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

Question

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

Accepted Answer

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

$\begin{gathered} {({A}^{T})}^{T} =\alpha {A}^{T}+I \ \ A =\alpha {A}^{T}+I \ \ A =\alpha (\alpha A+I)+I[\because {A}^{T}=\alpha A+I] \ \ A ={\alpha }^{2}A+(\alpha +1)I \ \ A (1-{\alpha }^{2})=(\alpha +1)I \ \ A =\frac{(\alpha +1)}{(1-\alpha )(1+\alpha )}I $

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

Also,

$ |{A}^{2}-A| =4 \ \ |A||A-I| =4 \ \ \frac{1}{(1-\alpha {)}^{2}}|(\frac{1}{1-\alpha }-1)I| =4 \ \ \frac{1}{(1-\alpha {)}^{2}}|(\frac{\alpha }{1-\alpha })I| =4 \ \ \frac{1}{(1-\alpha {)}^{2}}	imes \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{gathered}$

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

Sum of value of $\alpha =\frac{5}{2}\, [\because $ Sum of zero $=\frac{x}{{x}^{2}}]$

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

$	herefore $ No real value of $\alpha $

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

Solutions · 1