Maths

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

A² = AA = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 1*1+0*0+2*2\hfill & \hfill 1*0+2*0+2*0\hfill & \hfill 1*2+0*1+2*3\hfill \\ \hfill 0*1+2*0+1*2\hfill & \hfill a*0+2*2+1*0\hfill & \hfill 0*2+2*1+1*3\hfill \\ \hfill 2*1+0*0+3*2\hfill & \hfill 2*0+0*2+3*0\hfill & \hfill 2*2+0*1+3*3\hfill \end{array}\right]$

$=\left[\begin{array}{l}1+402+6\\ 242+3\\ 2+604+9\end{array}\right]=\left[\begin{array}{ccc}\hfill 5\hfill & \hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 8\hfill & \hfill 0\hfill & \hfill 13\hfill \end{array}\right]$

A³ = A²A = $\left[\begin{array}{ccc}\hfill 5\hfill & \hfill 0\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 8\hfill & \hfill 0\hfill & \hfill 13\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 5*1+0*0+8*2\hfill & \hfill 5*0+0*2+8*0\hfill & \hfill 5*2+0*1+8*3\hfill \\ \hfill 2*1+4*0+5*2\hfill & \hfill 2*0+4*2+5*0\hfill & \hfill 2*2+4*1+5*3\hfill \\ \hfill 8*1+0*0+13*2\hfill & \hfill 8*0+0*2+0*0\hfill & \hfill 8*2+0*1+13*3\hfill \end{array}\right]$

$\left[\begin{array}{ccc}\hfill 5+16\hfill & \hfill 0\hfill & \hfill 10+24\hfill \\ \hfill 2+10\hfill & \hfill 8\hfill & \hfill 4+4+15\hfill \\ \hfill 8+26\hfill & \hfill 0\hfill & \hfill 16+39\hfill \end{array}\right]$ $=\left[\begin{array}{lll}21\hfill & 0\hfill & 34\hfill \\ 12\hfill & 8\hfill & 23\hfill \\ 34\hfill & 0\hfill & 55\hfill \end{array}\right]$

So, A³-6A² + 7A + 2I

$=\left[\begin{array}{lll}21\hfill & 0\hfill & 34\hfill \\ 12\hfill & 8\hfill & 23\hfill \\ 34\hfill & 0\hfill & 55\hfill \end{array}\right]-6\left[\begin{array}{lll}5\hfill & 0\hfill & 8\hfill \\ 2\hfill & 4\hfill & 5\hfill \\ 8\hfill & 0\hfill & 13\hfill \end{array}\right]+7\left[\begin{array}{lll}1\hfill & 0\hfill & 2\hfill \\ 0\hfill & 2\hfill & 1\hfill \\ 2\hfill & 0\hfill & 3\hfill \end{array}\right]+2\left[\begin{array}{lll}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]$

$=\left[\begin{array}{lll}21\hfill & 0\hfill & 34\hfill \\ 12\hfill & 8\hfill & 23\hfill \\ 34\hfill & 0\hfill & 55\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 30\hfill & \hfill 0\hfill & \hfill 48\hfill \\ \hfill 12\hfill & \hfill 24\hfill & \hfill 30\hfill \\ \hfill 48\hfill & \hfill 0\hfill & \hfill 78\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 7\hfill & \hfill 0\hfill & \hfill 14\hfill \\ \hfill 0\hfill & \hfill 14\hfill & \hfill 7\hfill \\ \hfill 14\hfill & \hfill 0\hfill & \hfill 21\hfill \end{array}\right]+\left[\begin{array}{lll}2\hfill & 0\hfill & 0\hfill \\ 0\hfill & 2\hfill & 0\hfill \\ 0\hfill & 0\hfill & 2\hfill \end{array}\right]$

$=\left[\begin{array}{lll}21-30+7+2\hfill & 0-0+0+0\hfill & 34-48+14+0\hfill \\ 12-12+0+0\hfill & 8-24+14+2\hfill & 23-30+7+0\hfill \\ 34-48+14+0\hfill & 0-0+0+0\hfill & 55-78+21+2\hfill \end{array}\right]$

$=\left[\begin{array}{lll}0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill \end{array}\right]=0$ (2000 matrix )

Hence proud

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

A² = AA² = $\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]$

= $\left[\begin{array}{l}2*2+0*2+1*12*0+0*1+1*(1)2*1+0*3+1*0\\ 2*2+1*2+3*12*0+1*1+3*(-1)2*1+1*3+3*0\\ 1*2+(-1)*2+0*11*0+(-1)*1+0*(-1)1*1+(-1)*3+0*0\end{array}\right]$

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

Kindly go through the solution

Given,  $f(x)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

So, F (y) $=\left[\begin{array}{ccc}\hfill cosy\hfill & \hfill -siny\hfill & \hfill 0\hfill \\ \hfill siny\hfill & \hfill cosy\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\therefore f(x)f(y)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

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

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

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

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

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

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

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

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

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

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

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

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

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

(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 = $\left[\begin{array}{ll}2*2\hfill & 2*3\hfill \\ 2*4\hfill & 2*0\hfill \end{array}\right]$

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

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

4x+ 6y - (4x+ 6y) = $\left[\begin{array}{l}6-6\\ -315\end{array}\right]-$ $\left[\begin{array}{l}46\\ 80\end{array}\right]$

$\Rightarrow$ 5x = $\left[\begin{array}{l}6-4-6-6\\ -3-815-0\end{array}\right]=$ $\left[\begin{array}{l}2-12\\ -1115\end{array}\right]$

$\Rightarrow x=\frac{1}{5}\left[\begin{array}{rr}\hfill 2& \hfill -12\\ \hfill -11& \hfill 15\end{array}\right]=\left[\begin{array}{rr}\hfill 2*\frac{1}{5}& \hfill -12*\frac{1}{5}\\ \hfill -11*\frac{1}{5}& \hfill 15*\frac{1}{5}\end{array}\right]=$ $\left[\begin{array}{l}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$-12$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$-11$}\!\left/ \!\raisebox{-1ex}{$53$}\right.\end{array}\right]$