RD Sharma Solutions for Class 12 Maths Exercise 5.2 Chapter 5 Algebra of Matrices

RD Sharma Solutions for Class 12 Maths Exercise 5.2 Chapter 5 Algebra of Matrices are provided here for students to study and prepare for their board exams. Regular practice improves students to solve Mathematics problems accurately. Highly experienced subject experts having a vast knowledge of concepts prepare the answers based on the understanding ability of the students.

RD Sharma Solutions for Class 12 can be used by students in order to excel in the subject and obtain a good score. This exercise covers the topic addition of matrices with explanatory answers. RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices Exercise 5.2 are provided here.

RD Sharma Solutions for Class 12 Chapter 5 – Algebra of Matrices Exercise 5.2

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

Exercise 5.1 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 5.2

1. Compute the following sums:

Solution:

(i) Given

Corresponding elements of two matrices should be added.

Therefore, we get

Therefore,

(ii) Given

Therefore,

Find each of the following:

(i) 2A – 3B

(ii) B – 4C

(iii) 3A – C

(iv) 3A – 2B + 3C

Solution:

(i) Given

First, we have to compute 2A

Now by computing 3B, we get,

Now we have to compute 2A – 3B, and we get

Therefore,

(ii) Given

First, we have to compute 4C

Now,

Therefore, we get,

(iii) Given

First, we have to compute 3A

Now,

Therefore,

(iv) Given

First, we have to compute 3A

Now, we have to compute 2B

By computing 3C, we get,

Therefore,

(i) A + B and B + C

(ii) 2B + 3A and 3C – 4B

Solution:

(i) Consider A + B,

A + B is not possible because matrix A is an order of 2 x 2, and Matrix B is an order of 2 x 3, so the sum of the matrix is only possible when their order is the same.

Now consider B + C

(ii) Consider 2B + 3A

2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrices.

Now consider 3C – 4B,

Solution:

Given

Now we have to compute 2A – 3B + 4C

5. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find

(i) A – 2B

(ii) B + C – 2A

(iii) 2A + 3B – 5C

Solution:

(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

We have to find B + C – 2A

Here,

Now we have to compute B + C – 2A

(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

Now we have to find 2A + 3B – 5C

Here,

Now consider 2A + 3B – 5C

6. Given the matrices

Verify that (A + B) + C = A + (B + C)

Solution:

Given

Now we have to verify (A + B) + C = A + (B + C)

First consider LHS, (A + B) + C,

Now consider RHS, that is A + (B + C)

Therefore, LHS = RHS

Hence, (A + B) + C = A + (B + C)

7. Find the matrices X and Y,

Solution:

Consider,

Now by simplifying, we get,

Therefore,

Again consider,

Now by simplifying, we get,

Therefore,

Solution:

Given

Now by transposing, we get

Therefore,

Solution:

Given

Now by multiplying equations (1) and (2), we get,

Now by adding equations (2) and (3), we get,

Now by substituting X in equation (2), we get,

Solution:

Consider

Now, again consider

Therefore,

And