Question

If $A^2=I,$ then the value of $det(A-I)$ is (where $A$ has order 3)

• A. 1
• B. -1
• C. 0
• D. Cannot say anything

Answer 1

Answer: (D)

det (A-I)=det $\left(A-A^2\right)$ =det $A(I-A)$ =det A $\cdot$ det $(I-A)=-$det $A\cdot$ det $(A-I)$
Now $A^2=I$
$\Rightarrow$ det $\left(A^2\right)=$ det $(I)\Rightarrow ($ det $A{)}^2=1\Rightarrow$ det $(A)=\pm 1$
Thus, det $(A)$ can be $1$ or $-1,$ from which we cannot say anything about det $(A-I)$