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Class 12 NCERT Solutions – Mathematics Part I – Chapter 3 Matrices – Exercise 3.3
• Last Updated : 25 Jan, 2021

### Question 1. Find the transpose of each of the following matrices:

(i) (ii) (iii) Solution:

(i) Let A = ∴Transpose of A = A’ = A (ii) Let A = ∴Transpose of A = A’ = AT  = (iii) Let A = ∴Transpose of A = A’ = AT  = ### Question 2. If A = and B = then verify that:

(i) (A+B)’ = A’+B’

(ii) (A-B)’ = A’- B’

Solution:

(i) A+B = L.H.S. = (A+B)’ = R.H.S. = A’+B’ = ∴L.H.S = R.H.S.

Hence, proved.

(ii) A-B = L.H.S. = (A-B)’ R.H.S. = A’-B’ = ∴ L.H.S. = R.H.S.

Hence, proved.

### Question 3.  If A’ = and B = , then verify that:

(i) (A+B)’=A’+B’

(ii) (A-B)’=A’-B’

Solution:

Given A’= and B= then, (A’)’ = A = (i) A+B = ∴ L.H.S. =  (A+B)’= R.H.S.= A’+B’ = ∴ L.H.S. = R.H.S.

Hence, proved.

(ii) A-B = ∴ L.H.S. =  (A-B)’= R.H.S.= A’-B’ = ∴ L.H.S. = R.H.S.

Hence, proved.

### Question 4. If A’ = and B = then find (A+2B)’.

Solution:

Given: A’ = and B = then (A’)’ =A= Now, A+2B = ∴(A+2B)’ = ### Question 5. For the matrices A and B, verify that (AB)′ = B′A′, where

(i) A = and B = (ii) A = and B = Solution:

(i) AB = = ∴  L.H.S. = (AB)′ = R.H.S.= B′A’ = ∴ L.H.S. = R.H.S.

Hence, proved.

(ii) AB = ∴  L.H.S. = (AB)′ = Now, R.H.S.=B’A’ = ∴ L.H.S. = R.H.S.

Hence, proved.

### (ii) A = ,then verify that A′ A = I.

Solution:

(i) = I = R.H.S.

∴ L.H.S. = R.H.S.

(ii) = I = R.H.S.

∴ L.H.S. = R.H.S.

### (ii) Show that the matrix A = is a symmetric matrix.

(i) Given: A = Now, A’= ∵ A = A’

∴ A is a symmetric matrix.

(ii) Given: A = Now, A’= ∵ A = A’

∴ A is a symmetric matrix.

### Question 8.  For the matrix A = , verify that:

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

Solution:

(i) Given: A = Let B = (A+A’) = Now, B’ = (A+A’)’ = ∵ B = B’

∴ B=(A+A’) is a symmetric matrix.

(ii) Given: A = Let B = (A-A’) = Now, B’ = (A-A’)’ = ∵ -B = B’

∴ B=(A-A’) is a skew symmetric matrix.

### Question 9. Find 1/2(A+A’) and 1/2(A-A’) ,when A = .

Solution:

Given: A = ∴  A’ = Now,  A+A’ = + Now, A-A’ = ### Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) (ii) (iii) (iv) Solution:

(i) Given : A = ⇒ A’= Let P = and Q = Now, P = …..(1)

& P’ = ∵ P=P’

∴ P is a symmetric matrix.

Now, Q = …..(2)

& Q’ = ∵ -Q=Q’

∴ Q is a skew symmetric matrix.

By adding (1) and (2), we get, Therefore, A =P + Q

(ii) Given : ⇒ A’= P = …..(1)

Q = ……(2)

By adding (1) and (2), we get,

\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix} Therefore, A =P + Q

(iii) Given: A = ⇒ A’= P = } …..(1)

Q = ……(2)

By adding (1) and (2), we get

} Therefore, A =P + Q

(iv) Given: A = ⇒ A’= P = …..(1)

Q = …..(2)

By adding (1) and (2), we get Therefore, A =P + Q

### Question 11. If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix (B) Symmetric matrix

(C) Zero matrix (D) Identity matrix

Solution:

Given: A and B are symmetric matrices.

⇒ A=A’

⇒ B=B’

Now, ( AB – BA)’ =(AB)’-(BA)’              [∵ (X-Y)’=X’-Y’]

=B’A’-A’B’                [∵ (XY)’=Y’X’]

=BA-AB                   [∵ Given]

= -(AB-BA)

∴(AB-BA) is a skew symmetric matrix.

∴ The option (A) is correct.

### Question 12. If A = , and A + A′ = I, then the value of α is

(A)π/6    (B) π/3

(C) π    (D)3π/2

Solution:  On comparing both sides, we get

2cosα = 1

⇒      cosα = ⇒      cosα = cos ⇒      α = ∴ The option (B) is correct.

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