NCERT Solutions for Class 12 Maths Chapter 3 “Matrices” is an important chapter for students who want to pursue higher mathematics or science. Matrices are used in various fields such as physics, engineering, and computer science to represent data, perform operations on data, and solve complex problems.
The chapter begins by introducing the concept of matrices and its notation. Students will learn about different types of matrices such as square matrices, column matrices, row matrices, and their properties.
NCERT Solutions for Class 12 Maths Chapter 3 also provides detailed explanations and a variety of solved examples on how to perform matrix operations such as addition, subtraction, multiplication, and inverse of matrices. The solutions also cover the properties of these operations, such as commutative, associative, and distributive properties.
Additionally, the solutions offer step-by-step guidance on how to approach different types of problems and help students to develop problem-solving skills.

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Answers of Mathematics NCERT solutions for class 12 Chapter 3 Matrices

 

Exercise 3.1

Question 1:

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

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.

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=

*Question 2:

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 × 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.

Question 3:

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 × 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.

Question 5:

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

Question 6:

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

Question 7:

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

Question 8:

is a square matrix,if

1. m <n
2. m >n
3. m = n
4. 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.

Question 9:

Which of the given values of x and 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 

Equating the 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 x and y for which the given matrices are equal.

*Question 10:

The number of all possible matrices of order 3 × 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 × 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

Question 1:

Let 

Find each of the following

(i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA

Answer

(i)

(ii)

(iii)

Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is definedas:

Question 2:

Compute the following:

1. 
2. 

    (iii)

    (iv)    

Answer

(i)

(ii)

(iii)

(iv)

Question 3:

Compute the indicated products (i)

(ii)

(iii)

(iv)

(v)

Answer

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 4:

If

Compute and . Also, verify that

Answer

Question 5:

If   , then compute

Answer

Question 6:

Simplify

Answer

Question 7:

Find X and Y, if

1. 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,

Question 8:

Find X, if                   and

Answer

Question 9:

Find x and y, if 

Answer

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

∴x = 3 and y = 3

Question 10:

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

Answer

Comparing the corresponding elements of these two matrices, we get:

Question 11:

If ,       find values of x and 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

Question 12:

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

Answer

Comparing the corresponding elements of these two matrices, we get:

*Question 13:

If       , show that . 

Answer

Question 14:

Show that

(i)

(ii)

Answer

(i)

(ii)

Question 15:

Find if

Answer

We have A² = A × A

Question 16:

If , prove that

Answer

*Question 17:

If and     , find k so that

Answer

Comparing the corresponding elements, we have:

Thus, the value of k is 1.

*Question 18:

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

Answer

Question 19:

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 Rsx 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 x be 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.

Question 20:

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.

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.

Question 21:

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:

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

Question 22:

Assume X, Y, Z, W and P are matrices of order , and respectively. If n = p, then the order of the matrixis

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 × 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.

Exercise 3.3

Question 1:

Find the transpose of each of the following matrices:

(i)

(ii)

(iii)

Answer

(i)

(ii)

(iii)

Question 2:

If     and         , then verify that

(i)

(ii)

Answer

We have:

(i)

(ii)

Question 3:

If and   , then verify that 

(i)

(ii)

Answer

It is known that Therefore, we have:

(ii)

Question 4:

If  and   , then find 

Answer

We know that

Question 5:

For the matrices A and B, verify that (AB)′ =where

(i)

(ii)

Answer

(i)

(ii)

*Question 6:

If (i) , then verify that

, then verify that

Answer

(i)

(ii)

Question 7:

Show that the matrix is a symmetric matrix

Show that the matrix is a skew symmetric matrix

Answer

We have:

Hence, A is a symmetric matrix.

We have:

Hence, A is a skew-symmetric matrix.

*Question 8:

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.

Question 9:

Find and where  

Answer

The given matrix is

Question 10:

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:

Thus, is a symmetric matrix.

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

Question 11:

If A, B are symmetric matrices of same order, then AB − BA is a

A. Skew symmetric matrix B. Symmetric matrix

C. Zeromatrix D. Identitymatrix 

Answer

The correct answer is A.

A andB are symmetric matrices, therefore, we have:

Thus, (AB − BA) is a skew-symmetric matrix.

*Question 12:

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

Question 1:

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

Answer

We know that A = IA

Question 2:

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

Answer

We know that A = IA

Question 3:

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

Answer

We know that A = IA

Question 4:

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

Answer

We know that A = IA

Question 5:

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

Answer

We know that A = IA

Question 6:

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

Answer

We know that A = IA

Question 7:

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

Answer

We know that A = AI

Question 8:

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

Answer

We know that A = IA

Question 9:

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

Answer

We know that A = IA

Question 10:

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

Answer

We know that A = AI

Question 11:

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

Answer

We know that A = AI

Question 12:

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.

Question 13:

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

Answer

We know that A = IA

Question 14:

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⁻¹ does not exist.

Question 16:

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

Answer

We know that A = IA

Applying R₂ → R₂ + 3R₁ and R₃ → R₃ − 2R₁, we have:

Question 17:

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

Answer

We know that A = IA

Applying ,   we have:

*Question 18:

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

(a) AB = BA

(b) AB = 0, BA = I

(c) AB = BA =0

(d) AB = BA =I

Answer

The answer is d

We know that if A is a square matrix of order m, and if there exists another square matrix B of 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 A is the inverse of B.

