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Let A=[24−28]A = \begin{bmatrix} 2 & 4 \\ -2 & 8 \end{bmatrix}. If A−1=αI+βAA^{-1} = \alpha I + \beta A, α,β∈R\alpha,\beta \in R (Where R is set of real numbers) and I is a 2×22\times2 identity matrix, than the value of 4(α−2β)4(\alpha-2\beta) is equal to:

Solution

✅ Correct Option: 3

Trace is 1010 and ∣A∣=24|A| = 24, so by Cayley-Hamilton A2−10A+24I=0A^2 - 10A + 24I = 0. Multiplying by A−1A^{-1} gives A−1=1024I−124AA^{-1} = \frac{10}{24}I - \frac{1}{24}A, so α=512\alpha = \frac{5}{12} and β=−124\beta = -\frac{1}{24}. Then 4(α−2β)=4(512+112)=24(\alpha-2\beta) = 4\left(\frac{5}{12}+\frac{1}{12}\right) = 2.

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