# RD Sharma Solutions For Class 12 Maths Exercise 5.3 Chapter 5 Algebra of Matrices

RD Sharma Solutions for Class 12 are available at BYJU’S website in PDF format with solutions prepared by experienced faculty in a precise manner. It can mainly be used by the students as a study material to clear the Class 12 exams with a good score according to the CBSE syllabus. Exercise 5.3 of the fifth chapter contains problems, which are solved based on the transpose of a matrix. The solutions prepared are in an explanatory manner to bring about conceptual clarity among the students. RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices Exercise 5.3 are provided here.

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1. Compute the indicated products:

Solution:

(i) Consider

On simplification we get,

(ii) Consider

On simplification we get,

(iii) Consider

On simplification we get,

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

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(iii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

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

Solution:

(i) Consider,

BA does not exist

Because the number of columns in B is greater than the rows in A

(ii) Consider,

Again consider,

(iii) Consider,

AB = [0 + (-1) + 6 + 6]

AB = 11

Again consider,

(iv) Consider,

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

Solution:

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Again consider,

From equation (1) and (2) it is clear that,

AB ≠ BA

5. Evaluate the following:

Solution:

(i) Given

First we have to add first two matrix,

On simplifying, we get

(ii) Given,

First we have to multiply first two given matrix,

= 82

(iii) Given

First we have subtract the matrix which is inside the bracket,

Solution:

Given

We know that,

Again we know that,

Now, consider,

We have,

Now, from equation (1), (2), (3) and (4), it is clear that A2 = B2= C2= I2

Solution:

Given

Consider,

Now we have to find,

Solution:

Given

Consider,

Hence the proof.

Solution:

Given,

Consider,

Again consider,

Hence the proof.

Solution:

Given,

Consider,

Hence the proof.

Solution:

Given,

Consider,

We know that,

Again we have,

Solution:

Given,

Consider,

Again consider,

From equation (1) and (2) AB = BA = 03×3

Solution:

Given

Consider,

Again consider,

From equation (1) and (2) AB = BA = 03×3

Solution:

Given

Now consider,

Therefore AB = A

Again consider, BA we get,

Hence BA = B

Hence the proof.

Solution:

Given,

Consider,

Now again consider, B2

Now by subtracting equation (2) from equation (1) we get,

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

Solution:

(i) Given

Consider,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

(ii) Given,

Consider the LHS,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

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

Solution:

(i) Given

Consider LHS,

Now consider RHS,

From equation (1) and (2), it is clear that A (B + C) = AB + AC

(ii) Given,

Consider the LHS

Now consider RHS,

Solution:

Given,

Consider the LHS,

Now consider RHS

From the above equations LHS = RHS

Therefore, A (B – C) = AB – AC.

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

Solution:

Given

From the above matrix, a43 = 8and a22 = 0

Solution:

Given

Consider,

Again consider,

Now, consider the RHS

Therefore, A3 = p I + q A + rA2

Hence the proof.

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

Solution:

Given

It is also given that ω is a complex cube root of unity,

Consider the LHS,

We know that 1 + ω + ω2 = 0 and ω3 = 1

Now by simplifying we get,

Again by substituting 1 + ω + ω2 = 0 and ω3 = 1 in above matrix we get,

Therefore LHS = RHS

Hence the proof.

Solution:

Given,

Consider A2

Therefore A2 = A

Solution:

Given

Consider A2,

Hence A2 = I3

Solution:

(i) Given

= [2x + 1 + 2 + x + 3] = 0

= [3x + 6] = 0

= 3x = -6

x = -6/3

x = -2

(ii) Given,

On comparing the above matrix we get,

x = 13

Solution:

Given

⇒ [(2x + 4) x + 4 (x + 2) – 1(2x + 4)] = 0

⇒ 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

Hence, x = -1 or x = -2

Solution:

Given

By multiplying we get,

Solution:

Given

Now we have to prove A2 – A + 2 I = 0

Solution:

Given

Solution:

Given

Hence the proof.

Solution:

Given

Hence the proof.

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

Solution:

Given

I is identity matrix, so

Also given,

Now, we have to find A2, we get

Now, we will find the matrix for 8A, we get

So,

Substitute corresponding values from eqn (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

Therefore, the value of k is 7

Solution:

Given

To show that f (A) = 0

Substitute x = A in f(x), we get

I is identity matrix, so

Now, we will find the matrix for A2, we get

Now, we will find the matrix for 2A, we get

Substitute corresponding values from eqn (ii) and (iii) in eqn (i), we get

So,

Hence Proved

Solution:

Given

So

Now, we will find the matrix for A2, we get

Now, we will find the matrix for λ A, we get

But given, A2 = λ A + μ I

Substitute corresponding values from equation (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence, λ + 0 = 4 ⇒ λ = 4

And also, 2λ + μ = 7

Substituting the obtained value of λ in the above equation, we get

2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1

Therefore, the value of λ and μ are 4 and – 1 respectively

39. Find the value of x for which the matrix product

Solution:

We know,

is identity matrix of size 3.

So according to the given criteria

Now we will multiply the two matrices on LHS using the formula cij = ai1b1j + ai2b2j + … + ain bnj, we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

So we get

So the value of x is