If $\mathrm{A}=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$ and $\mathrm{A}^{-1}=\mathrm{KA}$,…
If $\mathrm{A}=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$ and $\mathrm{A}^{-1}=\mathrm{KA}$, then $\mathrm{K}$ is
- 19
- $\frac{-1}{19}$
- $-19$
- $\frac{1}{19}$
Solution
$\begin{aligned} & A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right] \\ & |A|-4-15=-19 \text { and adj } A=\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right] \\ & \therefore A^{-1}=\frac{-1}{19}\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]=\frac{1}{19}\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right] \Rightarrow \mathrm{k}=\frac{1}{19}\end{aligned}$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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