RD Sharma Class 12 Solutions Chapter 5 Algebra of Matrices Exercise 5.2 (Updated for 2024)

RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.2: The most important thing that students in the Mathematics domain should do is practice on a daily basis. RD Sharma Answers are prepared by highly experienced subject experts with extensive knowledge of concepts, and they are customized to the student’s understanding abilities. Students can use RD Sharma Solutions for Class 12 to succeed in the subject and achieve a high academic grade. With explanations, this exercise explains the concept of matrices addition. Here you will find RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices Exercise 5.2.

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RD Sharma Solutions Class 12 Maths Chapter 5 Exercise 5.2

Access answers to Maths RD Sharma Solutions For Class 12 Chapter 5 – Algebra of Matrices Exercise 5.2 Important Questions With Solution

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

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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 5.3 Solutions
• RD Sharma Class 12 Maths Exercise 5.4 Solutions
• RD Sharma Class 12 Maths Exercise 5.5 Solutions

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