Let \( \mathrm{I}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) and \( \mathrm{A}=\begin{bmatrix}1 & -2\\2 & 4\end{bmatrix} \) be two matrices such that \( 8\mathrm{A}^{-1}=k\mathrm{I}-\mathrm{A},\ k\in\mathrm{R} \), then the value of \( k \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
Since \( 8A^{-1} \) and kI − A are equal matrices, their corresponding entries are equal.
\( \displaystyle |A|=(1)(4)-(-2)(2) \)
\( \displaystyle =8 \)
\( \displaystyle A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A \)
\( \displaystyle =\frac{1}{8}\begin{bmatrix}4 & 2\\-2 & 1\end{bmatrix} \)
\( \displaystyle 8A^{-1}=\begin{bmatrix}4 & 2\\-2 & 1\end{bmatrix} \)
\( \displaystyle kI-A=\begin{bmatrix}k-1 & 2\\-2 & k-4\end{bmatrix} \)
Equating the diagonal entries:
\( k-1=4\ \Rightarrow\ k=5 \)
\( k-4=1\ \Rightarrow\ k=5 \)
The off-diagonal entries (2 and −2) already agree, so k = 5.
Shortcut: By the Cayley-Hamilton theorem, a 2 × 2 matrix satisfies \( A^{2}-(\mathrm{tr}\,A)A+|A|I=O \). Here tr A = 1 + 4 = 5 and |A| = 8:
\( A^{2}-5A+8I=O \)
Multiplying by \( A^{-1} \):
\( A-5I+8A^{-1}=O \)
\( 8A^{-1}=5I-A \)
So k = 5 = tr A.