Thus, matrices A and B will be inverses of each other only if AB = BA = I

Miscellaneous Solutions

Question 1:

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:

Question 2:

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:

*Question 3:

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:

Question 4:

If A and B are symmetric matrices, prove that AB − BA is a skew symmetric matrix. 

Answer

It is given that A and B are symmetric matrices. Therefore, we have:

Thus, (AB − BA) is a skew-symmetric matrix.

Question 5:

Show that the matrix is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer

We suppose that A is a symmetric matrix, then … (1) Consider

Thus, if A is a symmetric matrix, thenis a symmetric matrix. Now, we suppose that A is a skew-symmetric matrix.

Then,

Thus, if A is a skew-symmetric matrix, then is a skew-symmetric matrix.

Hence, if A is a symmetric or skew-symmetric matrix, then is a symmetric or skew- symmetric matrix accordingly.

*Question 6:

Solve system of linear equations, using matrix method.

Answer

The given system of equations can be written in the form of AX = B, where

Thus, A is non-singular. Therefore, its inverse exists.

Question 7:

For what values of

Answer 

We have:

∴4 + 4x = 0

⇒x = −1

Thus, the required value of x is −1.

Question 8:

If , show that

Answer

*Question 9:

Find x, if

Answer 

*Question 10:

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

 

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.

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

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 is Rs 46000 and the same in market II is Rs 53000.

The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and50 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 is Rs 46000, the gross profit in this market is (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 is Rs 53000, the gross profit in this market is (Rs 53000 − Rs 36000) Rs 17000.

Question 11:

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, X has 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 X is

Question 12:

If A and 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

A and 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 have Now, 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.

Question 13:

Choose the correct answer in the following questions:

If  is such that       then

1.  

B.

C.

D.

Answer

Answer: C

On comparing the corresponding elements, we have:

Question 14:

If the matrix A is both symmetric and skew symmetric, then

1. A is a diagonal matrix
2. A is a zero matrix
3. A is a square matrix
4. None of these 

Answer 

Answer:B

If A is both symmetric and skew-symmetric matrix, then we should have

Therefore, A is a zero matrix.

Question 15:

If A is square matrix such that then is equal to

1. A 
2. I –A
3. I 
4. 3A 

Answer

Answer: C

NCERT Solutions Math Class 12 Chapter 3 – Matrices

Students must manage a variety of tasks throughout their academic careers, making it necessary to seek expert assistance in order to get decent scores. Maths is a topic that can be applied if the fundamental principles are understood. Matrices is a crucial chapter in the NCERT Solutions for Class 12 Maths. It’s also a useful and important instrument in the field of mathematics. When compared to other direct ways, this Mathematical tool greatly simplifies our job. The development of the notion of matrices stems from a desire to find quick and easy solutions to solve a system of linear equations. Matrices aren’t just for displaying the coefficients of a system of linear equations.

 

Topics to study in Chapter 3 Class 12 Math

 

Section no	Topics
3.1	Introduction
3.2	Matrix
3.3	Types of Matrices
3.4	Operations on Matrices
3.5	Transpose of a Matrix
3.6	Symmetric and Skew Symmetric Matrices
3.7	Elementary Operation (Transformation) of a Matrix
3.8	Invertible Matrices

 

Weightage of Class 12 Math chapter 3 – Matrices

 

Chapter	Marks
Matrices	10 Marks

 

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Conclusion                                                                                  NCERT Solutions for Class 12 Maths Chapter 3 “Matrices” provide students with a comprehensive understanding of the concept of matrices and their operations. The chapter covers a variety of topics such as types of matrices, matrix operations such as addition, subtraction, multiplication, and inverse of matrices, and properties of matrix operations.
The solutions offered by Swastik Classes provide detailed explanations and numerous solved examples to help students understand the concepts more effectively. The step-by-step guidance provided in the solutions also assists students in developing problem-solving skills and building a strong foundation in matrix algebra.

Videos on Class 12 Math Matrices

 

What are the main topics discussed in NCERT Solutions for Class 12 Maths Chapter 3?

In Mathematics, matrices are one of the easiest chapters which when understood would be fun to solve. Matrix, types of matrices, operations on matrices, transpose of a matrix, symmetric and skew symmetric matrices, elementary operation on matrix and invertible matrices are the main topics discussed in this chapter. These topics are explained in a simple language to help students score well in the board exams irrespective of their intelligence quotient.

 

Why should we learn about matrices in NCERT Solutions for Class 12 Maths Chapter 3?

Matrices are rectangular arrays of numbers which are represented in rows and columns. Various mathematical operations like multiplication, addition, subtraction and division can be performed using matrices. Representing the data related to infant mortality rate, population etc. are the widely used areas where matrices are used to simplify the calculation of complex data. The other substantial use of matrices are statistics, plotting graphs and various scientific research purposes. The method of solving difficult linear equations are also made simple using the matrices.

 

Does NCERT Solutions for Class 12 Maths Chapter 3 help you to score well in the board exams?

Students should first solve the easier problems and then move on to the problems of higher level difficulty. After completing each exercise, students will be able to analysis the areas in which they are lagging behind. By practising the weaker concepts numerous times, students will be able to perform well in the board exams. Short cut tips are also highlighted to help students understand the easier way of solving complex problems effortlessly.