Question

If $a$, $b$ and $c$ are negative and different real numbers, then $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$ is}

• A. $a^2+b^2+c^2$
• B. $-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
• C. $0$
• D. $abc$

Hint:

Circulant determinant identity: $\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = -(a^3+b^3+c^3-3abc)$. Remember $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Answer 1

The Correct Option is B

Step 1: Conceptual Understanding:
This is a circulant determinant. It factors using the identity for $a^3+b^3+c^3-3abc$.
Step 2: Explanation in Detail:
The determinant equals $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
With the sign from the determinant expansion: $-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
Step 3: Therefore, Stating the Final Answer
The determinant $= -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.