# Class 12 NCERT Solutions- Mathematics Part I – Chapter 3 Matrices – Exercise 3.2 | Set 1

• Last Updated : 05 Apr, 2021

(i) A + B

(ii) A – B

(iii) 3A – C

(iv) AB

(v) BA

Solution:

(i)

(ii)

(iii)

(iv)

(v)

Solution:

(i)

(ii)

(iii)

(iv)

Solution:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

### Question 4. If , then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

Solution:

Now we have to show A + (B – C) = (A + B) – C

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

Hence, Proved

Solution:

### Question 6.  Simplify

Solution:

= 1 = identity matrix

### (ii)

Solution:

(i) Given:

Adding (1) and (2), we get

(ii) Given:

Now, multiply equation (1) by 2 and equation (2) by 3 we get

Subtracting equation (4) from (3), we get,

Solution:

### Question 9. Find X and Y, if

Solution:

Given:

Equating corresponding entries, we have

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

y = 5 – 2 and 2(x + 1) = 8

y = 3 and x + 1 = 4

Therefore, y = 3 and x = 3

### Question 10. Solve the equation for x, y, z and t, if

Solution:

Given:

On comparing both sides, we have

2x + 3 = 9 ⇒ 2x = 9 – 3 ⇒ 2x = 6 ⇒ x = 3

2z – 3 = 15 ⇒ 2z = 15 + 3 ⇒ 2z = 18 ⇒ z = 9

2y = 12 ⇒ y = 6

2t + 6 = 18 ⇒ 2t = 18 – 6 ⇒ 2t = 12 ⇒ t = 6

Therefore, x = 3, y = 6, z = 9, t = 6

### Chapter 3 Matrices – Exercise 3.2 | Set 2

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