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# If $A=\begin{bmatrix}0&2\\3&-4\end{bmatrix}$ and $kA=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$ then the values of $k,a,b$ are respectively

$\begin{array}{1 1}(A)\;-6,-12,-18&(B)\;-6,4,9\\(C)\;-6,-4,-9&(D)\;-6,12,18\end{array}$

Given $kA=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$
$\Rightarrow k\begin{bmatrix}0&2\\3&-4\end{bmatrix}=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$
$\Rightarrow 2k=3a$
$\Rightarrow 3k=2b$
$\Rightarrow -4k=24$
$2k=3a$
$a=\large\frac{2k}{3}$
$3k=2b$
$b=\large\frac{3k}{2}$
$-4k=24$
$k=-6$
$a=\large\frac{2(-6)}{3}$$=-4 b=\large\frac{3(-6)}{2}$$=-9$
Hence (C) is the correct answer.