Ncert Solutions Maths class 12th

Given A and B are symmetric matrices,

(E) Then, A’ = A and B’ = B.

Now, (AB - BA)’ = (AB)’-  (BA)’

= B’A’ - A’B’.

= BA - AB

(AB - BA)’ = - (AB - BA)

AB - BA is a skew symmetric matrix

∴ Option A is correct.

(i) Let A = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right].$

Then, A’ = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -1\hfill \end{array}\right].$

Let P = $\frac{1}{2}$ (A + A’) = $\frac{1}{2}$ $\left\{\left[\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -1\hfill \end{array}\right]\right\}=\frac{1}{2}\left[\begin{array}{cc}\hfill 3+3\hfill & \hfill 5+1\hfill \\ \hfill 1+5\hfill & \hfill -1+\hfill \end{array}\right]$

= $\frac{1}{2}\left[\begin{array}{cc}\hfill 6\hfill & \hfill 6\hfill \\ \hfill 6\hfill & \hfill -2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -1\hfill \end{array}\right].$

Then, P’ = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -1\hfill \end{array}\right]$ = P.

∴ P = $\frac{1}{2}$ (A + A’) is symmetric matrix

Let Q = $\frac{1}{2}$ (A + A’) = $\frac{1}{2}\left\{\left[\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -1\hfill \end{array}\right]\right\}=\frac{1}{2}\left[\begin{array}{cc}\hfill 3-3\hfill & \hfill 5-1\hfill \\ \hfill 1-5\hfill & \hfill -1-(-1)\hfill \end{array}\right]$

= $\frac{1}{2}\left[\begin{array}{cc}\hfill 0\hfill & \hfill 4\hfill \\ \hfill -4\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right].$

Then Q.’ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$ = (-1) $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right]$ = (-1) Q.

$\Rightarrow$ Q.’ = Q,

∴ Q = $\frac{1}{2}$ (A - A’) is a symmetric matrix

Now, P + Q

Given, A = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill -a\hfill & \hfill 0\hfill & \hfill c\hfill \\ \hfill -b\hfill & \hfill -c\hfill & \hfill 0\hfill \end{array}\right]$

Then, A’ = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill -b\hfill \\ \hfill a\hfill & \hfill 0\hfill & \hfill -c\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill 0\hfill \end{array}\right]$

So, A + A’ = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill -a\hfill & \hfill 0\hfill & \hfill c\hfill \\ \hfill -b\hfill & \hfill -c\hfill & \hfill 0\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill -b\hfill \\ \hfill a\hfill & \hfill 0\hfill & \hfill -c\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a-a\hfill & \hfill b-b\hfill \\ \hfill -a+a\hfill & \hfill 0\hfill & \hfill c-c\hfill \\ \hfill -b+b\hfill & \hfill -c+c\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore \frac{1}{2}$ (A + A’) = $\frac{1}{2}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right].$

And A - A’ = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill -a\hfill & \hfill 0\hfill & \hfill c\hfill \\ \hfill -b\hfill & \hfill -c\hfill & \hfill 0\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill -b\hfill \\ \hfill a\hfill & \hfill 0\hfill & \hfill -c\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a-(-a)\hfill & \hfill b-(-b)\hfill \\ \hfill -a-a\hfill & \hfill 0\hfill & \hfill c-(-c)\hfill \\ \hfill -b-b\hfill & \hfill -c-c\hfill & \hfill 0\hfill \end{array}\right]$

= $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2a\hfill & \hfill 2b\hfill \\ \hfill -2a\hfill & \hfill 0\hfill & \hfill 2c\hfill \\ \hfill -2b\hfill & \hfill -2c\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore \frac{1}{2}$ (A - A’) = $\frac{1}{2}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2a\hfill & \hfill 2b\hfill \\ \hfill -2a\hfill & \hfill 0a\hfill & \hfill 2c\hfill \\ \hfill -2b\hfill & \hfill -2c\hfill & \hfill 0\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill -a\hfill & \hfill 0\hfill & \hfill c\hfill \\ \hfill -b\hfill & \hfill -c\hfill & \hfill 0\hfill \end{array}\right].$

Given, A = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 5\hfill \\ \hfill 6\hfill & \hfill 7\hfill \end{array}\right]$

Then, A’ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 6\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]$

Let P = A + A’ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 5\hfill \\ \hfill 6\hfill & \hfill 7\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 1\hfill & \hfill 6\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1+1\hfill & \hfill 5+6\hfill \\ \hfill 6+5\hfill & \hfill 7+7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 11\hfill \\ \hfill 11\hfill & \hfill 14\hfill \end{array}\right]$

So, P’ = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 11\hfill \\ \hfill 11\hfill & \hfill 14\hfill \end{array}\right]$ = P

i e, ( A + A’ )’ = A + A’.

Hence, A + A’ is symmetric matrix.

Let Q = A A’ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 5\hfill \\ \hfill 6\hfill & \hfill 7\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 1\hfill & \hfill 6\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1-1\hfill & \hfill 5-6\hfill \\ \hfill 6-5\hfill & \hfill 7-7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right].$

So,Q¹ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$ = (1) $\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$ = (1) Q.

$\Rightarrow$ Q¹ = Q.

i e, (A A’)’ = -(A - A’).

Have, A - A’ is a show symmetric matrix

(i) Given A = ${\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 5\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]}_{\cdot }$

Then, A’ = $\left[\begin{array}{ccc}\hfill 1\cdot \hfill & \hfill -\cdot 1\hfill & \hfill \cdot 5\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right]$

∴A’ = A.

Here, A is symmetric matrix

(ii) Given, A = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$

Then, A’ = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]=(-1)\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$

$\Rightarrow$ A’ = (1) A.

