Find x, y satisfying the matrix equations
(i) $\left[\begin{array}{crr}x-y & 2 & -2 \\ 4 & x & 6\end{array}\right]+\left[\begin{array}{rrr}3 & -2 & 2 \\ 1 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}6 & 0 & 0 \\ 5 & 2 x+y & 5\end{array}\right]$
(ii) $\left[\begin{array}{lll}x & y+2 & z-3\end{array}\right]+\left[\begin{array}{lll}y & 4 & 5\end{array}\right]=\left[\begin{array}{lll}4 & 9 & 12\end{array}\right]$
(iii) $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=0$
(i)
Given : $\left[\begin{array}{ccc}x-y & 2 & -2 \\ 4 & x & 6\end{array}\right]+\left[\begin{array}{ccc}3 & -2 & 2 \\ 1 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}6 & 0 & 0 \\ 5 & 2 x+y & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}x-y+3 & 2-2 & -2+2 \\ 4+1 & x+0 & 6-1\end{array}\right]=\left[\begin{array}{ccc}6 & 0 & 0 \\ 5 & 2 x+y & 5\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}x-y+3 & 0 & 0 \\ 5 & x & 5\end{array}\right]=\left[\begin{array}{ccc}6 & 0 & 0 \\ 5 & 2 x+y & 5\end{array}\right]$
$\Rightarrow x-y+3=6$
$\Rightarrow x-y=6-3$
$\Rightarrow x-y=3$ ...(1)
Also,
$x=2 x+y$
$\Rightarrow-x=y$ ...(2)
Putting the value of $y$ in eq. (1), we get
$x-(-x)=3$
$\Rightarrow 2 x=3$
$\Rightarrow x=\frac{3}{2}$
Putting the value of $x$ in eq. (2), we get
$-\left(\frac{3}{2}\right)=y$
$\Rightarrow y=-\frac{3}{2}$
(ii)
$\left[\begin{array}{lll}x & y+2 & z-3\end{array}\right]+\left[\begin{array}{lll}y & 4 & 5\end{array}\right]=\left[\begin{array}{lll}4 & 9 & 12\end{array}\right]$
$\Rightarrow\left[\begin{array}{lll}x+y & y+2+4 & z-3+5\end{array}\right]=\left[\begin{array}{lll}4 & 9 & 12\end{array}\right]$
$\Rightarrow\left[\begin{array}{lll}x+y+6 & z+2\end{array}\right]=\left[\begin{array}{lll}4 & 9 & 12\end{array}\right]$
$\therefore x+y=4$ $\ldots(1)$
Also,
$y+6=9$
$\Rightarrow y=3$
$z+2=12$
$\Rightarrow z=10$
Putting the value of $y$ in eq. (1), we get
$x+3=4$
$\Rightarrow x=4-3$
$\Rightarrow x=1$
$\therefore x=1, y=3$ and $z=10$
(iii)
Given : $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}2 x+3 y-8 \\ x+5 y-11\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]$
$\Rightarrow 2 x+3 y-8=0$
$\Rightarrow 2 x+3 y=8$ ...(1)
Also,
$x+5 y-11=0$
$\Rightarrow x+5 y=11$
$\Rightarrow x=11-5 y$ ...(2)
Putting the value of $x$ in $(1)$, we get
$2(11-5 y)+3 y=8$
$\Rightarrow 22-10 y+3 y=8$
$\Rightarrow-7 y=8-22$
$\Rightarrow-7 y=-14$
$\Rightarrow y=2$
Putting the value of $y$ in $(2)$, we get
$x=11-5(2)$
$\Rightarrow x=11-10$
$\Rightarrow x=1$
$\therefore x=1$ and $y=2$
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