Question

If $A=\left[\begin{array}{cc}0& 2\\ 3& -4\end{array}\right]$ and $kA=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right]$, then the values of k, a, b, are respectively
(a) −6, −12, −18
(b) −6, 4, 9
(c) −6, −4, −9
(d) −6, 12, 18

Answer 1

(c) −6, −4, −9


$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}0& 2\\ 3& -4\end{array}\right] \\ \mathrm{Here}, \\ kA=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right] \\ \Rightarrow k\left[\begin{array}{cc}0& 2\\ 3& -4\end{array}\right]=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}0& 2k\\ 3k& -4k\end{array}\right]=\left[\begin{array}{cc}0& 3a\\ 2b& 24\end{array}\right] \\ \mathrm{The}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{equal}\mathrm{matrices}\mathrm{are}\mathrm{equal}. \\ \Rightarrow 2k=3a,3k=2b\mathrm{and}-4k=24 \\ \mathrm{Now}, \\ -4k=24 \\ \Rightarrow k=-6 \\ \mathrm{Also}, \\ 2k=3a\mathrm{and}3\mathrm{k}=2b \\ \Rightarrow 2\left(-6\right)=3a\mathrm{and}3\left(-6\right)=2b\left[\mathrm{Using}k=-6\right] \\ \Rightarrow -12=3a\mathrm{and}-18=2b \\ \therefore a=-4\mathrm{and}b=-9 \end{gathered}$$