JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 8)
Let $$\alpha$$ and $$\beta$$ be real numbers. Consider a 3 $$\times$$ 3 matrix A such that $$A^2=3A+\alpha I$$. If $$A^4=21A+\beta I$$, then
$$\alpha=1$$
$$\alpha=4$$
$$\beta=8$$
$$\beta=-8$$
Explanation
$\mathrm{A}^{2}=3 \mathrm{~A}+\alpha \mathrm{I}$
$A^{3}=3 A^{2}+\alpha A$
$\mathrm{A}^{3}=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A}$
$\mathrm{A}^{3}=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I}$
$\mathrm{A}^{4}=(9+\alpha) \mathrm{A}^{2}+3 \alpha \mathrm{A}$
$=(9+\alpha)(3 \mathrm{~A}+\alpha \mathrm{I})+3 \alpha \mathrm{A}$
$=\mathrm{A}(27+6 \alpha)+\alpha(9+\alpha)$
$\Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1$
$\Rightarrow \beta=\alpha(9+\alpha)=-8$
$A^{3}=3 A^{2}+\alpha A$
$\mathrm{A}^{3}=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A}$
$\mathrm{A}^{3}=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I}$
$\mathrm{A}^{4}=(9+\alpha) \mathrm{A}^{2}+3 \alpha \mathrm{A}$
$=(9+\alpha)(3 \mathrm{~A}+\alpha \mathrm{I})+3 \alpha \mathrm{A}$
$=\mathrm{A}(27+6 \alpha)+\alpha(9+\alpha)$
$\Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1$
$\Rightarrow \beta=\alpha(9+\alpha)=-8$
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