# Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.3 | Set 1

• Last Updated : 21 Jul, 2021

(i)

Solution:

We have,

=

=

(ii)

Solution:

We have,

=

=

(iii)

Solution:

We have,

=

=

### Question 2. Show that AB ≠ BA in each of the following cases:

(i) and

Solution:

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence, proved.

(ii) and

Solution:

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

(iii) and

Solution:

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

### Question 3. Compute the products AB and BA whichever exists in each of the following cases:

(i) and

Solution:

We have,

A =and B =

As A is of order 2 × 2 and B is of order 2 × 3, AB is possible but BA is not possible.

So, we get

AB =

=

=

(ii) and

Solution:

We have,

A =and B =

As A is of order 3 × 2 and B is of order 2 × 3, AB and BA both are possible.

So, we get,

AB =

=

=

Also we have,

BA =

=

=

(iii) and

Solution:

We have,

A =and B =

As A is of order 1 × 4 and B is of order 4 × 1, AB and BA both are possible.

So, we get,

AB =

=

=

=

Also, we have,

BA =

=

=

(iv)

Solution:

We have,

=

=

=

### Question 4. Show that AB ≠ BA in each of the following cases:

(i) and

Solution:

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

(ii) and

Solution:

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

(i)

Solution:

We have,

=

=

=

=

=

(ii)

Solution:

We have,

=

=

=

=

=

(iii)

Solution:

We have,

=

=

=

=

=

### Question 6. If A =, B =and C =, then show that A2 = B2 = C2 = I2.

Solution:

We have,

A =, B =and C =

A2 =

=

=

Therefore, A2 = I2

B2 =

=

=

Therefore, B2 = I2

C2 =

=

=

Therefore, C2 = I2

So, we get A2 = B2 = C2 = I2

Hence proved.

Solution:

We are given,

A =and B =

So, we get,

3A2 – 2B + I =

=

=

=

=

=

### Question 8. If A =, prove that (A – 2I) (A – 3I) = 0.

Solution:

We are given,

A =

L.H.S. = (A – 2I) (A – 3I)

=

=

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 9. If A =, show that A2 =and A3 =.

Solution:

We have,

A =

So, A2 =

=

=

Hence, A3 = A2 . A

=

=

=

Hence proved.

Solution:

We have,

A =

So, we get

L.H.S. = A2 =

=

=

= 0

= R.H.S.

Hence proved.

Solution:

We have,

A =

So, we get

A2 =

=

=

=

### Question 12. If A =and B =, show that AB = BA = O3×3.

Solution:

We have,

A =and B =

So, we get

AB =

=

=

= O3×3

And we have,

BA =

=

=

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

### Question 13. If A =and B =, show that AB = BA = O3×3.

Solution:

We have,

A =and B =

So, we have,

AB =

=

=

And we have,

BA =

=

=

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

Solution:

We have,

A =and B =

AB =

=

=

= A

And we have,

BA =

=

=

= B

Hence proved.

Solution:

We have,

A =and B =

A2 =

=

=

And we have,

B2 =

=

=

So, we get

A2 – B2 =

=

=

### Question 16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC).

(i) A =, B =, C =

Solution:

We are given,

A =, B =, C =

L.H.S. = (AB) C

=

=

=

=

And R.H.S. = A (BC)

=

=

=

=

=

= L.H.S.

Hence proved.

(ii) A =, B =, C =

Solution:

We are given,

A =, B =, C =

L.H.S. = (AB) C

=

=

=

=

=

And R.H.S. = A (BC)

=

=

=

=

=

= L.H.S.

Hence proved.

### Question 17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.

(i) A =, B =, C =

Solution:

We have,

A =, B =, C =

L.H.S. = A (B + C)

=

=

=

=

=

R.H.S. = AB + AC

=

=

=

=

=

= L.H.S.

Hence proved.

(ii) A =, B =, C =

Solution:

We have,

A =, B =, C =

L.H.S. = A (B + C)

=

=

=

=

=

R.H.S. = AB + AC

=

=

=

=

=

= L.H.S.

Hence proved.

### Question 18. If A =, B =and C =, show that A (B – C) = AB –AC.

Solution:

We have,

A =, B =and C =

L.H.S. = A (B – C)

=

=

=

=

=

R.H.S. = AB – AC

=

=

=

=

=

### Question 19. Compute the elements a43 and a22 of the matrix:

A =

Solution:

We are given,

A =

=

=

=

=

Therefore, a43 = 8 and a22 = 0.

### Question 20. If A =and I is the identity matrix of order 3, show that A3 = pI + qA + rA2.

Solution:

We have,

A =

L.H.S. = A3

=

=

=

=

=

And R.H.S. = pI + qA + rA2

=

=

=

=

=

=

= L.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

=

=

=

= R.H.S.

Hence proved.

Solution:

We have,

A =

So, A2 =

=

=

= A

Hence proved.

Solution:

We have,

A =

So, A2 =

=

=

= I3

Hence proved.

### Question 24.

(i) If= 0, find x.

Solution:

We have,

=>= 0

=>= 0

=>= 0

=>= 0

=> [3x + 6] = 0

=> 3x = –6

=> x = –6/3

=> x = –2

Therefore, the value of x is –2.

(ii) If= 0, find x.

Solution:

We have,

=>

=>

=>

On comparing the above matrix we get,

x = 13

Therefore, the value of x is –13.

### Question 25. If, find x.

Solution:

We have,

=>

=>

=>

=>

=> 2x2 + 4x + 4x + 8 – 2x – 4 = 0

=> 2x2 + 6x + 4 = 0

=> 2x2 + 2x + 4x + 4 = 0

=> 2x (x + 1) + 4 (x + 1) = 0

=> (x + 1) (2x + 4) = 0

=> x = –1 or x = –2

Therefore, the value of x is –1 or –2.

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