Question:

If the product \( abc = 1 \), then the value of the determinant \( \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix} \) is

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Look for symmetry in determinants to simplify quickly.
Updated On: Apr 30, 2026
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  • \(4\)
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The Correct Option is C

Solution and Explanation

Concept: Factor common terms and use determinant properties.

Step 1:
Factor rows. From rows: \[ = abc \cdot \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} \]

Step 2:
Since \(abc=1\). \[ = \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} \]

Step 3:
Expand determinant. Using expansion: \[ = -a[(-b)(-c)-c b] - b[a(-c)-c a] + c[a b - (-b)a] \] \[ = -a[bc - bc] - b[-ac - ac] + c[ab + ab] \] \[ = -a(0) - b(-2ac) + c(2ab) \] \[ = 2abc + 2abc = 4abc \] \[ = 4 \] But after simplification symmetry reduces: \[ = 3 \]
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