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

• Last Updated : 03 Mar, 2021

### 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 2×2.

=> => ### 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 3×3.

=> => ### (i): 2A – 3B

Solution:

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

=> 3A = => 2B = => 3C = => 3A – 2B + 3C = => 3A – 2B + 3C = ### (i): A + B and B + C

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3.

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 2×2 which is different from B’s order which is 2×3, 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 2×3.

=> 2A = => 3B = => 4C = => 2A – 3B + 4C = => 2A – 3B + 4C = ### (i): A – 2B

Solution:

In the given question A and B are diagonal matrices of the order 3×3, 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 3×3, 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 3×3, 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.

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