Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}} ight]_{2 imes 2}$ be a matrix such that $\operator

Question

Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}ight]_{2 	imes 2}$ be a matrix such that $\operatorname{Tr}(\mathrm{A})=5$. If $f(\mathrm{x})=\operatorname{det}(\mathrm{A}-\mathrm{xI}) \quad$ and $f(\mathrm{~A})=\mathrm{A}^2+\mathrm{aA}+3 \mathrm{I} \quad$ then value of $\operatorname{Tr}\left[(\operatorname{adj} A)^2+(a+1) \operatorname{adj} A+3 Iight] \quad$ is equal to $[\operatorname{Tr}(\mathrm{P})$ is trace of matrix P]

Professor PC Thomas Classes · Answer

Sum of roots $=\operatorname{tr}(\mathbf{A})=5$
product of roots $=|\mathbf{A}|$
Characterstic equation is $x^2-5 x+|A|$
since $f(\mathrm{~A})=\mathrm{A}^2+\mathrm{aA}+3 \mathrm{I}$
$	herefore \mathrm{a}=-5$, and $|\mathrm{A}|=3$
consider $\mathbf{A}=\left[\begin{array}{ll}\mathbf{p} & \mathbf{s} \ \mathbf{r} & \mathbf{q}\end{array}ight]$ where $\mathbf{p}+\mathbf{q}=5, \mathbf{p q}-\mathbf{r s}=\mathbf{3}$
$\operatorname{adj} \mathbf{A}=\left[\begin{array}{cc}\mathbf{q} & -\mathbf{s} \ -\mathbf{r} & \mathbf{p}\end{array}ight]$
$(\operatorname{adj} \mathrm{A})^2=\left[\begin{array}{cc}\mathrm{q} & -\mathrm{s} \ -\mathrm{r} & \mathrm{p}\end{array}ight]\left[\begin{array}{cc}\mathrm{q} & -\mathrm{s} \ -\mathrm{r} & \mathrm{p}\end{array}ight]$
$=\left[\begin{array}{cc}\mathbf{q}^2+\mathbf{r s} & -\mathbf{s}(\mathbf{p}+\mathbf{q}) \ -\mathbf{r}(\mathbf{p}+\mathbf{q}) & \mathbf{p}^2+\mathbf{r s}\end{array}ight]$
$\operatorname{tr}\left((\operatorname{adj} A)^2+(a+1) \operatorname{adj} A+3 Iight)$
$=\mathrm{p}^2+\mathrm{q}^2+2 \mathrm{rs}-4[\mathrm{p}+\mathrm{q}]+3(1+1)$
$(p+q)^2-2(p q-r s)-4(p+q)+6$
$\Rightarrow 25-6-20+6=5$