Solve the following equations

It is given that, $A=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]$ and $k A=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$.

$\therefore k\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}0 & 2 k \\ 3 k & -4 k\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$

$\Rightarrow 3 a=2 k, 2 b=3 k$ and $-4 k=24$

Now,

$-4 k=24 \Rightarrow k=-6$

$\therefore 3 a=2 k$

$\Rightarrow 3 a=2 \times(-6)=-12$

$\Rightarrow a=-4$

Also,

$2 b=3 k$

$\Rightarrow 2 b=3 \times(-6)=-18$

$\Rightarrow b=-9$

Thus, $(k, a, b)=(-6,-4,-9)$

If $A=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]$ and $k A=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$, then $(k, a, b)=\underline{-(-6,-4,-9)}$.