JEE Main 2024 — 27 January, Shift 2 — Question 23

$|A-x I|=0$ Roots are $-1$ and $3$

Sum of roots $=\operatorname{tr}(A)=2$

Product of roots $=|\mathrm{A}|=-3$

Let $A=\left[\begin{array}{ll}a & b\\ c & d\end{array}\right]$

We have $\mathrm{a}+\mathrm{d}=2$

$\mathrm{ad}-\mathrm{bc}=-3$

$A^{2}=\left[\begin{array}{ll}a & b\\ c & d\end{array}\right] \times\left[\begin{array}{ll}a & b \\c & d\end{array}\right]=\left[\begin{array}{ll}a^{2}+b c & a b+b d\\ a c+c d & b c+d^{2}\end{array}\right]$

We need $a^{2}+b c+b c+d^{2}$

$=a^{2}+2 b c+d^{2}$

$=(a+d)^{2}-2 a d+2 b c$

$=4-2(\mathrm{ad}-\mathrm{bc})$

$=4-2(-3)$

$=4+6$

$=10$