Let A = begin{bmatrix} 1 & 2 -2 & -5 end{bma | vedclass

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(D) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. $\begin{vmatrix} 1 - \lambda & 2 \ -2 & -5 - \lambda \end{vmatrix} =

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$10$

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(D) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$. $\begin{vmatrix} 1 - \lambda & 2 \ -2 & -5 - \lambda \end{vmatrix} = 0$ $(1 - \lambda)(-5 - \lambda) - (2)(-2) = 0$ $-5 - \lambda + 5\lambda + \lambda^{2} + 4 = 0$ $\lambda^{2} + 4\lambda - 1 = 0$ By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation,so $A^{2} + 4A - I = 0$,which implies $A^{2} + 4A = I$. Multiplying by $2$,we get $2A^{2} + 8A = 2I$. We are given $\alpha A^{2} + \beta A = 2I$. Comparing the coefficients,we get $\alpha = 2$ and $\beta = 8$. Therefore,$\alpha + \beta = 2 + 8 = 10$.

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

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