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

Chapter 5 of RD Sharma's Class 12 Mathematics textbook on the Algebra of Matrices continues with Exercise 5.3. This exercise typically focuses on more advanced matrix operations and properties, including determinants, adjoint matrices, and inverse matrices. Set 1 of this exercise often introduces these concepts and provides problems that help students understand their applications and relationships.

Question 1. Compute the indicated products:

(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.

Question 5. Evaluate the following:

(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.

Question 7. If A =and B =, find 3A² – 2B + I.

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.

Question 10. If A =, show that A² = 0.

Solution:

We have,

A =

So, we get

L.H.S. = A² =

=

=

= 0

= R.H.S.

Hence proved.

Question 11. If A =, find A².

Solution:

We have,

A =

So, we get

A² =

=

=

=

Question 12. If A =and B =, show that AB = BA = O_3×3.

Solution:

We have,

A =and B =

So, we get

AB =

=

=

= O_3×3

And we have,

BA =

=

=

= O_3×3

Therefore, AB = BA = O_3×3.

Hence proved.

Question 13. If A =and B =, show that AB = BA = O_3×3.

Solution:

We have,

A =and B =

So, we have,

AB =

=

=

And we have,

BA =

=

=

= O_3×3

Therefore, AB = BA = O_3×3.

Hence proved.

Question 14. If A =and B =, show that AB = A and BA = B.

Solution:

We have,

A =and B =

AB =

=

=

= A

And we have,

BA =

=

=

= B

Hence proved.

Question 15. If A =and B =, compute A² – B².

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.

Question 21. If ω is a complex cube root of unity, show that

Solution:

We have,

L.H.S. =

=

=

=

=

=

=

= R.H.S.

Hence proved.

Question 22. If A =, prove that A² = A.

Solution:

We have,

A =

So, A² =

=

=

= A

Hence proved.

Question 23. If A =, show that A² = I₃.

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.