Question

Find the values of x and y if
$\left[\begin{array}{cc}x+10& y^2+2y\\ 0& -4\end{array}\right]=\left[\begin{array}{cc}3x+4& 3\\ 0& y^2-5y\end{array}\right]$

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ x+10=3x+4\left[\because \mathrm{All}\mathrm{the}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{matrix}\mathrm{are}\mathrm{equal}\right] \\ \Rightarrow x-3x=4-10 \\ \Rightarrow -2x=-6 \\ \therefore x=3 \\ \mathrm{Also}, \\ {y}^2+2y=3 \\ \Rightarrow y^2+2y-3=0 \\ \Rightarrow y^2+3y-y-3=0 \\ \Rightarrow y\left(y+3\right)-1\left(y+3\right)=0 \\ \Rightarrow \left(y+3\right)\left(y-1\right)=0 \\ \Rightarrow y+3=0\mathrm{or}y-1=0 \\ \Rightarrow y=-3\mathrm{or}y=1 \\ \mathrm{Now}, \\ -4=y^2-5y \\ \Rightarrow y^2-5y+4=0 \\ \Rightarrow y^2-4y-y+4=0 \\ \Rightarrow y\left(y-4\right)-1\left(y-4\right)=0 \\ \Rightarrow \left(y-4\right)\left(y-1\right)=0 \\ \Rightarrow y-4=0\mathrm{or}y-1=0 \\ \Rightarrow y=4\mathrm{or}y=1 \end{gathered}$$

Since $y^2+2y=3\mathrm{and}{y}^2-5y=-4$ must hold good simultaneously, we take the common solution of these
two equations.

Thus,
y = 1, x = 3 and y = 1