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

**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 A**^{2} = B^{2} = C^{2} = I_{2}.

^{2}= B

^{2}= C

^{2}= I

_{2}.

**Solution:**

We have,

A =, B =and C =

A

^{2}==

=

Therefore, A

^{2}= I_{2}B

^{2}==

=

Therefore, B

^{2}= I_{2}C

^{2}==

=

Therefore, C

^{2}= I_{2}So, we get A

^{2}= B^{2}= C^{2}= I_{2}

Hence proved.

**Question 7. If A =****and B =****, find 3A**^{2} – 2B + I.

^{2}– 2B + I.

**Solution:**

We are given,

A =and B =

So, we get,

3A

^{2}– 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 A**^{2} =**and A**^{3} =**.**

^{2}=

^{3}=

**Solution:**

We have,

A =

So, A

^{2}==

=

Hence, A

^{3}= A^{2}. A=

=

=

Hence proved.

**Question 10. If A =****, show that A**^{2} = 0.

^{2}= 0.

**Solution:**

We have,

A =

So, we get

L.H.S. = A

^{2 }==

=

= 0

= R.H.S.

Hence proved.

**Question 11. If A =****, find A**^{2}.

^{2}.

**Solution:**

We have,

A =

So, we get

A

^{2}==

=

=

**Question 12. If A =****and B =****, show that AB = BA = O**_{3×3}.

_{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}.

_{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 A**^{2} – B^{2}.

^{2}– B

^{2}.

**Solution:**

We have,

A =and B =

A

^{2}==

=

And we have,

B

^{2}==

=

So, we get

A

^{2}– B^{2}==

=

**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 a**_{43} and a_{22} of the matrix:

_{43}and a

_{22}of the matrix:

**A =**

**Solution:**

We are given,

A =

=

=

=

=

Therefore, a_{43}= 8 and a_{22}= 0.

**Question 20. If A =****and I is the identity matrix of order 3, show that A**^{3} = pI + qA + rA^{2}.

^{3}= pI + qA + rA

^{2}.

**Solution:**

We have,

A =

L.H.S. = A

^{3}=

=

=

=

=

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

^{2}=

=

=

=

=

=

= 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 A**^{2} = A.

^{2}= A.

**Solution:**

We have,

A =

So, A

^{2}==

=

= A

Hence proved.

**Question 23. If A =****, show that A**^{2} = I_{3}.

^{2}= I

_{3}.

**Solution:**

We have,

A =

So, A

^{2}==

=

= I

_{3}

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,

=>

=>

=>

=>

=> 2x

^{2}+ 4x + 4x + 8 – 2x – 4 = 0=> 2x

^{2}+ 6x + 4 = 0=> 2x

^{2}+ 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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