If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$ be such that $A^{-1}=k A$, then find the value of $k$.
$A=\left[\begin{array}{ll}2 & 3\end{array}\right.$
$5-2]$
$\therefore|A|=\mid 2 \quad 3$
$5-2 \mid=-14-15=-19$
The value is non-zero, so $A^{-1}$ exists.
By definition, we have
$A^{-1} A=I$ $[I$ is the identity matrix $]$
$\Rightarrow k A \cdot A=I$ [Substituting $\left.A^{-1}=k A\right]$
$\Rightarrow k\left[\begin{array}{ll}2 & 3\end{array}\right.$
$5-2]\left[\begin{array}{ll}2 & 3\end{array}\right.$
$5-2]=\left[\begin{array}{ll}1 & 0\end{array}\right.$
$\left.\begin{array}{ll}0 & 1\end{array}\right]$
$\Rightarrow k[4+15 \quad 6-6$
$10-10 \quad 15+4]=\left[\begin{array}{ll}1 & 0\end{array}\right.$
$\left.\begin{array}{ll}0 & 1\end{array}\right]$
$\Rightarrow k\left[\begin{array}{ll}19 & 0\end{array}\right.$
$0 \quad 19]=\left[\begin{array}{ll}1 & 0\end{array}\right.$
$\left.\begin{array}{ll}0 & 1\end{array}\right]$
$\Rightarrow k=\frac{1}{19}$
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