Question

Let $A,B,C$ be three square matrices of order $3$ such that $A^2+2I=O,$ $det(2C-A^2)=32$ and $A^5-2A^{3}C+B{A}^2-2BC=O.$ Then the absolute value of $(det(B-A){)}^2$ is

Answer 1

Given : $A^2+2I=O$
$\Rightarrow A^2=-2I$
$\Rightarrow |A{|}^2=-8$
Also, $det(2C-A^2)=32$
$\Rightarrow det(2C+2I)=32$
$\Rightarrow 2^3\cdot det(C+I)=32$
$\Rightarrow det(C+I)=4$

Now, $A^5-2A^{3}C+B{A}^2-2BC=O$
$\Rightarrow 4A+4AC-2B-2BC=O(\because A^2=-2I)$
$\Rightarrow 4A(I+C)-2B(I+C)=O$
$\Rightarrow (I+C)(4A-2B)=O$
Since $I+C$ is non-singular,
$\therefore 4A-2B=O$
$\Rightarrow B=2A$
$\therefore {|B-A|}^2={|2A-A|}^2={\left|A\right|}^2=-8$