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.
Answer & explanation
Correct answer: option 3
We have,
$A =\begin{bmatrix}0&2\\3&-4\end{bmatrix}⇒kA =\begin{bmatrix}0&2k\\3k&-4k\end{bmatrix}$
But, $kA =\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$
$∴\begin{bmatrix}0&2k\\3k&-4k\end{bmatrix}=\begin{bmatrix}0&3a\\2b&24\end{bmatrix}$
$⇒2k = 3a, 3k = 2b, - 4k = 24 ⇒k=-6, a=-4, b = -9$