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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) \begin{bmatrix}5 \\\frac{1}{2} \\-1 \end{bmatrix}    

(ii) \begin{bmatrix}1 & -1 \\2 & 3 \\\end{bmatrix}

(iii) \begin{bmatrix}-1 & 5 & 6\\ \sqrt{3} & 5 & 6\\2 & 3 & -1\end{bmatrix}

Solution:

(i) Let A =\begin{bmatrix}5 \\\frac{1}{2} \\-1 \end{bmatrix}



∴Transpose of A = A’ = A\begin{bmatrix}5&\frac{1}{2} &-1\\\end{bmatrix}

(ii) Let A =\begin{bmatrix}1 & -1 \\2 & 3 \\\end{bmatrix}

∴Transpose of A = A’ = AT  =\begin{bmatrix}1 & 2 \\-1 & 3 \\\end{bmatrix}

(iii) Let A =\begin{bmatrix}-1 & 5 & 6\\ \sqrt{3} & 5 & 6\\2 & 3 & -1\end{bmatrix}

∴Transpose of A = A’ = AT  =\begin{bmatrix}-1 & \sqrt{3} & 2\\ 5 & 5 & 3\\6 & 6 & -1\end{bmatrix}

Question 2. If A =\begin{bmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{bmatrix}     and B = \begin{bmatrix}-4 & 2 & -5\\1 & 2 & 0\\1 & 3 & 1\end{bmatrix}  then verify that:

(i) (A+B)’ = A’+B’

(ii) (A-B)’ = A’- B’

Solution:



(i) A+B =\begin{bmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & -5\\1 & 2 & 0\\1 & 3 & 1\end{bmatrix}=\begin{bmatrix}-1-4 & 2=1 & 3-5\\5+1 & 7+2 & 9+0\\-2+1 & 1+3 & 1+1\end{bmatrix}=\begin{bmatrix}-5 & 3 & -2\\6 & 9 & 9\\-1 & 4 & 2\end{bmatrix}

L.H.S. = (A+B)’ = \begin{bmatrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{bmatrix}

R.H.S. = A’+B’ = \begin{bmatrix}-1 & 5 & -2\\2 & 7 & 1\\-2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & 1\\1 & 2 & 3\\-5 & 0 & 1\end{bmatrix}=\begin{bmatrix}-1-4 & 5+1 & -2+1\\2+1 & 7+2 & 1+3\\-2-5 & 1+0 & 1+1\end{bmatrix}=\begin{bmatrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{bmatrix}

∴L.H.S = R.H.S.

Hence, proved.

(ii) A-B = \begin{bmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{bmatrix}-\begin{bmatrix}-4 & 2 & -5\\1 & 2 & 0\\1 & 3 & 1\end{bmatrix}=\begin{bmatrix}3 & 1 & 8\\4 & 5 & 9\\-3 & -2 & 0\end{bmatrix}

L.H.S. = (A-B)’=\begin{bmatrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{bmatrix}

R.H.S. = A’-B’ =\begin{bmatrix}-1 & 5 & -2\\2 & 7 & 1\\-2 & 1 & 1\end{bmatrix}-\begin{bmatrix}-4 & 1 & 1\\1 & 2 & 3\\-5 & 0 & 1\end{bmatrix}=\begin{bmatrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{bmatrix}

∴ L.H.S. = R.H.S.

Hence, proved.

Question 3.  If A’ =\begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}     and B = \begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix} , then verify that:

(i) (A+B)’=A’+B’

(ii) (A-B)’=A’-B’

Solution:

Given A’=\begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}    and B=\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix}

then, (A’)’ = A =\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\\\end{bmatrix}

(i) A+B =\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\\\end{bmatrix}+\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix}=\begin{bmatrix}2 &1 & 1\\5& 4 & 4\\\end{bmatrix}

∴ L.H.S. =  (A+B)’=\begin{bmatrix}2 & 5\\1 & 4 \\1 & 4 \end{bmatrix}

R.H.S.= A’+B’ = \begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}+\begin{bmatrix}-1 & 1  \\2 & 2 \\1 & 3 \end{bmatrix}=\begin{bmatrix}2 & 5\\1 & 4 \\1 & 4 \end{bmatrix}

∴ L.H.S. = R.H.S.

Hence, proved.

