RD Sharma Class 12 Solutions Chapter 5 Algebra of Matrices Exercise 5.3 (Updated for 2024)

RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.3: RD Sharma Solutions for Class 12 Ex 5.3 are available in PDF format on the Kopykitab website, with solutions carefully compiled by an expert team. It is primarily intended for students to utilize as study material in order to pass the Class 12 exams with a decent grade in accordance with the CBSE syllabus. The fifth chapter’s Exercise 5.3 presents questions that are solved using the transpose of a matrix. The RD Sharma Solutions are written in an explanatory style to help students understand the concepts. RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices Exercise 5.3 are provided in this article.

Download RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.3 Free PDF

RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.3

Access answers to RD Sharma Solutions For Chapter 5 – Algebra of Matrices Ex 5.3 Class 12 Maths Important Questions With Solution

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 equations (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Now again, consider,

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

AB ≠ BA

(iii) Consider,

Now again, consider,

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

AB ≠ BA

(ii) Consider,

Again consider,

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

AB ≠ BA

5. Evaluate the following:

Solution:

(i) Given

First, we have to add the first two matrices.

On simplifying, we get

(ii) Given,

First, we have to multiply the first two given matrices.

= 82

(iii) Given

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

Solution:

Given

We know that,

Again we know that,

Now, consider,

We have,

Now, from equations (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 equations (1) and (2), AB = BA = 0_3×3

Solution:

Given

Consider,

Again consider,

From equations (1) and (2), AB = BA = 0_3×3

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

(ii) Given,

Consider the LHS,

Now consider RHS,

From equations (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 equations (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 the identity matrix, so

Also given,

Now, we have to find A², we get

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

So,

Substitute corresponding values from eqns (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 the identity matrix, so

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

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

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

So,

Hence Proved

Solution:

Given

So

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

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

But given, A² = λ A + μ I

Substitute corresponding values from equations (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 values of λ and μ are 4 and – 1, respectively

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

Solution:

We know,

is the identity matrix of size 3.

So, according to the given criteria

Now we will multiply the two matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_in b_nj, 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

Access other exercises of RD Sharma Solutions For Class 12 Chapter 5 

• RD Sharma Class 12 Maths Exercise 5.1 Solutions
• RD Sharma Class 12 Maths Exercise 5.2 Solutions
• RD Sharma Class 12 Maths Exercise 5.4 Solutions
• RD Sharma Class 12 Maths Exercise 5.5 Solutions

We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.3. If you have any queries related to the CBSE Class 12 Exam, please ask us in the comment section below.

FAQs on RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.3