Question:

Let \(A=\begin{bmatrix}1&2\\-1&4\end{bmatrix}\). If \(A^{-1}=\alpha I+\beta A,\ \alpha,\beta\in\mathbb{R}\), \(I\) is a \(2\times2\) identity matrix then \(4(\alpha-\beta)\) is equal to :

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Use the Cayley-Hamilton theorem (A^2 - 5A + 6I = 0) to express A^{-1} as a combination of I and A, or compute A^{-1} directly and compare entries.
Updated On: Nov 4, 2026
  • \(5\)
  • \(\frac{8}{3}\)
  • \(2\)
  • \(4\)
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The Correct Option is D

Solution and Explanation

For \(A=\begin{bmatrix}1&2\\-1&4\end{bmatrix}\): trace \(=5\), \(|A|=4+2=6\).
Cayley-Hamilton: \(A^2-5A+6I=O\).
Multiply by \(A^{-1}\): \(A-5I+6A^{-1}=O\Rightarrow A^{-1}=\dfrac{5I-A}{6}=\dfrac{5}{6}I-\dfrac{1}{6}A\).
So \(\alpha=\dfrac56,\ \beta=-\dfrac16\), and \(\alpha-\beta=1\).
\(4(\alpha-\beta)=4\).
Answer: (4) \(4\)
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