Question

If \( \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \), then the value of \( k \) is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 4 \)

Hint:

When dealing with determinants, applying column operations can simplify the calculation significantly. Watch for symmetry to identify potential simplifications.

Answer 1

The Correct Option is D

We are given the following determinant and are tasked with finding its value:

\[ \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \]
Taking \(a\), \(b\), and \(c\) out of the matrix from columns \(C_1\), \(C_2\), and \(C_3\), respectively:

\[ abc \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = kabc \]
Dividing both sides by \(abc\), we get:

\[ \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = k \]
Using column operations \(C_2 \to C_2 + C_1\) and \(C_3 \to C_3 + C_1\), the determinant simplifies to:

\[ \begin{vmatrix} -1 & 0 & 0 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{vmatrix} = k \]
Expanding the determinant along the first row:

\[ -1(0 \times 0 - 2 \times 2) = k \]
Simplifying further:

\[ -1(-4) = k \]
\[ k = 4 \]
\(\therefore k = 4\)

Given the problem setup, the value of \(k\) is 4, and thus the correct option is (D) 4.