NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Matrices |

Exercise |
Ex 3.1 |

Number of Questions Solved |
10 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 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.

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