If \(A = \begin{bmatrix}3&2\\4&-3\end{bmatrix}\) and \(A^{-1} = kA\), then the value of k is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{17}\)
Squaring A by row-column multiplication:
\( \displaystyle A^2 = \begin{bmatrix}9 + 8&6 - 6\\12 - 12&8 + 9\end{bmatrix} = \begin{bmatrix}17&0\\0&17\end{bmatrix} = 17I \)
So
\( \displaystyle A\left(\frac{1}{17}A\right) = \frac{1}{17}A^2 = I \)
and in the same way \(\left(\displaystyle\frac{1}{17}A\right)A = I\). By the uniqueness of the inverse,
\( \displaystyle A^{-1} = \frac{1}{17}A \), so \( k = \displaystyle\frac{1}{17} \)