Question

Find x, y satisfying the matrix equations

(i) $\left[\begin{array}{ccc}x-y& 2& -2\\ 4& x& 6\end{array}\right]+\left[\begin{array}{ccc}3& -2& 2\\ 1& 0& -1\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2x+y& 5\end{array}\right]$

(ii) $\left[\begin{array}{ccc}x& y+2& z-3\end{array}\right]+\left[\begin{array}{ccc}y& 4& 5\end{array}\right]=\left[\begin{array}{ccc}4& 9& 12\end{array}\right]$

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Given}:\hspace{0.17em}\left[\begin{array}{ccc}x-y& 2& -2\\ 4& x& 6\end{array}\right]+\left[\begin{array}{ccc}3& -2& 2\\ 1& 0& -1\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2x+y& 5\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}x-y+3& 2-2& -2+2\\ 4+1& x+0& 6-1\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2x+y& 5\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}x-y+3& 0& 0\\ 5& x& 5\end{array}\right]=\left[\begin{array}{ccc}6& 0& 0\\ 5& 2x+y& 5\end{array}\right] \\ \Rightarrow x-y+3=6 \\ \Rightarrow x-y=6-3 \\ \Rightarrow x-y=3...\left(1\right) \\ \mathrm{Also}, \\ x=2x+y \\ \Rightarrow -x=y...\left(2\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ x-\left(-x\right)=3 \\ \Rightarrow 2x=3 \\ \Rightarrow x=\frac{3}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{eq}.\left(2\right),\mathrm{we}\mathrm{get} \\ -\left(\frac{3}{2}\right)=y \\ \Rightarrow y=-\frac{3}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ \left[\begin{array}{ccc}x& y+2& z-3\end{array}\right]+\left[\begin{array}{ccc}y& 4& 5\end{array}\right]=\left[\begin{array}{ccc}4& 9& 12\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}x+y& y+2+4& z-3+5\end{array}\right]=\left[\begin{array}{ccc}4& 9& 12\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}x+y& y+6& z+2\end{array}\right]=\left[\begin{array}{ccc}4& 9& 12\end{array}\right] \\ \therefore x+y=4...\left(1\right) \\ \mathrm{Also}, \\ y+6=9 \\ \Rightarrow y=3 \\ z+2=12 \\ \Rightarrow z=10 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ x+3=4 \\ \Rightarrow x=4-3 \\ \Rightarrow x=1 \\ \therefore x=1,y=3\mathrm{and}z=10 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{Given}:\hspace{0.17em}x\left[\begin{array}{c}2\\ 1\end{array}\right]+y\left[\begin{array}{c}3\\ 5\end{array}\right]+\left[\begin{array}{c}-8\\ -11\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}2x+3y-8\\ x+5y-11\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \Rightarrow 2x+3y-8=0 \\ \Rightarrow 2x+3y=8...\left(1\right) \\ \mathrm{Also}, \\ x+5y-11=0 \\ \Rightarrow x+5y=11 \\ \Rightarrow x=11-5y...\left(2\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ 2\left(11-5y\right)+3y=8 \\ \Rightarrow 22-10y+3y=8 \\ \Rightarrow -7y=8-22 \\ \Rightarrow -7y=-14 \\ \Rightarrow y=2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ x=11-5\left(2\right) \\ \Rightarrow x=11-10 \\ \Rightarrow x=1 \\ \therefore x=1\mathrm{and}y=2 \end{gathered}$$