Question

Let $A=\left[a_{i j}\right]_{2 \times 2}$ be a matrix such that $Tr\left(A\right)=5$ . If $f\left(x\right)=det\left(A - x I\right)$ and $f\left(A\right)=A^{2}+aA+3I$ then value of $Tr\left[\left(adj A\right)^{2} + \left(a + 1\right) adj A + 3 I\right]$ 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 $A=\begin{bmatrix} p & s \\ r & q \end{bmatrix}$ where $p+q=5$ , $pq-rs=3$
$adjA=\begin{bmatrix} q & -s \\ -r & p \end{bmatrix}$
$\left(adj A\right)^{2}=\begin{bmatrix} q & -s \\ -r & p \end{bmatrix}\begin{bmatrix} q & -s \\ -r & p \end{bmatrix}=\begin{bmatrix} q^{2}+rs & -rs+p^{2} \\ -r\left(p + q\right) & p^{2}+r8 \end{bmatrix}$
$tr\left(\left(adj A\right)^{2} + \left(a + 1\right) adj A + 3 I\right)$
$=p^{2}+q^{2}+2rs-4\left[p + q\right]+3\left(1 + 1\right)$
$\left(p + q\right)^{2}-2\left(p q - r s\right)-4\left(p + q\right)+6$
$\Rightarrow 25-6-20+6=5$