Ncert Solutions Maths class 12th

(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 = $\frac{1}{2}$ (A + A') + $\frac{1}{2}$ (A - A')

$\Rightarrow$ P + Q = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -1\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3+0\hfill & \hfill 3+2\hfill \\ \hfill 3-2\hfill & \hfill -1+0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$ = A.

This A is represented as a sun of symmetric and skew symmetric matrix

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

Then A' = $\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right].$

Now, A + A' = $\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]$

= $\left[\begin{array}{ccc}\hfill 6+6\hfill & \hfill -2+(-2)\hfill & \hfill 2+2\hfill \\ \hfill -2+(-2)\hfill & \hfill 3+3\hfill & \hfill -1+(-1)\hfill \\ \hfill 2+2\hfill & \hfill -1+(-1)\hfill & \hfill 3+3\hfill \end{array}\right]$ = $\left[\begin{array}{ccc}\hfill 12\hfill & \hfill -4\hfill & \hfill 4\hfill \\ \hfill -4\hfill & \hfill 6\hfill & \hfill -2\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 6\hfill \end{array}\right]$

Let P = $\frac{1}{2}$ (A + A') = $\frac{1}{2}\left[\begin{array}{ccc}\hfill 12\hfill & \hfill -4\hfill & \hfill 4\hfill \\ \hfill -4\hfill & \hfill 6\hfill & \hfill -2\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 6\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]$

Then, P' = $\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]$ = P'

∴ P = $\frac{1}{2}$ (A + A') is asyntri matrix.

A - A' = $\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

Let Q = $\frac{1}{2}$ (A - A') = $\frac{1}{2}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sigma \hfill & \hfill 0\hfill \\ \hfill \sigma \hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right].$

Q' = (-1) $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$ = -Q.

$\Rightarrow$ Q' = -Q.

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

Have P + Q = $\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 6\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right]$ = A.

Then A is represented as a sum of symmetric & skew symmetric matrix

(iii) Let A = $\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -2\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill -5\hfill & \hfill 2\hfill \end{array}\right]$

Then, A' = $\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -4\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$

Let P = $\frac{1}{2}$ (A + A') = $\frac{1}{2}\left\{\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -2\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill -5\hfill & \hfill 2\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -4\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]\right\}$

= $\frac{1}{2}\left[\begin{array}{ccc}\hfill 3+3\hfill & \hfill 3-2\hfill & \hfill -1-4\hfill \\ \hfill -2+3\hfill & \hfill -2-2\hfill & \hfill 1-5\hfill \\ \hfill -4-1\hfill & \hfill -5+1\hfill & \hfill 2+2\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}\hfill 6\hfill & \hfill 1\hfill & \hfill -5\hfill \\ \hfill 1\hfill & \hfill -4\hfill & \hfill -4\hfill \\ \hfill -5\hfill & \hfill -4\hfill & \hfill 4\hfill \end{array}\right]$

= $\left[\begin{array}{ccc}\hfill 3\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill -2\hfill & \hfill -2\hfill \\ \hfill \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill -2\hfill & \hfill 2\hfill \end{array}\right]$

Then P' = $\left[\begin{array}{ccc}\hfill 3\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill -2\hfill & \hfill -2\hfill \\ \hfill \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill -2\hfill & \hfill 2\hfill \end{array}\right]$ = P.

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

Let Q = $\frac{1}{2}$ (A - A') = $\frac{1}{2}\left\{\left[\begin{array}{ccc}\hfill 3\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -2\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 45\hfill & \hfill 2\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -4\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]\right\}$

= $\frac{1}{2}\left[\begin{array}{ccc}\hfill 3-3\hfill & \hfill 3-(-2)\hfill & \hfill -1-(-4)\hfill \\ \hfill -2-3\hfill & \hfill -2-(-2)\hfill & \hfill 1-(-5)\hfill \\ \hfill -4-(-1)\hfill & \hfill -5-1\hfill & \hfill 2-2\hfill \end{array}\right].$

= $\frac{1}{2}\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 5\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill -3\hfill & \hfill -6\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right]$

Then Q' = $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 3\hfill & \hfill 0\hfill \end{array}\right]$ = (-1) $\left[\begin{array}{ccc}\hfill 0\hfill & \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 3\hfill & \hfill 0\hfill \end{array}\right]$ = (-1) Q.

$\Rightarrow$ Q' = Q.

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

∴ P + Q = symmetric

= .

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