Question

Let $A$ being $3\times 3$ matrix and $x$ is a real number such that $det\left(xA\right)=|x|.det\left(A\right)$, then $x$ is

• A. $0$ or $1$
• B. $0$ or $-1$
• C. $1$ or $-1$
• D. $0$ or $\pm 1$

Answer 1

Answer: (A)

$0$ or $1$

Given, $A$ being $3\times 3$ matrix and $x$ is a real number such that $det\left(xA\right)=|x|.det\left(A\right)$.

Now $det\left(xA\right)=x^3.det\left(A\right)$ [ Since $A$ is of $3\times 3$ matrix.]

According to the problem,

$x^3=|x|$

It is possible only when $x=0$ or $1$.