Question

Find the value of \( \left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \end{matrix} \right|^2 \)

• A. \( a^2b^2c^2 \)
• B. \( 4a^2b^2c^2 \)
• C. \( 2a^2b^2c^2 \)
• D. \( (a+b+c)^2 \)

Hint:

For any skew-symmetric-like structure with zeros on the diagonal and symmetric off-diagonal terms, the determinant often simplifies to \( 2abc \). Memorizing this standard form helps in solving such determinant questions instantly.

Answer 1

The Correct Option is B

Step 1: Name the determinant.
Let $D = \begin{vmatrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \end{vmatrix}$; we want $D^2$.
Step 2: Expand along the first column instead of the first row.
\[ D = 0\cdot M_{11} - c\cdot\begin{vmatrix} c & b \\ a & 0 \end{vmatrix} + b\cdot\begin{vmatrix} c & b \\ 0 & a \end{vmatrix} \]
Step 3: Evaluate the two $2\times 2$ minors.
$\begin{vmatrix} c & b \\ a & 0 \end{vmatrix} = (c)(0)-(b)(a) = -ab$, and $\begin{vmatrix} c & b \\ 0 & a \end{vmatrix} = (c)(a)-(b)(0) = ac$.
Step 4: Combine the terms.
\[ D = -c(-ab) + b(ac) = abc + abc = 2abc \]
Step 5: Square the determinant.
\[ D^2 = (2abc)^2 \]
Step 6: Expand the square.
\[ D^2 = 4a^2b^2c^2 \] Expanding down the first column gives the same $2abc$ as a row expansion would, confirming the answer. \[ \boxed{4a^2b^2c^2} \]