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