Let A=\left(\begin{array}{cc}1 & 2 \ -2 & -5\end{array} ight). Let \alpha, \beta \in \mathbb{R} be such that \alpha A^{2}+\beta A=2 I. Then \alpha+\beta is equal to

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Let A=\left(\begin{array}{cc}1 & 2 \ -2 & -5\end{array}ight). Let \alpha, \beta \in \mathbb{R} be such that \alpha A^{2}+\beta A=2 I. Then \alpha+\beta is equal to

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Answer: D. 10 {A^2} = \left[ {\begin{matrix} 1 & 2 \ { - 2} & { - 5} \ \end{matrix} } ight]\left[ {\begin{matrix} 1 & 2 \ { - 2} & { - 5} \ \end{matrix} } ight] = \left[ {\begin{matrix} { - 3} & { - 8} \ 8 & {21} \ \end{matrix} } ight] \alpha {A^2} + \beta A = \left[ {\begin{matrix} { - 3\alpha } & { - 8\alpha } \ {8\alpha } & {21\alpha } \ \end{matrix} } ight] + \left[ {\begin{matrix} \beta & {2\beta } \ { - 2\beta } & { - 5\beta } \ \end{matrix} } ight] = \left[ {\begin{matrix} { - 3\alpha + \beta } & { - 8\alpha + 2\beta } \ {8\alpha - 2\beta } & {21\alpha - 5\beta } \ \end{matrix} } ight] = \left[ {\begin{matrix} 2 & 0 \ 0 & 2 \ \end{matrix} } ight] On Comparing 8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2 \Rightarrow \alpha = 2,\,\beta = 8 So, \alpha + \beta = 10

Let $$A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right)$$. Let $$\alpha, \beta \in \mathbb{R}$$ be such that $$\alpha A^{2}+\beta A=2 I$$. Then $$\alpha+\beta$$ is equal to

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