If \( \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \), then the value of \( k \) is:
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 <strong>(D) 4</strong>.
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is: