JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 4)

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
$$-$$10
$$-$$6
6
10

Explanation

$${A^2} = \left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right]\left[ {\matrix{ 1 & 2 \cr { - 2} & { - 5} \cr } } \right] = \left[ {\matrix{ { - 3} & { - 8} \cr 8 & {21} \cr } } \right]$$

$$\alpha {A^2} + \beta A = \left[ {\matrix{ { - 3\alpha } & { - 8\alpha } \cr {8\alpha } & {21\alpha } \cr } } \right] + \left[ {\matrix{ \beta & {2\beta } \cr { - 2\beta } & { - 5\beta } \cr } } \right]$$

$$ = \left[ {\matrix{ { - 3\alpha + \beta } & { - 8\alpha + 2\beta } \cr {8\alpha - 2\beta } & {21\alpha - 5\beta } \cr } } \right] = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]$$

On Comparing

$$8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2$$

$$ \Rightarrow \alpha = 2,\,\beta = 8$$

So, $$\alpha + \beta = 10$$

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