Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.2 | Set 1

Chapter 5 of RD Sharma's Class 12 Mathematics textbook focuses on the Algebra of Matrices. Exercise 5.2 specifically deals with operations on matrices, including addition, subtraction, and multiplication. This exercise helps students understand how to perform these operations and apply them to solve various problems involving matrices.

Question 1(i): Compute the following sum:.

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 2x2.

=>

=>

Question 1(ii): Compute the following sum:.

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 3x3.

=>

=>

Question 2: Let A =, B =and C =. Find each of the following:

(i): 2A - 3B

Solution:

Both the matrices A and B are of the same order which is 2x2, hence the operation can be performed.

=> 2A =

=> 3B =

=> 2A - 3B =

=> 2A - 3B =

(ii): B - 4C

Solution:

Both the matrices B and C are of the same order which is 2x2, hence the operation can be performed.

=> B =

=> 4C =

=> B - 4C =

=> B - 4C =

(iii): 3A - C

Solution:

Both the matrices A and C are of the same order which is 2x2, hence the operation can be performed.

=> 3A =

=> C =

=> 3A - C =

=> 3A - C =

(iv): 3A -2B + 3C

Solution:

The matrices A, B and C are of the same order which is 2x2, hence the operation can be performed.

=> 3A =

=> 2B =

=> 3C =

=> 3A - 2B + 3C =

=> 3A - 2B + 3C =

Question 3: If A =, B =, C =, find:

(i): A + B and B + C

Solution:

A and B can not be added since A's order is 2x2 which is different from B's order which is 2x3.

B+C can be computed and is solved as follows:

=> B + C =

=> B + C =

(ii): 2B + 3A and 3C - 4B

Solution:

A and B can not be added since A's order is 2x2 which is different from B's order which is 2x3, and thus 2B + 3A can not be calculated.

3C - 4B can be computed and is solved as follows:

=> 3C =

=> 4B =

=> 3C - 4B =

=> 3C - 4B =

Question 4: Let A =, B =and C =. Compute 2A - 3B + 4C.

Solution:

The result can be computed since A, B and C are of the same order which is 2x3.

=> 2A =

=> 3B =

=> 4C =

=> 2A - 3B + 4C =

=> 2A - 3B + 4C =

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

(i): A - 2B

Solution:

In the given question A and B are diagonal matrices of the order 3x3, thus the only non-zero elements are present in the diagonal.

=> A = diag(2, -5, 9)

=> 2B = 2. diag(1, 1, -4) = diag(2, 2, -8)

=> A - 2B = diag(2-2, -5-2, 9+8)

=> A - 2B = diag(0, -7, 17)

(ii): B + C - 2A

Solution:

In the given question A, B and C are diagonal matrices of the order 3x3, thus the only non-zero elements are present in the diagonal.

=> B = diag(1, 1, -4)

=> C = diag(-6, 3, 4)

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> B + C - 2A = diag(1-6-4, 1+3+10, -4+4-18)

=> B + C - 2A = diag(-9, 14, -18)

(iii): 2A + 3B - 5C

Solution:

In the given question A, B and C are diagonal matrices of the order 3x3, thus the only non-zero elements are present in the diagonal.

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> 3B = 3. diag(1, 1, -4) = diag(3, 3, -12)

=> 5C = 5. diag(-6, 3, 4) = diag(-30, 15, 20)

=> 2A + 3B - 5C = diag(4+3+30, -10+3-15, 18-12-20)

=> 2A + 3B - 5C = diag(37, -22, -14)

Question 6: Given the matrices A =, B =and C =. Verify that (A + B) + C = A + (B + C).

Solution:

Given L.H.S :

=> (A + B) = 

=> (A + B) =

=> (A + B) =

=> (A + B) + C =

=> (A + B) + C =

=> (A + B) + C =

Given R.H.S :

=> (B + C) =

=> (B + C) =

=> (B + C) =

=> A + (B + C) =

=> A + (B + C) =

=> A + (B + C) =

Hence R.H.S = L.H.S has been verified.

Question 7: Find the matrices X and Y, if X + Y =and X - Y =.

Solution:

We know that (X + Y) + (X - Y) = 2X.

=> (X + Y) + (X - Y) =

=> 2X =

=> 2X =

=> X =

=> X =

Now Y = (X + Y) - X

=> Y =

=> Y =

=> Y =

Question 8: Find X, if Y =and 2X + Y =.

Solution:

Given 2X + Y =

=> 2X +=

=> 2X =

=> 2X =

=> 2X =

=> X =

=> X =

Question 9: Find matrices X and Y, if 2X - Y =and X + 2Y =.

Solution:

We know that 2 (2X - Y) + (X + 2Y) = 4X - 2Y + X + 2Y = 5X .

=> 2 (2X - Y) =

=> 2 (2X -Y) =

=> 2 (2X - Y) + (X + 2Y) =

=> 5X =

=> 5X =

=> X =

=> X =

As (X + 2Y) =

=>

=> 2Y =

=> 2Y =

=> Y =

=> Y =

Question 10: If X - Y =and X + Y =, find X and Y.

Solution:

We know that (X + Y) + (X - Y) = 2X.

=> 2X =

=> 2X =

=> 2X =

=> X =

=> X =

Also (X + Y) - (X -Y) = 2Y.

=> 2Y =

=> 2Y =

=> 2Y =

=> Y =

=> Y =