Question

Find the value of $\left|\frac{1}{xyz}\right|$ if the matrix $A=\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$ satisfies the equation $A^{T}A=I$.

Answer 1

It is given that $A=\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$
$\therefore A^T=\left[\begin{array}{ccc}0& x& x\\ 2y& y& -y\\ z& -z& z\end{array}\right]$

Now, $A^{T}A=I$

$\Rightarrow \left[\begin{array}{ccc}0& x& x\\ 2y& y& -y\\ z& -z& z\end{array}\right]\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\left[\begin{array}{ccc}0+x^2+x^2& 0+xy-xy& 0-xz+xz\\ 0+xy-xy& 4y^2+y^2+y^2& 2yz-yz-yz\\ 0-xz+xz& 2yz-yz-yz& z^2+z^2+z^2\end{array}\right]$
$\left[\begin{array}{ccc}2x^2& 0& 0\\ 0& 6y^2& 0\\ 0& 0& 3z^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

On comparing the corresponding elements, we have

$2x^2=1$ or $x=\pm \frac{1}{\sqrt{2}}$

$6y^2=1$ or $y=\pm \frac{1}{\sqrt{6}}$

$3z^2=1$ or $z=\pm \frac{1}{\sqrt{3}}$

$\therefore \left|\frac{1}{xyz}\right|=6$