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QJEEM23Ma0aea: JEE Main 2023 Mathematics - Class 12 JEEM23Ma0aea Solution: Let be a matrix with real entries such that , where . If , then …

Let A\mathrm{A} be a 2×22 \times 2 matrix with real entries such that A′=αA+I\mathrm{A}'=\alpha \mathrm{A}+\mathrm{I} , where α∈R−{−1,1}\alpha \in \mathbb{R}-\{-1,1\} . If det⁡(A2−A)=4\operatorname{det}\left(A^{2}-A\right)=4 , then the sum of all possible values of α\alpha is equal to :

Multiple Choice Options

  • A 2
  • B 32\frac{3}{2}
  • C 0
  • D 52\frac{5}{2}

Step-by-Step Verified Solution

We have, AT=αA+IA^T=\alpha A+I , where AA is 2×22 \times 2 matrix and α∈R−{−1,1}\alpha \in R-\{-1,1\}

(AT)Tamp;=αAT+IAamp;=αAT+IAamp;=α(αA+I)+I[∵AT=αA+I]Aamp;=α2A+(α+1)IAamp;(1−α2)=(α+1)IAamp;=(α+1)(1−α)(1+α)I\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}

Aamp;=11−αI..........(i)∣A∣amp;=1(1−α)2...........(ii)\begin{aligned}A & =\frac{1}{1-\alpha} I ..........(i)\\\\|A| & =\frac{1}{(1-\alpha)^2} ...........(ii)\end{aligned}

Also,

∣A2−A∣amp;=4∣A∣∣A−I∣amp;=41(1−α)2∣(11−α−1)I∣amp;=41(1−α)2∣(α1−α)I∣amp;=41(1−α)2×α2(1−α)2amp;=4α2(1−α)4amp;=4α(1−α)2amp;=±22(1−α)2amp;=±α\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−α)2=α2(1-\alpha)^2=\alpha , then 2α2−5α+2=02 \alpha^2-5 \alpha+2=0

Sum of value of α=52[∵\alpha=\frac{5}{2} \quad\left[\because\right. Sum of zero =-Coefficient of x Coefficient of x2]\left.=\frac{\text {-Coefficient of } x}{\text { Coefficient of } x^2}\right]

If 2(1−α)2=−α2(1-\alpha)^2=-\alpha , then 2α2−3α+2=02 \alpha^2-3 \alpha+2=0

∴\therefore No real value of α\alpha

Hence, sum of all values of α=52\alpha=\frac{5}{2}

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