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

First compute

$$A^2=\begin{pmatrix}1&2\\-2&-5\end{pmatrix}\begin{pmatrix}1&2\\-2&-5\end{pmatrix}$$

$$=\begin{pmatrix}-3&-8\\8&21\end{pmatrix}$$

Now,

$$\alpha A^2+\beta A=2I$$

gives

$$\alpha\begin{pmatrix}-3&-8\\8&21\end{pmatrix}+\beta\begin{pmatrix}1&2\\-2&-5\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}$$

Comparing the off-diagonal entries,

$$-8\alpha+2\beta=0$$

$$-4\alpha+\beta=0$$

$$\beta=4\alpha$$

Now compare the

$$(1,1)$$

entry:

$$-3\alpha+\beta=2$$

Substituting

$$\beta=4\alpha$$

gives

$$\alpha=2$$

Hence

$$\beta=8$$

Therefore,

$$\alpha+\beta=2+8$$

$$=10$$

Final Answer :

$$10$$