Question

Let A,B,C be three square matrices of order $3\times 3$ such that $A^2+2I=0$, $|(2C-A^2)|=30$ and satisfying the equation $A^5-2A^{3}C+B{A}^2-2BC=0$

Value of $(det.(B-A){)}^2$ is

• A. 8
• B. -8
• C. 64
• D. -64

Answer 1

The correct option is B -8

$(A^3+B)A^2-2(A^3+B)C=0$
$(A^3+B)(A^2-2C)=0$
Since $|2C-A^2|=30$, so inverse of $(2C-A^2)$ exists.
So $A^3+B=0$
$B=-A^3=-A\left(A^2\right)=-A(-2I)=2A$
B =2A
$|B-A{|}^2=|2A-A{|}^2=|A{|}^2$
We know that $A^2=-2I$
$|A{|}^2=|-2I|\Rightarrow |A{|}^2=-8\times 1=-8$
$|B{|}^2=64|A{|}^2=64\times -8=-512$
Therefore $|B-A{|}^2=|A{|}^2=-8$