If $A=\left[\begin{array}{cc}1 & 2 \ -2 & -5\end{array} ight]$ and $\alpha A^2+\beta A=2 I$ for some $\alpha, \beta, \in R$ then $\alpha+\beta=$

Question

If $A=\left[\begin{array}{cc}1 & 2 \ -2 & -5\end{array}ight]$ and $\alpha A^2+\beta A=2 I$ for some $\alpha, \beta, \in R$ then $\alpha+\beta=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } A=\left[\begin{array}{cc} 1 & 2 \ -2 & -5 \end{array}ight] 	ext {, So, } A^2-(-5+1) A+(-5+4) I=0 \ \Rightarrow & A^2+4 A=I \Rightarrow 2 A^2+8 A=2 I \end{aligned}$ If we compare with $\alpha A^2+\beta A=2 I$, we get $\alpha=2, \beta=8$ $\Rightarrow \alpha+\beta=10$.

If $A=\left[\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right]$ and $\alpha A^2+\beta A=2 I$ for some…

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