Q47 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
If $\mathrm{A}=\left[\begin{array}{cc}0 & 2 \\ 3 & -4\end{array}\right]$ and $k \mathrm{~A}=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$, then the values of $k, a$ and $b$ respectively are:
- A$-6,-12,-18$
- Correct answerB$-6,-4,-9$
- C$-6,4,9$
- D$-6,12,18$
Solution
$k A=\left[\begin{array}{cc}0 & 2 k \\ 3 k & -4 k\end{array}\right]=\left[\begin{array}{cc}0 & 3 a \\ 2 b & 24\end{array}\right]$
$\Rightarrow \quad k=-6, a=-4$ and $b=-9$
$\Rightarrow \quad k=-6, a=-4$ and $b=-9$
Hence, the correct answer is option: 'B'.