Let [A] be a [2 2] matrix with real entries such that [A'= A+I], where [ R--1,1].

Question

Let $$\mathrm{A}$$ be a $$2 \times 2$$ matrix with real entries such that $$\mathrm{A}'=\alpha \mathrm{A}+\mathrm{I}$$, where $$\alpha \in \mathbb{R}-\{-1,1\}$$. If $$\operatorname{det}\left(A^{2}-A\right)=4$$, then the sum of all possible values of $$\alpha$$ is equal to :

Accepted Answer

We have, $A^T=\alpha A+I$, where $A$ is $2 \times 2$ matrix and $\alpha \in R-\{-1,1\}$

$$
\begin{aligned}
\left(A^T\right)^T & =\alpha A^T+I \\
A & =\alpha A^T+I \\
A & =\alpha(\alpha A+I)+I  \left[\because A^T=\alpha A+I\right]\\
A & =\alpha^2 A+(\alpha+1) I \\
A & \left(1-\alpha^2\right)=(\alpha+1) I \\
A & =\frac{(\alpha+1)}{(1-\alpha)(1+\alpha)} I
\end{aligned}
$$

$$
\begin{aligned}
A & =\frac{1}{1-\alpha} I ..........(i)\\
A| & =\frac{1}{(1-\alpha)^2} ...........(ii)
\end{aligned}
$$

Also,

$$
\begin{aligned}
\left|A^2-A\right| & =4 \\
A||A-I| & =4 \\
\frac{1}{(1-\alpha)^2}\left|\left(\frac{1}{1-\alpha}-1\right) I\right| & =4 \\
\frac{1}{(1-\alpha)^2}\left|\left(\frac{\alpha}{1-\alpha}\right) I\right| & =4 \\
\frac{1}{(1-\alpha)^2} \times \frac{\alpha^2}{(1-\alpha)^2} & =4 \\
\frac{\alpha^2}{(1-\alpha)^4} & =4 \\
\frac{\alpha}{(1-\alpha)^2} & = \pm 2 \\
2(1-\alpha)^2 & = \pm \alpha
\end{aligned}
$$

If $2(1-\alpha)^2=\alpha$, then $2 \alpha^2-5 \alpha+2=0$

Sum of value of $\alpha=\frac{5}{2} \quad\left[\because\right.$ Sum of zero $\left.=\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\right]$

If $2(1-\alpha)^2=-\alpha$, then $2 \alpha^2-3 \alpha+2=0$

$\therefore$ No real value of $\alpha$

Hence, sum of all values of $\alpha=\frac{5}{2}$

Solutions · 1