BA = I 

(D) AB = BA = I

Solution :
By definition of inverse of square matrix,

Option (A) is correct.