(ii) A-B = \begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\\\end{bmatrix}-\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix}=\begin{bmatrix}4 &-3 & -1\\3 & 0 & -2\\\end{bmatrix}

∴ L.H.S. =  (A-B)’=\begin{bmatrix}4 & 3 \\-3 & 0 \\-1 & -2 \end{bmatrix}

R.H.S.= A’-B’ = \begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}-\begin{bmatrix}-1 & 1  \\2 & 2 \\1 & 3 \end{bmatrix}=\begin{bmatrix}4 & 3\\-3 & 0 \\-1 & -2 \end{bmatrix}

∴ L.H.S. = R.H.S.

Hence, proved.

Question 4. If A’ = \begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}     and B = \begin{bmatrix}-1 & 0 \\1 & 2 \\\end{bmatrix}  then find (A+2B)’.

Solution:

Given: A’ =\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}     and B =\begin{bmatrix}-1 & 0 \\1 & 2 \\\end{bmatrix}

then (A’)’ =A=\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}

Now, A+2B = \begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}+2\begin{bmatrix}-1 & 0 \\1 & 2 \\\end{bmatrix}=\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}+\begin{bmatrix}-2 & 0 \\2 & 4 \\\end{bmatrix}=\begin{bmatrix}-2-2 & 1+0 \\3+2 & 2+4 \\\end{bmatrix}=\begin{bmatrix}-4 & 1 \\5 & 6 \\\end{bmatrix}

∴(A+2B)’ = \begin{bmatrix}-4 & 5 \\1 & 6 \\\end{bmatrix}

Question 5. For the matrices A and B, verify that (AB)′ = B′A′, where

(i) A =\begin{bmatrix}1 \\-4 \\3 \end{bmatrix} and B = \begin{bmatrix}-1 & 2 & 1\\\end{bmatrix}

(ii) A =\begin{bmatrix}0 \\1 \\2 \end{bmatrix} and B =\begin{bmatrix}1 & 5 & 7\\\end{bmatrix}

Solution:

(i) AB = =\begin{bmatrix}1 \\-4 \\3 \end{bmatrix}\begin{bmatrix}-1 & 2 & 1\\\end{bmatrix}=\begin{bmatrix}-1 & 2 & 1\\4 & -8 & -4\\-3 & 6 & 3\end{bmatrix}

∴  L.H.S. = (AB)′ =\begin{bmatrix}-1 & 4 & -3\\2 & -8 & 6\\1 & -4 & 3\end{bmatrix}

R.H.S.= B′A’ = \begin{bmatrix}-1 \\2 \\1 \end{bmatrix}\begin{bmatrix}1 & -4 & 3\\\end{bmatrix}=\begin{bmatrix}-1 & 4 & -3\\2 & -8 & 6\\1 & -4 & 3\end{bmatrix}

∴ L.H.S. = R.H.S.

Hence, proved.

(ii) AB =\begin{bmatrix}0 \\1 \\2 \end{bmatrix}\begin{bmatrix}1 & 5 & 7\\\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\1 & 5 & 7\\2 & 10 & 14\end{bmatrix}

∴  L.H.S. = (AB)′ =\begin{bmatrix}0 & 1 & 2\\0 & 5 & 10\\0 & 7 & 14\end{bmatrix}

Now, R.H.S.=B’A’ = \begin{bmatrix}1 \\5 \\7 \end{bmatrix}\begin{bmatrix}0 & 1 & 2\\\end{bmatrix}=\begin{bmatrix}0 & 1 & 2\\0 & 5 & 7\\0 & 7 & 14\end{bmatrix}

∴ L.H.S. = R.H.S.

Hence, proved.

Question 6. If (i) A =\begin{bmatrix}cosα & sinα \\-sinα & cosα \\\end{bmatrix}      , then verify that A′ A = I.

(ii) A =\begin{bmatrix}sinα & cosα \\-cosα & sinα \\\end{bmatrix}      ,then verify that A′ A = I.

Solution:

(i) \begin{bmatrix}cosα &-sinα \\sinα & cosα \\\end{bmatrix}\begin{bmatrix}cosα & sinα \\-sinα & cosα \\\end{bmatrix}=\begin{bmatrix}cos ^{2}α+sin^{2}α & cosαsinα-sinαcosα\\sinαcosα-cosαsinα & sin ^{2}α+cos^{2}α \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}

= I = R.H.S.

