If \(\begin{bmatrix} 2x - 1 & -3 & 6\\  3 & 3y - 2 & 4\\ -6 & -4 & 4z - 2\\ \end{bmatrix}\) is skew symmetric matrix, then xyz is equal to

\(\frac{1}{6}\)

\(A=\begin{bmatrix} 2x - 1 & -3 & 6\\  3 & 3y - 2 & 4\\ -6 & -4 & 4z - 2\\ \end{bmatrix}⇒A^T=\begin{bmatrix} 2x - 1 & 3 & -6\\  -3 & 3y - 2 & -4\\ 6 & 4 & 4z - 2\\ \end{bmatrix}\)

A is symmetric matrix. So, (A^T) = -A

\(\begin{bmatrix} 2x - 1 & 3 & -6\\  -3 & 3y - 2 & -4\\ 6 & 4 & 4z - 2\\ \end{bmatrix}=-\begin{bmatrix} 2x - 1 & -3 & 6\\  3 & 3y - 2 & 4\\ -6 & -4 & 4z - 2\\ \end{bmatrix}\)

∴ corresponding elements of two equal matrices are equal.

\(\begin{bmatrix} 2x - 1 & 3 & -6\\  -3 & 3y - 2 & -4\\ 6 & 4 & 4z - 2\\ \end{bmatrix}=-\begin{bmatrix} -(2x - 1) & 3 & -6\\  -3 & -(3y - 2) & -4\\ 6 & 4 & -(4z - 2)\\ \end{bmatrix}\)

$⇒ 2x - 1 = -(2x-1)$

$2x-1=-2x+1⇒4x=2⇒x=\frac{1}{2}$

$3y-2=-3y+2⇒6y=4⇒y=\frac{2}{3}$

$4z-2=-4z+2⇒8z=4⇒z = \frac{1}{2}$

$x×y×z=\frac{1}{2}×\frac{2}{3}×\frac{1}{2}=\frac{1}{6}$