If $a, b$ and $c$ are distinct prime numbers then the value of $\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Let the matrix be
$M=\begin{bmatrix} a-b & b-c & c-a\\ b-c & c-a & a-b\\ c-a & a-b & b-c \end{bmatrix}$
Set $x=a-b,\;y=b-c,\;z=c-a$. Then $x+y+z=0$ and
$M=\begin{bmatrix} x&y&z\\ y&z&x\\ z&x&y \end{bmatrix}$
$M$ is a $3\times3$ circulant matrix. Its determinant equals
$(x+y+z)\,(x+\omega y+\omega^{2}z)\,(x+\omega^{2}y+\omega z)$, where $\omega$ is a cube root of unity.
Since $x+y+z=0$,
$\det M=0$.
Answer: $0$.