∴ L.H.S. = R.H.S.

(ii) \begin{bmatrix}sinα &-cosα \\cosα & sinα \\\end{bmatrix}\begin{bmatrix}sinα & cosα \\-cosα & sinα \\\end{bmatrix}=\begin{bmatrix}sin ^{2}α+sin^{2}α & sinαcosα-cosαsinα\\cosαsinα-sinαcosα & cos ^{2}α+sin^{2}α \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}

= I = R.H.S.

∴ L.H.S. = R.H.S.

Question 7. (i) Show that the matrix A\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}     = is a symmetric matrix.

(ii) Show that the matrix A\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}     = is a symmetric matrix.

(i) Given: A =\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}         

Now, A’=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}'=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}     

∵ A = A’

∴ A is a symmetric matrix.

(ii) Given: A = \begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}

Now, A’=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}'=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}

∵ A = A’

∴ A is a symmetric matrix.

Question 8.  For the matrix A =\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}    , verify that:

(i) (A + A′) is a symmetric matrix

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

Solution:

(i) Given: A =\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}

Let B = (A+A’) = \begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}+\begin{bmatrix}1 & 6 \\5 & 7 \\\end{bmatrix}=\begin{bmatrix}1+1 & 5+6 \\6+5 & 7+7 \\\end{bmatrix}=\begin{bmatrix}2 & 11 \\11 & 14 \\\end{bmatrix}

Now, B’ = (A+A’)’ = \begin{bmatrix}2 & 11 \\11 & 14 \\\end{bmatrix}'=\begin{bmatrix}2 & 11 \\11 & 14 \\\end{bmatrix}

∵ B = B’

∴ B=(A+A’) is a symmetric matrix.

(ii) Given: A =\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}

Let B = (A-A’) =\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}-\begin{bmatrix}1 & 6 \\5 & 7 \\\end{bmatrix}=\begin{bmatrix}1-1 & 5-6 \\6-5 & 7-7 \\\end{bmatrix}=\begin{bmatrix}0 & -1 \\1 & 0 \\\end{bmatrix}

Now, B’ = (A-A’)’ =\begin{bmatrix}0 & -1 \\1 & 0 \\\end{bmatrix}'=\begin{bmatrix}0 & 1 \\-1 & 0 \\\end{bmatrix}=-\begin{bmatrix}0 & -1 \\1 & 0 \\\end{bmatrix}

∵ -B = B’

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

Question 9. Find 1/2(A+A’) and 1/2(A-A’) ,when A =\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}    .

Solution:

Given: A = \begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}

∴  A’ = \begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}'=\begin{bmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{bmatrix}

Now,  A+A’ = +\frac{1}{2}\{\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}+\begin{bmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}0 +0& a-a & b-b\\-a+a & 0+0 & c-c\\-b+b & -c+c & 0+0\end{bmatrix}=\frac{1}{2}\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

Now, A-A’ =\frac{1}{2}\{\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}-\begin{bmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}0 -0& a+a & b+b\\-a-a & 0-0 & c+c\\-b-b & -c-c & 0-0\end{bmatrix}=\frac{1}{2}\begin{bmatrix}0 & 2a & 2b\\-2a & 0 & 2c\\-2b & -2c & 0\end{bmatrix}=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) \begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}

(ii) \begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}

(iii) \begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}

(iv) \begin{bmatrix}1 & 5 \\-1& 2 \\\end{bmatrix}

Solution:

(i) Given : A =\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}

⇒ A’=\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}

Let P = \frac{1}{2}(A+A')

and Q = \frac{1}{2}(A-A')

Now, P =\frac{1}{2} \{\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}+\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}6 & 6 \\6 & -2 \\\end{bmatrix}=\begin{bmatrix}3 & 3 \\3 & -1 \\\end{bmatrix}    …..(1)

& P’ = \begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}'=\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}

∵ P=P’

∴ P is a symmetric matrix.

Now, Q =\frac{1}{2}(A-A')=\frac{1}{2} \{\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}-\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}0 & 4 \\-4 & 0 \\\end{bmatrix}=\begin{bmatrix}0 & 2 \\-2 & 0 \\\end{bmatrix}    …..(2)

& Q’ = \begin{bmatrix}0 & 2 \\-2 & 0 \\\end{bmatrix}'=-\begin{bmatrix}0 & -2 \\2 & 0 \\\end{bmatrix}

∵ -Q=Q’ 

∴ Q is a skew symmetric matrix.

