RD Sharma Solutions for Class 12 Maths Exercise 5.3 Chapter 5 Algebra of Matrices are available on BYJU’S website in PDF format with solutions prepared by experienced faculty in a precise manner. It can be used by the students as study materials to clear the Class 12 exam 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.

RD Sharma Solutions for Class 12 Chapter 5 – Algebra of Matrices Exercise 5.3

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Access other exercises of RD Sharma Solutions for Class 12 Chapter 5 – Algebra of Matrices

Exercise 5.1 Solutions

Exercise 5.2 Solutions

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Exercise 5.5 Solutions

Access answers to Maths RD Sharma Solutions for Class 12 Chapter 5 – Algebra of Matrices Exercise 5.3

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 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 the identity matrix of size 3.

So, according to the given criterias

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