# NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 3: Matrices (Part II)

NCERT Solutions for CBSE 12th Maths, Chapter 3: Matrices are available here. In this article, you will find solutions of exercise 3.1 (from question number 6 to question number 10). These questions are mainly based on equality of Matrices. These questions are also important for CBSE Class 12 board exam 2018 and other engineering entrance exams.

**Question 6:**

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**Solution 6:**

(*i*)

In the given matrix, if we compare the corresponding elements, *x* = 1, *y* = 4, and* z* = 3.

(*ii*)

In the given matrix, on comparing the corresponding elements, we have,

*x* + *y* = 6

⇒ *xy* = 8

⇒ 5 + z = 5 ⇒ z = 0

Now, (*x* − *y*)^{2} = (*x* + *y*)^{2} − 4*xy*

⇒ (*x* − *y*)^{2} = 36 − 32 = 4

⇒ *x* − *y* = ±2

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

(*iii*)

Comparing the corresponding elements,

*x* + *y* + *z* = 9 … (1)

*x* + *z* = 5 …..... (2)

*y* + *z* = 7 …..... (3)

From …(1) and ….(2)

*y* + 5 = 9 ⇒ *y* = 4

Then, from (3) 4 + *z* = 7

⇒ *z* = 3 ∴ *x* + *z* = 5

⇒ *x* = 2

∴ *x* = 2, *y* = 4, and *z* = 3

**NCERT Exemplar Class 12 Mathematics – Chapter 3: Matrics**

**Question 7:**

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**Solution 7:**

Comparing the corresponding elements,

*a* − *b* = −1 … (1)

2*a* − *b* = 0 … (2)

2*a* + *c* = 5 … (3)

3*c* + *d* = 13 … (4)

From equation …(2)

*b* = 2*a*

From (1), *a* − 2*a* = −1

⇒ a = 1 ⇒ b =** 2**

From (3), 2 ×1 + *c *= 5

⇒ *c* = 3** **

From equation (4), 3 ×3 + *d* = 13

⇒ 9 + *d* = 13

⇒ *d* = 4

∴ *a* = 1, *b* = 2, *c* = 3, and *d* = 4.

**Question 8: **A = [a_{ij}]m × n is a square matrix, if

(A) *m* < *n*

(B) *m* > *n *

(C) *m* = *n*

(D) None of these

**Solution 8:**

We know that, A given matrix is said to be a square matrix if the number of rows is equal to the number of columns. The correct answer is C.

A = [*a _{ij}*]

_{m }_{× n}is a square if

*m*=

*n*.

**Question 9:**

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**Solution 9:**

On comparing the corresponding elements, we have,

3 *x* + 7 = 0

⇒ *x* = ‒7/3

Also, 5 = *y* ‒ 2 ⇒ *y* = 7

Again, *y* + 1 = 8

⇒ *y* = 7.

Also,

2 ‒ 3*x* = 4

⇒ *x *= ‒2/3.

**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 10:**

Elements of a 3 × 3 matrix = 9.

Each of these elements can be either 0 or 1.

Each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is 2^{9} = 512.

**Download NCERT Solutions for Class 12 Maths (Chapter 3 - Matrices) in PDF format**