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

\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}+\begin{bmatrix}0 & 2 \\-2 & 0 \\\end{bmatrix}=\begin{bmatrix}3 & 5\\1 & -1 \\\end{bmatrix}

Therefore, A =P + Q

(ii) Given : \begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}

⇒ A’=\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}

P = \frac{1}{2}(A+A')=\frac{1}{2}(\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}+\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix})=\frac{1}{2}\begin{bmatrix}12 & -4 & 4\\-4 & 6 & -2\\4 & -2 & 6\end{bmatrix}=\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}

…..(1)

Q = \frac{1}{2}(A-A')=\frac{1}{2}(\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}-\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix})=\frac{1}{2}\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

……(2)

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

\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}+\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}

Therefore, A =P + Q

(iii) Given: A =\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}

⇒ A’=\begin{bmatrix}3 & -2 & -4\\3 & -2 & -5\\-1 & 1 & 2\end{bmatrix}

P = }\frac{1}{2}(A+A')=\frac{1}{2}(\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}+\begin{bmatrix}3 & -2 & -4\\3 & -2 & -5\\-1 & 1 & 2\end{bmatrix})=\begin{bmatrix}6 & 1 & -5\\1 & -4 & -4\\-5 & -4 & 4\end{bmatrix}=\begin{bmatrix}3 & 1/2 & -5/2\\1/2 & -2 & -2\\-5/2 & -2 & 2\end{bmatrix}    …..(1)

Q = \frac{1}{2}(A-A')=\frac{1}{2}(\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}-\begin{bmatrix}3 & -2 & -4\\3 & -2 & -5\\-1 & 1 & 2\end{bmatrix})=\frac{1}{2}\begin{bmatrix}0 & 5 & 3\\-5 & 0 & 6\\-3 & -6 & 0\end{bmatrix}=\begin{bmatrix}0 & 5/2 & 3/2\\-5/2 & 0 & 3\\-3/2 & -3 & 0\end{bmatrix}    ……(2)

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

}\begin{bmatrix}3 & 1/2 & -5/2\\1/2 & -2 & -2\\-5/2 & -2 & 2\end{bmatrix}+\begin{bmatrix}0 & 5/2 & 3/2\\-5/2 & 0 & 3\\-3/2 & -3 & 0\end{bmatrix}=\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}

Therefore, A =P + Q

(iv) Given: A = \begin{bmatrix}1 & 5 \\-1& 2 \\\end{bmatrix}

⇒ A’= \begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}

P =\frac{1}{2} \{\begin{bmatrix}1 & 5 \\-1 & 2 \\\end{bmatrix}+\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}2 & 4 \\4 & 4 \\\end{bmatrix}=\begin{bmatrix}1 & 2 \\2 & 2 \\\end{bmatrix}

…..(1)

Q = \frac{1}{2}(A-A')=\frac{1}{2} \{\begin{bmatrix}1 & 5 \\-1 & 2 \\\end{bmatrix}-\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}=\begin{bmatrix}0 & 3 \\-3 & 0 \\\end{bmatrix}

…..(2)

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

\begin{bmatrix}1 & 2 \\2 & 2 \\\end{bmatrix}+\begin{bmatrix}0 & 3 \\-3 & 0 \\\end{bmatrix}=\begin{bmatrix}1 & 5 \\-1 & 2 \\\end{bmatrix}

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 =\begin{bmatrix}cosα & -sinα \\sinα & cosα \\\end{bmatrix}   , and A + A′ = I, then the value of α is

(A)π/6    (B) π/3

(C) π    (D)3π/2

Solution:

\begin{bmatrix}cosα & -sinα \\sinα & cosα \\\end{bmatrix}+\begin{bmatrix}cosα & sinα \\-sinα & cosα \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}

\begin{bmatrix}2cosα & 0 \\0 & 2cosα \\\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1 \\\end{bmatrix}

On comparing both sides, we get

           2cosα = 1

⇒      cosα = \frac{1}{2}

⇒      cosα = cos\frac{π}{3}

⇒      α = \frac{π}{3}

∴ The option (B) is correct.

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