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
Answer & explanation
Correct answer: option 3
\(A=\begin{bmatrix} 2x - 1 & -3 & 6\\ 3 & 3y - 2 & 4\\ -6 & -4 & 4z - 2\\ \end{bmatrix}⇒A^T=\begin{bmatrix} 2x - 1 & 3 & -6\\ -3 & 3y - 2 & -4\\ 6 & 4 & 4z - 2\\ \end{bmatrix}\)
A is symmetric matrix. So, (AT) = -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.
\(\begin{bmatrix} 2x - 1 & 3 & -6\\ -3 & 3y - 2 & -4\\ 6 & 4 & 4z - 2\\ \end{bmatrix}=-\begin{bmatrix} -(2x - 1) & 3 & -6\\ -3 & -(3y - 2) & -4\\ 6 & 4 & -(4z - 2)\\ \end{bmatrix}\)
$⇒ 2x - 1 = -(2x-1)$
$2x-1=-2x+1⇒4x=2⇒x=\frac{1}{2}$
$3y-2=-3y+2⇒6y=4⇒y=\frac{2}{3}$
$4z-2=-4z+2⇒8z=4⇒z = \frac{1}{2}$
$x×y×z=\frac{1}{2}×\frac{2}{3}×\frac{1}{2}=\frac{1}{6}$