Question

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

• A. $0$
• B. $1$
• C. $7$
• D. $5$

Answer 1

Answer: (D)

Sum of roots $=tr\left(A\right)=5$
product of roots $=\left|A\right|$
Characterstic equation is $x^2-5x+\left|A\right|$
Since $f\left(A\right)=A^2+aA+3I$
$\therefore a=-5,$ and $\left|A\right|=3$
Consider $A=\left[\begin{array}{cc}p& s\\ r& q\end{array}\right]$ where $p+q=5$ , $pq-rs=3$
$adjA=\left[\begin{array}{cc}q& -s\\ -r& p\end{array}\right]$
${\left(adjA\right)}^2=\left[\begin{array}{cc}q& -s\\ -r& p\end{array}\right]\left[\begin{array}{cc}q& -s\\ -r& p\end{array}\right]=\left[\begin{array}{cc}{q}^2+rs& -rs+p^2\\ -r\left(p+q\right)& p^2+r8\end{array}\right]$
$tr\left({\left(adjA\right)}^2+\left(a+1\right)adjA+3I\right)$
$=p^2+q^2+2rs-4\left[p+q\right]+3\left(1+1\right)$
${\left(p+q\right)}^2-2\left(pq-rs\right)-4\left(p+q\right)+6$
$\Rightarrow 25-6-20+6=5$