JEE Main 2023 — Mathematics Algebra

Given,

$A$A be a $2\times 2$2×2 matrix with real entries such that ${A}^{'}=\alpha A+1,$A′=αA+1,

Now let $A=[\begin{matrix}a & b \\ c & d\end{matrix}]$A=[ac​bd​]

Now putting the value in ${A}^{'}=\alpha A+I$A′=αA+I we get,

$$\Rightarrow [\begin{matrix}a & c \\ b & d\end{matrix}]=[\begin{matrix}\alpha a+1 & \alpha b \\ \alpha c & \alpha d+1\end{matrix}]$$⇒[ab​cd​]=[αa+1αc​αbαd+1​]

Now equating both side we get,

$$a=\alpha a+1\Rightarrow a=\frac{1}{1-\alpha }.........(i)$$a=αa+1⇒a=1−α1​.........(i)

$b=\alpha c....(ii)$b=αc....(ii) and $c=\alpha b........(iii)$c=αb........(iii)

Now from equation (ii)&(iii) we get,

$c=0$c=0 or $\alpha =\pm 1$α=±1 (not possible as mentioned in question)

$$\therefore c=0$$∴c=0

$$\Rightarrow c=0,b=0$$⇒c=0,b=0

Also $d=\alpha d+1\Rightarrow d=\frac{1}{1-\alpha }$d=αd+1⇒d=1−α1​

Also given, $|{A}^{2}-A|=4$∣A2−A∣=4

$$\Rightarrow |A||A-I|=4$$⇒∣A∣∣A−I∣=4

$$\Rightarrow {(\frac{1}{1-\alpha })}^{2}{(\frac{1}{1-\alpha }-1)}^{2}=4$$⇒(1−α1​)2(1−α1​−1)2=4

Now let $\frac{1}{1-\alpha }=t$1−α1​=t

So, ${t}^{2}{(t-1)}^{2}=4$t2(t−1)2=4

$$\Rightarrow t(t-1)=\pm 2$$⇒t(t−1)=±2

$$\Rightarrow t=2\text{or}t=-1$$⇒t=2ort=−1

$$\Rightarrow \alpha =\frac{1}{2},2$$⇒α=21​,2

So, sum of all possible value will be $\frac{1}{2}+2=\frac{5}{2}$21​+2=25​