Contents

These Solutions are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths Chapter 3 Matrices

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Matrices |

Exercise |
Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4 |

Number of Questions Solved |
62 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 3 Matrices

### Chapter 3 Matrices Exercise 3.1

**Question 1.**

**In the matrix **

**(i) The order of the matrix**

**(ii) The number of elements**

**(iii) Write the elements a _{13}, a_{21}, a_{33}, a_{24}, a_{23}**

**Solution:**

(i) The matrix A has three rows and 4 columns.

The order of the matrix is 3 x 4.

(ii) There are 3 x 4 = 12 elements in the matrix A

(iii) a

_{13}= 19, a

_{21}= 35, a

_{33}= – 5, a

_{24}= 12, a

_{23}=

**Question 2.**

**If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?**

**Solution:**

(i) 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6

Thus there are 8 matrices having 24 elements their order are (1 x 24), (24 x 1), (2 x 12), (12 x 2),(3 x 8), (8 x 3), (4 x 6), (6 x 4).

(ii) 13 = 1 x 13,

There are 2 matrices of 13 elements of order (1 x 13) and (13 x 1).

**Question 3.**

**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 order m × n, it has mn elements.

=> 18 = 1 x 18 = 2 x 9 = 3 x 6

Thus, all possible ordered pairs of the matrix

having 18 elements are:

(1,18), (18,1), (2,9), (9,2), (3,6), (6,3)

If it has 5 elements, then possible order are: (1,5), (5,1)

**Question 4.**

**Construct a 2 x 2 matrix, A= [a _{ij}] whose elements are given by:**

**Solution:**

**Question 5.**

**Construct a 3 x 4 matrix , whose elements are given by:**

**Solution:**

**Question 6.**

**Find the values of x, y, z from the following equations:**

**Solution:**

Clearly x = 1,y = 4,z = 3

Now 5 + z = 5 => z = 0

Now x + y = 6 and xy = 8

∴ y = 6 – x and x(6 – x) = 8

6x – x² = 8

x² – 6x + 8 = 0

(x – 4)(x – 2) = 0

=>x = 2,4

When x = 2, y = 6 – 2 = 4

and when x = 4,y = 6 – 4 = 2

Hence x = 2,y = 4,z = 0 or x = 4,y = 2,z = 0.

(iii) Equating the corresponding elements.

=> x+y+z=9 …..(i)

x+z = 5 …(ii)

y+ z = 7 …(iii)

Adding eqs. (ii) & (iii)

x + y + 2z = 12

=> (x+y+z) + z = 12,

9+z = 12 (from equ (i))

z = 3

x + z = 5

=>x + 3 = 5 => x = 2

and y+z = 7

=>y+3 = 7

=> y = 4

=> x = 2, y = 4 and z = 3

**Question 7.**

**Find the values of a,b,c and d from the equation:**

**Solution:**

**Question 8.**

**A = [a _{ij}]_{m×n} is a square matrix, if**

**(a) m < n (b) n > n**

**(c) m = n**

**(d) none of these**

**Solution:**

For a square matrix m=n.

Thus option (c) m = n, is correct.

**Question 9.**

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

**(a) **

**(b) Not possible to find**

**(c) **

**(d) **

**Solution:**

(a)

**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**

**Solution:**

There are 3 x 3 matrix or 9 entries in matrix each place can be filled with 0 or 1

∴ 9 Places can be filled in 2^{9} = 512 ways

Number of such matrices = 512

Option (d) is correct.

### Chapter 3 Matrices Exercise 3.2

**Question 1.**

**Let **

**Find each of the following:**

**(i) A + B**

**(ii) A – B**

**(iii) 3A – C**

**(iv) AB**

**(v) BA**

**Solution:**

Let

(i) A + B

**Question 2.**

**Compute the following:**

**Solution:**

**Question 3.**

**Compute the indicated products.**

**(i) **

**(ii) **

**(iii) **

**(iv) **

**(v) **

**(vi) **

**Solution:**

(i)

=

**Question 4.**

**If **

**then compute (A + B) and (B – C). Also verify that A + (B – C) = (A + B) – C.**

**Solution:**

Given

**Question 5.**

**If **

**then compute 3A – 5B.**

**Solution:**

=

**Question 6.**

**Simplify:**

**Solution:**

**Question 7.**

**Find X and Y if**

**Solution:**

**Question 8.**

**Find**

**Solution:**

We are given that

**Question 9.**

**Find x and y, if **

**Solution:**

=>

=> 2+y = 5 and 2x+2 = 8

=> y=3 and x=3

Hence x=3 and y=3

**Question 10.**

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

**Solution:**

**Question 11.**

**If then find the values of x and y**

**Solution:**

=>

**Question 12.**

**Given**

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

**Solution:**

=>

=> 3x = x + 4 => x = 2

and 3y = 6 + x + y => y = 4

Also, 3w = 2w + 3 => w = 3

Again, 3z = – 1 + z + w

=> 2z = – 1 + 3

=> 2z = 2

=> z = 1

Hence x = 2 ,y = 4, z = 1, w = 3.

**Question 13.**

**If F(x) = **

**then show that F(x).F(y) = F(x+y)**

**Solution:**

F(x) =

∴ F(y) =

**Question 14.**

**Show that**

**Solution:**

L.H.S≠R.H.S

**Question 15.**

**Find A² – 5A + 6I, if A = **

**Solution:**

A² – 5A + 6I =

**Question 16.**

**If A = Prove that A³-6A²+7A+2I = 0**

**Solution:**

We have

A² = A x A

=

**Question 17.**

**If then ****find k so that A²=kA-2I**

**Solution:**

Given

Required: To find the value of k

Now A²=kA-2I

**Question 18.**

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

**Solution:**

L.H.S=

**Question 19.**

**A trust has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bond if the trust fund obtains an annual total interest of**

**(a) Rs 1800**

**(b) Rs 2000**

**Solution:**

Let Rs 30,000 be divided into two parts and Rs x and Rs (30,000-x)

Let it be represented by 1 x 2 matrix [x (30,000-x)]

Rate of interest is 005 and 007 per rupee.

