Ncert Solutions Maths class 12th

Given,  $x\left[\begin{array}{ll}x\hfill & y\hfill \\ z\hfill & w\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill x\hfill & \hfill 6\hfill \\ \hfill -1\hfill & \hfill 2w\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 4\hfill & \hfill x+y\hfill \\ \hfill z+w\hfill & \hfill 3\hfill \end{array}\right].$

$(B)\Rightarrow \left[\begin{array}{cc}\hfill 3x\hfill & \hfill 3y\hfill \\ \hfill 3z\hfill & \hfill 3w\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill x+4\hfill & \hfill 6+x+y\hfill \\ \hfill -1+2+w\hfill & \hfill 2w+3\hfill \end{array}\right]$

3x + y = 5 _____ (ii)

Adding (i) and (i) we get,

2x - y + 3x + y = 10 + 5.

5x = 15

$\Rightarrow x=\frac{15}{5}$ => x = 3

5x = 15

$\Rightarrow x=\frac{15}{5}$ => x = 3

Putting x = 3 in (i) we gel,

2 × 3 - y = 10

6 - 10 = y

y = -4

$\begin{array}{l}\\ 2\left[\begin{array}{ll}x\hfill & z\hfill \\ y\hfill & t\hfill \end{array}\right]+3\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]=3\left[\begin{array}{ll}3\hfill & 5\hfill \\ 4\hfill & 6\hfill \end{array}\right]\end{array}$

$\Rightarrow \left[\begin{array}{cc}\hfill 2\times x\hfill & \hfill 2\times 2\hfill \\ \hfill 2\times y\hfill & \hfill 2\times t\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 3\times 1\hfill & \hfill 3\times (-1)\hfill \\ \hfill 3\times 0\hfill & \hfill 3\times 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\times 3\hfill & \hfill 3\times 5\hfill \\ \hfill 3\times 4\hfill & \hfill 3\times 6\hfill \end{array}\right]$

$$ $\Rightarrow \left[\begin{array}{cc}\hfill 2x\hfill & \hfill 2z\hfill \\ \hfill 2y\hfill & \hfill 21\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 3\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 6\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 9\hfill & \hfill 15\hfill \\ \hfill 12\hfill & \hfill 18\hfill \end{array}\right].$

$\Rightarrow \left[\begin{array}{cc}\hfill 2x+3\hfill & \hfill 2z-3\hfill \\ \hfill 2y+0\hfill & \hfill 21+6\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 9\hfill & \hfill 15\hfill \\ \hfill 12\hfill & \hfill 18\hfill \end{array}\right]$

Equating the corresponding elements of the matrices $u$ get,

2x + 3 = 9

2x = 9 - 3

$\Rightarrow x=\frac{6}{2}$

x = 3

2z - 3 = 15

2z = 15 + 3

$\Rightarrow z=\frac{18}{2}$

z = 9

2y = 12

$\Rightarrow y=\frac{12}{2}$ => y = 6

2t + 6 = 18

2t = 18 - 6

2t = 12 => t = $\frac{12}{2}$ t = 6

Given, $2\left[\begin{array}{l}13\\ 0x\end{array}\right]+$ $\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$ $=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\hfill 2\times 1\hfill & \hfill 2\times 3\hfill \\ \hfill 2\times 0\hfill & \hfill 2\times x\hfill \end{array}\right]$ $+\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\hfill 2+y\hfill & \hfill 6+0\hfill \\ \hfill \hfill & \hfill 2x+2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

Equating the corresponding elements of the matrices we get,

2 + y = 5

y = 5 - 2 = 3

And 2x + 2 = 8

2x = 8 - 2

$\Rightarrow x=\frac{6}{2}$ $$

x = 3

∅x = 3, y - 3.

Give, 2x + 4 = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]$

$\Rightarrow 2x=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]-$ y $=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]-\left[\begin{array}{ll}3\hfill & 2\hfill \\ 1\hfill & 4\hfill \end{array}\right]$ $=\left[\begin{array}{l}1-30-2\\ -3-12-4\end{array}\right]$

$\Rightarrow x=\frac{1}{2}\left[\begin{array}{ll}-2\hfill & -2\hfill \\ -4\hfill & -2\hfill \end{array}\right]=\left[\begin{array}{ll}-2\times \frac{1}{2}\hfill & -2\times \frac{1}{2}\hfill \\ -4\times \frac{1}{2}\hfill & -2\times \frac{1}{2}\hfill \end{array}\right]=$ $\left[\begin{array}{l}-1-1\\ -2-1\end{array}\right]$

(i) x + y = $\left[\begin{array}{ll}7\hfill & 0\hfill \\ 2\hfill & 5\hfill \end{array}\right]$

x - y = $\left[\begin{array}{l}30\\ 03\end{array}\right]$

$\Rightarrow Y=\frac{1}{2}\left[\begin{array}{ll}4\hfill & 0\hfill \\ 2\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ll}\frac{1}{2}\times 4\hfill & \frac{1}{2}\times 0\hfill \\ \frac{1}{2}\times 2\hfill & \frac{1}{2}\times 2\hfill \end{array}\right]$ $=\left[\begin{array}{l}20\\ 11\end{array}\right]$

(ii) 2x + 3y = $\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]$ ___ (1)

2x + 2y = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill -1\hfill & \hfill 5\hfill \end{array}\right]$ _____ (2)

