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Let A be a 2 × 2 matrix with real entries such that A' = \alpha A + I, where a \epsilon \mathbb{R} – {–1, 1}. If det (A2 – A) = 4,
then the sum of all possible values of \alpha is equal to :

Option: 1

0


Option: 2

\frac{5}{2}


Option: 3

3


Option: 4

\frac{3}{2}


Answers (1)

best_answer

\begin{aligned} & \mathrm{A}^{\mathrm{T}}=\alpha \mathrm{A}+\mathrm{I} \\ & \mathrm{A}=\alpha \mathrm{A}^{\mathrm{T}}+\mathrm{I} \\ & \mathrm{A}=\alpha(\alpha \mathrm{A}+\mathrm{I})+\mathrm{I} \\ & \mathrm{A}=\alpha^2 \mathrm{~A}+(\alpha+1) \mathrm{I} \\ & \mathrm{A}\left(1-\alpha^2\right)=(\alpha+1) \mathrm{I} \\ & \mathrm{A}=\frac{\mathrm{I}}{1-\alpha} \ldots(1) \\ & |\mathrm{A}|=\frac{1}{(1-\alpha)^2} \ldots(2) \\ & \left|\mathrm{A}^2-\mathrm{A}\right|=|\mathrm{A}||\mathrm{A}-\mathrm{I}| \ldots(3) \\ & \mathrm{A}-\mathrm{I}=\frac{\mathrm{I}}{\mathrm{I}-\alpha}-\mathrm{I}=\frac{\alpha}{1-\alpha} \mathrm{I} \\ & |\mathrm{A}-\mathrm{I}|=\left(\frac{\alpha}{1-\alpha}\right)^2 \ldots(4) \\ & \text { Now }\left|\mathrm{A}^2-\mathrm{A}\right|=4 \\ & |\mathrm{~A}||\mathrm{A}-\mathrm{I}|=4 \end{aligned}

\begin{aligned} & \Rightarrow \frac{1}{(1-\alpha)^2} \frac{\alpha^2}{\left(1-\alpha^2\right)}=4 \\ & \Rightarrow \frac{\alpha}{(1-\alpha)^2}= \pm 2 \\ & \Rightarrow 2(1-\alpha)^2= \pm \alpha \\ & \left(C_1\right) 2(1-\alpha)^2=\alpha \\ & \left(C_2\right) 2(1-\alpha)^3=-\alpha \\ & 2 \alpha^2-5 \alpha+2=0 \alpha_1 \\ & 2 \alpha^2-3 \alpha+2=0 \\ & \alpha_1+\alpha_2=\frac{5}{2} \\ & \alpha \notin \mathrm{R} \\ & \end{aligned}

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