ISI2015-DCG-31

Question

Let $A$ be an $n \times n$ matrix such that $\mid A^{2} \mid\: =1$. Here $\mid A \mid $ stands for determinant of matrix $A$. Then

• A. $\mid A \mid =1$
• B. $\mid A \mid =0 \text{ or } 1$
• C. $\mid A \mid =-1, 0 \text{ or } 1$
• D. $\mid A \mid =-1  \text{ or }  1$

Comments

Ans will be D)

We can check with two matrices

$\begin{bmatrix} 1 & 0\\ 0 & 1  \end{bmatrix}$

$\begin{bmatrix} -1 & 0\\ 0 & -1  \end{bmatrix}$

both are giving $|A^{2}|=1$

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Yeah.

$|A^n| = |A|^n$ because $|A*A*.....*A|=|A|*|A|*....$

Answer 1

$|A^n| = |A|^n$ for any determinant.

$|A^2|=1 \implies |A|^2 =1 \implies |A|*|A| =1$

So $|A|$ can be either $1$ or $-1$

$\therefore$ Option $D.$ is correct answer.