# Find x, y satisfying the matrix equations

Question:

Find xy 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) $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=0$

Solution:

(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 $y=2$