Solve the following Question

Since the corresponding elements of two equal matrices are equal,

$\left[\begin{array}{ccc}3 x+4 y & 2 & x-2 y \\ a+b & 2 a-b & -1\end{array}\right]=\left[\begin{array}{ccc}2 & 2 & 4 \\ 5 & -5 & -1\end{array}\right]$

$\Rightarrow 3 x+4 y=2$             ....(1)

$\Rightarrow x-2 y=4$

$\Rightarrow x=4+2 y$              .....(2)

Putting the value of $x$ in eq. $(1)$, we get

$3(4+2 y)+4 y=2$

$\Rightarrow 12+6 y+4 y=2$

$\Rightarrow 12+10 y=2$

$\Rightarrow 10 y=2-12$

$\Rightarrow 10 y=-10$

$\Rightarrow y=\frac{-10}{10}=-1$

Putting the value of $y$ in eq. (2), we get

$x=4+2(-1)$

$\Rightarrow x=4-2=2$

$a+b=5$

$\Rightarrow a=5-b$       .....(3)

$\Rightarrow 2 a-b=-5$   .....(4)

Putting the value $a$ in eq. (4), we get

$2(5-b)-b=-5$

$\Rightarrow 10-2 b-b=-5$

$\Rightarrow 10-3 b=-5$

$\Rightarrow-3 b=-15$

$\Rightarrow b=\frac{-15}{-3}$

$\Rightarrow b=5$

Putting the value of $b$ in eq. (3), we get

$a=5-5$

$\Rightarrow a=0$

$\therefore x=2, y=-1, a=0$ and $b=5$