If $A = \left[\begin{array}{cc}0& 2\\ 3& -4\end{array}\right]$ and $k A = \left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right]$ then find the value of $b-a -k$

We have, $A = \left[\begin{array}{cc}0& 2\\ 3& -4\end{array}\right] \Rightarrow k A = \left[\begin{array}{cc}0& 2k\\ 3k& -4k\end{array}\right]$

But $$\begin{gathered} k A = \left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right] \\ \therefore \left[\begin{array}{cc}0& 2k\\ 3k& -4k\end{array}\right] = \left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right] \\ \Rightarrow 2k = 3a, 3k = 2b, -4k = 24 \\ \Rightarrow k = -6, a= -4, b = -9 \end{gathered}$$

Hence, $b-a-k =-9-(-4) -(-6) = -9 +4 +6 = 1$