Skip to content
Related Articles
Get the best out of our app
GeeksforGeeks App
Open App
geeksforgeeks
Browser
Continue

Related Articles

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

Improve Article
Save Article
Like Article
author
kavyagupta0098
contributor
22 published articles
Improve Article
Save Article
Like Article

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’ = AT  = \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.


My Personal Notes arrow_drop_up
Last Updated : 25 Jan, 2021
Like Article
Save Article
Similar Reads
1. Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Miscellaneous Exercise on Chapter 3
2. Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2 | Set 1
3. Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 1
4. Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 2
5. Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.1
6. Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2 | Set 2
7. Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5
8. Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 1
9. Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 2
10. Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 2
Related Tutorials
1. Linear Algebra
2. Maths
3. Physics
4. Chemistry
5. CBSE Class 12 Syllabus 2023-2024
Previous
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5
Next
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.1
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
kavyagupta0098
kavyagupta0098
Vote for difficulty
Article Tags :
Report Issue
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox