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Class 12 NCERT Solutions – Mathematics Part I – Chapter 3 Matrices – Exercise 3.3

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.

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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