Question

Consider $A=\left[\begin{array}{ccc}a& 2& 1\\ 0& b& 0\\ 0& -3& c\end{array}\right]$ where $a,b,c$ are the roots of the equation $x^3-3x^2+2x-1=0.$ If matrix $B$ is such that $AB=BA,|A+B-2I|\ne 0$ and $A^2-B^2=4I-4B,$ where $I$ is the $3\times 3$ identity matrix, then the value of $\sqrt{det\left(B\right)}$ is

Answer 1

$|A|=\left|\begin{array}{ccc}a& 2& 1\\ 0& b& 0\\ 0& -3& c\end{array}\right|=abc=1$
where $x^3-3x^2+2x-1=(x-a)(x-b)(x-c)\cdots \left(1\right)$

Given, $A^2-B^2=4I-4B$
$\Rightarrow A^2=B^2-4B+4I\Rightarrow A^2-(B-2I{)}^2=O\Rightarrow (A+B-2I\left)\right(A-B+2I)=O(\because AB=BA)$
Since $|A+B-2I|\ne 0,$
$\therefore A-B+2I=0\Rightarrow B=A+2I=\left|\begin{array}{ccc}a+2& 2& 1\\ 0& b+2& 0\\ 0& -3& c+2\end{array}\right|$

$|B|=(a+2)(b+2)(c+2)$
Putting $x=-2$ in equation $\left(1\right),$ we get
$-8-12-4-1\equiv (-2-a)(-2-b)(-2-c)\Rightarrow 25=(a+2)(b+2)(c+2)\therefore |B|=25\therefore \sqrt{|B|}=5$