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Question:

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 $k$.

Solution:

$A=\left[\begin{array}{ll}2 & 3\end{array}\right.$

$5-2]$

$\therefore|A|=\mid 2 \quad 3$

$5-2 \mid=-14-15=-19$

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 $\left.A^{-1}=k A\right]$

$\Rightarrow k\left[\begin{array}{ll}2 & 3\end{array}\right.$

$5-2]\left[\begin{array}{ll}2 & 3\end{array}\right.$

$5-2]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow k[4+15 \quad 6-6$

$10-10 \quad 15+4]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow k\left[\begin{array}{ll}19 & 0\end{array}\right.$

$0 \quad 19]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow k=\frac{1}{19}$