$\Rightarrow$ A’ = A.

Hers A is a show symmetric matrix.

(i) Given, A = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$

Then, A’ = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$

∴A’ A = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cot\alpha \hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill cos\alpha \cdot cos\alpha +(-sin\alpha )(-sin\alpha )\hfill & \hfill cos\alpha \cdot sin\alpha +(-sin\alpha )cos\alpha \hfill \\ \hfill sin\alpha \cdot cos\alpha +cos\alpha (-sin\alpha )\hfill & \hfill sin\alpha \cdot sin\alpha +cos\alpha cos\alpha \hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill co{s}^2\alpha +si{n}^2\alpha \hfill & \hfill cos\alpha sin\alpha -sin\alpha cos\alpha \hfill \\ \hfill sin\alpha cos\alpha -cos\alpha sin\alpha \hfill & \hfill si{n}^2\alpha +co{s}^2\alpha \hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ $\left\{?co{s}^{2}x+si{n}^{2}x=2\right\}$

= A ’ A = 1.

Given,

(ii) 1 A = $\left[\begin{array}{cc}\hfill sin\alpha \hfill & \hfill cos\alpha \hfill \\ \hfill -cot\alpha \hfill & \hfill sin\alpha \hfill \end{array}\right]$

Then, A’ = $\left[\begin{array}{cc}\hfill sin\alpha \hfill & \hfill -cos\alpha \hfill \\ \hfill cot\alpha \hfill & \hfill sin\alpha \hfill \end{array}\right]$

∴A’ A = $\left[\begin{array}{cc}\hfill sin\alpha \hfill & \hfill -cos\alpha \hfill \\ \hfill cot\alpha \hfill & \hfill sin\alpha \hfill \end{array}\right]\left[\begin{array}{cc}\hfill sin\alpha \hfill & \hfill cos\alpha \hfill \\ \hfill -cos\alpha \hfill & \hfill sin\alpha \hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill sin\cdot \alpha sin\alpha +(cos\alpha )(-cos\alpha )\hfill & \hfill sin\alpha \cdot cos\alpha +(-cos\alpha )cos\alpha \hfill \\ \hfill cos\alpha \cdot sin\alpha +sin\alpha (-cos\alpha )\hfill & \hfill cos\alpha \cdot cos\alpha +sin\alpha sin\alpha \hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill si{n}^2\alpha +co{s}^2\alpha \hfill & \hfill sin\alpha cot\alpha -cos\alpha sin\alpha \hfill \\ \hfill cos\alpha sin\alpha -sin\alpha cos\alpha \hfill & \hfill co{s}^2\alpha +si{n}^2\alpha \hfill \end{array}\right]$

= $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ $\left\{?co{s}^{2}x+cosco{s}^{2}x=1\right\}.$

$$ $\Rightarrow$ A’ A. = I

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

(i) Given, A = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill & \hfill 9\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ and B = $\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Then, A’ = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 1\hfill \end{array}\right]$ and B = $\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

A + B = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill & \hfill 9\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ + $\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill -1-4\hfill & \hfill 2+1\hfill & \hfill 3-5\hfill \\ \hfill 5+1\hfill & \hfill 7+2\hfill & \hfill 9+0\hfill \\ \hfill -2+1\hfill & \hfill 1+3\hfill & \hfill 1+1\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill -5\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill 6\hfill & \hfill 9\hfill & \hfill 9\hfill \\ \hfill -1\hfill & \hfill 4\hfill & \hfill 2\hfill \end{array}\right]$

LHS. = (A + B)’ = $\left[\begin{array}{ccc}\hfill -5\hfill & \hfill 6\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 9\hfill & \hfill 2\hfill \end{array}\right]$

RHS = A’+ B’ = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -1-4\hfill & \hfill 5+1\hfill & \hfill -2+1\hfill \\ \hfill 2+1\hfill & \hfill 7+2\hfill & \hfill 1+3\hfill \\ \hfill 3-5\hfill & \hfill 9+0\hfill & \hfill 1+1\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill -5\hfill & \hfill 6\hfill & \hfill -1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 4\hfill \\ \hfill -2\hfill & \hfill 9\hfill & \hfill 2\hfill \end{array}\right]$ = L H S

$\therefore$ (A + B)’ = A’+ B’

(ii) A – B = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill & \hfill 9\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill -1-\left(-4\right)\hfill & \hfill 2-1\hfill & \hfill 3-(5)\hfill \\ \hfill 5-1\hfill & \hfill 7-2\hfill & \hfill 9-0\hfill \\ \hfill -2-1\hfill & \hfill 1-3\hfill & \hfill 1-1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 1\hfill & \hfill 8\hfill \\ \hfill 4\hfill & \hfill 5\hfill & \hfill 9\hfill \\ \hfill -3\hfill & \hfill -2\hfill & \hfill 0\hfill \end{array}\right]$

L.H.S. = (A − B)’ = $\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 4\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 8\hfill & \hfill 9\hfill & \hfill 0\hfill \end{array}\right]$

R.H.S. = A’ − B’ = $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 9\hfill & \hfill 1\hfill \end{array}\right]- \left[\begin{array}{ccc}\hfill -4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=$ $\left[\begin{array}{ccc}\hfill -1-(-4)\hfill & \hfill 5-1\hfill & \hfill -2-1\hfill \\ \hfill 2-1\hfill & \hfill 7-2\hfill & \hfill 1-3\hfill \\ \hfill 3-(-5)\hfill & \hfill 9-0\hfill & \hfill 1-1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 4\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 5\hfill & \hfill -2\hfill \\ \hfill 8\hfill & \hfill 9\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore$ (A − B)’