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:

For determinants, use row expansion and carefully simplify the minors. Pay attention to symmetry in scalar matrices.

Answer 1

The Correct Option is D

Step 1: Expand the determinant.
The determinant of the given matrix is: \[ \begin{vmatrix} -a & b & c
a & -b & c
a & b & -c \end{vmatrix}. \] Expanding along the first row: \[ -a \begin{vmatrix} -b & c
b & -c \end{vmatrix} - b \begin{vmatrix} a & c
a & -c \end{vmatrix} + c \begin{vmatrix} a & -b
a & b \end{vmatrix}. \] Step 2: Simplify the minors.
- First minor: \[ \begin{vmatrix} -b & c
b & -c \end{vmatrix} = (-b)(-c) - (b)(c) = bc - bc = -2bc. \] - Second minor: \[ \begin{vmatrix} a & c
a & -c \end{vmatrix} = (a)(-c) - (a)(c) = -ac - ac = -2ac. \] - Third minor: \[ \begin{vmatrix} a & -b
a & b \end{vmatrix} = (a)(b) - (a)(-b) = ab + ab = 2ab. \] Step 3: Substitute back and simplify.
\[ -a(-2bc) - b(-2ac) + c(2ab) = 2abc + 2abc + 2abc = 4abc. \] Step 4: Conclusion.
The value of \( k \) is:4