Q. 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
Explanation
(d)
det$(A - I) = \det\left( A - A^{2} \right)$
$= \det{A(I - A)}$
$= \det{A \times \det(I - A)}$
$= {- det}{A \times \det(I - A)}$
Now,
$A^{2} = I$
$\Rightarrow \det\left( A^{2} \right) = \det(I)$
$\Rightarrow \left( \det A \right)^{2} = 1$
$\Rightarrow$det$(A) = \pm 1$
Thus, det$(A)$ can be 1 or $- 1$, which we cannot say anything about det$(A - I)$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.