It is denoted by the matrix R of order 2 x 1.

**Question 20.**

**The book-shop of a particular school has 10 dozen Chemistry books, 8 dozen Physics books, 10 dozen Economics books. Their selling price are Rs 80, Rs 60 and Rs 40 each respectively. Find die total amount the book-shop will receive from selling all the books using matrix algebra.**

**Solution:**

Number of Chemistry books = 10 dozen books

= 120 books

Number of Physics books = 8 dozen books = 96 books

Number of Economics books = 10 dozen books

= 120 books

**Assuming X, Y, Z, W and P are the matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Choose the correct answer in Question 21 and 22.**

**Question 21.**

**The restrictions on n, k and p so that PY + WY will be defined are**

**(a) k = 3 ,p = n**

**(b) k is arbitrary,p = 2**

**(c) pis arbitrary, k = 3**

**(d) k = 2,p = 3**

**Solution:**

Given : x_{2xn,} y_{3xn}, z_{2xp,} w_{nx3, }P_{pxk}

Now py +wy = P_{pxk} x y_{3+k} x w_{nx3} x y_{3xk}

Clearly, k = 3 and p = n

Hence, option (a) is correct p x 2.

**Question 22.**

**If n = p, then the order of the matrix 7X – 5Z is:**

**(a) p x 2**

**(b) 2 x n**

**(c) n x 3**

**(d) p x n.**

**Solution:**

7X – 5Z = 7X_{2xn} – 5X_{2xp}

∴ We can add two matrices if their order is same n = P

∴ Order of 7X – 5Z is 2 x n.

Hence, option (b) is correct 2 x n.

### Chapter 3 Matrices Exercise 3.3

**Question 1.**

**Find the transpose of each of the following matrices:**

**(i) **

**(ii) **

**(iii) **

**Solution:**

(i) let A =

∴ transpose of A = A’ =

**Question 2.**

**If **

**then verify that:**

**(i) (A+B)’=A’+B’**

**(ii) (A-B)’=A’-B’**

**Solution:**

**Question 3.**

**If **

**then verify that:**

**(i) (A+B)’ = A’+B’**

**(ii) (A-B)’ = A’-B’**

**Solution:**

**Question 4.**

**If **

**then find (A+2B)’**

**Solution:**

**Question 5.**

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

**Solution:**

**Question 6.**

**If (i) ,the verify that A’A=I**

**If (ii) ,the verify that A’A=I**

**Solution:**

(i)

**Question 7.**

**(i) Show that the matrix is a symmetric matrix.**

**(ii) Show that the matrix is a skew-symmetric matrix.**

**Solution:**

(i) For a symmetric matrix a_{ij} = a_{ji}

Now,

**Question 8.**

**For the matrix, **

**(i) (A+A’) is a symmetric matrix.**

**(ii) (A-A’) is a skew-symmetric matrix.**

**Solution:**

=>

**Question 9.**

**Find and ,when**

**Solution:**

**Question 10.**

**Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.**

**(i)**

**(ii)**

**(iii)**

**(iv)**

**Solution:**

(i) let

=>

**Question 11.**

**Choose the correct answer in the following questions:**

**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:**

Now A’ = B, B’ = B

(AB-BA)’ = (AB)’-(BA)’

= B’A’ – A’B’

= BA-AB

= – (AB – BA)

AB – BA is a skew-symmetric matrix Hence, option (a) is correct.

**Question 12.**

**If then A+A’ = I, if the**

**value of α is**

**(a) **

**(b) **

**(c) π**

**(d) **

**Solution:**

Now

Thus option (b) is correct.

### Chapter 3 Matrices Exercise 3.4

**Using Elementary transformation, find the inverse each of matrices, if it exists in ques 1 to 17.**

**Question 1.**

**Solution:**

Let

We know that

A = IA

**Question 2.**

**Solution:**

Let

We know that

A = IA

**Question 3.**

**Solution:**

Let

We know that

A = IA

**Question 4.**

**Solution:**

Let

We know that

A = IA

**Question 5.**

**Solution:**

Let

We know that

A = IA

**Question 6.**

**Solution:**

Let

We know that

A = IA

**Question 7.**

**Solution:**

Let

We know that

A = IA

**Question 8.**

**Solution:**

Let

We know that

A = IA

**Question 9.**

**Solution:**

Let

We know that

A = IA

**Question 10.**

**Solution:**

Let

**Question 11.**

**Solution:**

Let

We know that

A = IA

**Question 12.**

**Solution:**

Let

We know that

A = IA

**Question 13.**

**Solution:**

Let

We know that

A = IA

**Question 14.**

**Solution:**

Let

We know that

A = IA

**Question 15.**

**Solution:**

Let

**Question 16.**

**Solution:**

Let

We know that

A = IA

**Question 17.**

**Solution:**

Let

We know that

A = IA

**Question 18.**

**Choose the correct answer in the following question:**

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

**(a) AB = BA**

**(b) AB = BA = 0**

**(c) AB = 0,BA = 1**

**(d) AB = BA = I**

**Solution:**

Choice (d) is correct

i.e., AB = BA = I

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