If \(A=\begin{pmatrix}2&2\\9&4\end{pmatrix}\) and \(I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\), then \(10A^{-1}\) is equal to :
Show Hint
- \(6I-A\)
- \(4I-A\)
- \(A-6I\)
- \(A-4I\)
The Correct Option is C
Solution and Explanation
\(\operatorname{adj}A=\begin{pmatrix}4&-2\\-9&2\end{pmatrix}\).
\(A^{-1}=\dfrac{1}{-10}\begin{pmatrix}4&-2\\-9&2\end{pmatrix}\), so \(10A^{-1}=\begin{pmatrix}-4&2\\9&-2\end{pmatrix}\).
Now \(A-6I=\begin{pmatrix}2-6&2\\9&4-6\end{pmatrix}=\begin{pmatrix}-4&2\\9&-2\end{pmatrix}\), which equals \(10A^{-1}\).
Answer: (3) \(A-6I\)
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