Let \( A=\begin{bmatrix}2 & 4\\-2 & 8\end{bmatrix} \). If \( \mathrm{A}^{-1}=\alpha\mathrm{I}+\beta\mathrm{A} \), \( \alpha,\beta\in R \) (Where \( R \) is set of real numbers) and I is a \( 2\times 2 \) identity matrix, then the value of \( 4(\alpha-2\beta) \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
For a 2 × 2 matrix, the Cayley-Hamilton theorem gives \( A^2-(\mathrm{tr}\,A)A+|A|\,\mathrm{I}=\mathrm{O} \).
\( \displaystyle \mathrm{tr}\,A=2+8=10 \)
\( \displaystyle |A|=16-(-8)=24 \)
\( \displaystyle A^2-10A+24\mathrm{I}=\mathrm{O} \)
Multiplying by \( A^{-1} \):
\( \displaystyle A-10\mathrm{I}+24A^{-1}=\mathrm{O} \)
\( \displaystyle A^{-1}=\frac{10}{24}\mathrm{I}-\frac{1}{24}A \)
\( \displaystyle =\frac{5}{12}\mathrm{I}-\frac{1}{24}A \)
So \( \alpha=\frac{5}{12} \) and \( \beta=-\frac{1}{24} \).
\( \displaystyle 4(\alpha-2\beta)=4\left(\frac{5}{12}+\frac{2}{24}\right) \)
\( \displaystyle =4\left(\frac{5}{12}+\frac{1}{12}\right) \)
\( \displaystyle =4\times\frac{1}{2}=2 \)
Tip: For every 2 × 2 invertible matrix, \( A^{-1}=\frac{1}{|A|}\left[(\mathrm{tr}\,A)\mathrm{I}-A\right] \); α and β can be read off at once.