Multiplying  eqⁿ by 2 and $q^n(u)$

2 × (2x+ 3y) = 2 × $\left[\begin{array}{l}23\\ 40\end{array}\right]$

$\Rightarrow$ 4x + 6y = $\left[\begin{array}{ll}2\times 2\hfill & 2\times 3\hfill \\ 2\times 4\hfill & 2\times 0\hfill \end{array}\right]$

$\Rightarrow$ 4x + 6y = $\left[\begin{array}{ll}4\hfill & 6\hfill \\ 8\hfill & 0\hfill \end{array}\right]$ $\to (iii).$

3 × (3x+ 2y) = 3 × $\left[\begin{array}{l}2-2\\ -15\end{array}\right]$

$\Rightarrow$ 9x + 6y = $\left[\begin{array}{cc}\hfill 3\times (2)\hfill & \hfill 3\times (-2)\hfill \\ \hfill 3\times (-1)\hfill & \hfill 3\times 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 6\hfill & \hfill -6\hfill \\ \hfill -3\hfill & \hfill 15\hfill \end{array}\right]$ $\to (iv)$

Subtracting  eqⁿ  (iii) from (iv) we get,

4x+ 6y

$cos\theta \left[\begin{array}{cc}\hfill cos\theta \hfill & \hfill sin\theta \hfill \\ \hfill -sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right]+sin\theta \left[\begin{array}{cc}\hfill sin\theta \hfill & \hfill -cos\theta \hfill \\ \hfill cos\theta \hfill & \hfill sin\theta \hfill \end{array}\right]$

(E) $=\left[\begin{array}{l}co{r}^2\theta cot\theta sin\theta \\ -cos\theta sin\theta co{s}^2\theta \end{array}\right]$ $+\left[\begin{array}{l}sin2\theta -sin\theta cos\theta .\\ sin\theta cos\theta si{n}^2\theta \end{array}\right]$ $$

$=\left[\begin{array}{l}co{s}^2\theta +si{n}^2\theta cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta co{s}^2\theta +si{n}^2\theta \end{array}\right]$

$=\left[\begin{array}{l}10\\ 01\end{array}\right]$

3a - 5b = 3 × $\left[\begin{array}{ccc}\hfill \frac{2}{3}\hfill & \hfill 1\hfill & \hfill \frac{5}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill \frac{2}{3}\hfill & \hfill \frac{4}{3}\hfill \\ \hfill \frac{7}{3}\hfill & \hfill 2\hfill & \hfill \frac{2}{3}\hfill \end{array}\right]$ - $5\times \left[\begin{array}{ccc}\hfill \frac{2}{5}\hfill & \hfill \frac{3}{5}\hfill & \hfill 1\hfill \\ \hfill \frac{1}{5}\hfill & \hfill \frac{2}{5}\hfill & \hfill \frac{4}{5}\hfill \\ \hfill \frac{7}{5}\hfill & \hfill \frac{6}{5}\hfill & \hfill \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 3\times \frac{2}{3}\hfill & \hfill 3\times 1\hfill & \hfill 3\times \frac{5}{3}\hfill \\ \hfill 3\times \frac{1}{3}\hfill & \hfill 3\times \frac{2}{3}\hfill & \hfill 3\times \frac{4}{3}\hfill \\ \hfill 3\times \frac{7}{3}\hfill & \hfill 3\times 2\hfill & \hfill 3\times \frac{2}{3}\hfill \end{array}\right]-$ $\left[\begin{array}{ccc}\hfill 5\times \frac{2}{5}\hfill & \hfill 5\times \frac{3}{5}\hfill & \hfill 5\times 1\hfill \\ \hfill 5\times \frac{1}{5}\hfill & \hfill 5\times \frac{2}{5}\hfill & \hfill 5\times \frac{4}{5}\hfill \\ \hfill 5\times \frac{7}{5}\hfill & \hfill 5\times \frac{6}{5}\hfill & \hfill 5\times \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2-2\hfill & \hfill 3-3\hfill & \hfill 5-5\hfill \\ \hfill 1-1\hfill & \hfill 2-2\hfill & \hfill 4-4\hfill \\ \hfill 7-7\hfill & \hfill 6-6\hfill & \hfill 2-2\hfill \end{array}\right]$ = $\left[\begin{array}{l}000\\ 600\\ 000\end{array}\right]$

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

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

$B-C=\left[\begin{array}{lll}3\hfill & -1\hfill & 2\hfill \\ 4\hfill & 2\hfill & 5\hfill \\ 2\hfill & 0\hfill & 3\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$ $=\left[\begin{array}{l}3-4-1-12-2\\ 4-02-35-2\\ 2-10-(2)33\end{array}\right]$

$B-C=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

Now

L.H.S. = a+ (b - c) = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$ $=\left[\begin{array}{l}1-12-2-3+0\\ 5+40-12+3\\ 1+1-1+21+0\end{array}\right]$

$=\left[\begin{array}{l}00-3\\ 9-15\\ 211\end{array}\right]$

R.H.S. = (a+ b) - c = $\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 9\hfill & \hfill 2\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$ $=\left[\begin{array}{l}4-41-1-1-3\\ 9-02-37-2\\ 3-1-1-(2)4-3\end{array}\right]$

$=\left[\begin{array}{l}00-3\\ 9-15\\ 211\end{array}\right]$ = L.H.S.

∴A + (B - C) = (A + B) - C

Kindly go through the solution