Question

Find the values of $x,y,z$ if the matrix $A=\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$ satisfy the equation $A^{\prime }A=I.$

Answer 1

Given,

$A=\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$

$A^{\prime }=\left[\begin{array}{ccc}0& x& x\\ 2y& y& -y\\ z& -z& z\end{array}\right]$

$I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Now, $A^{\prime }A=I$

Putting values

$\left[\begin{array}{ccc}0& x& x\\ 2y& y& -y\\ z& -z& z\end{array}\right]\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}0\left(0\right)+x\left(x\right)+x\left(x\right)& 0\left(2y\right)+x\left(y\right)+x(-y)& 0\left(z\right)+x(-z)+x\left(z\right)\\ 2y\left(0\right)+y\left(x\right)-y\left(x\right)& 2y\left(2y\right)+y\left(y\right)-y(-y)& 2y\left(z\right)+y(-z)-y\left(z\right)\\ z\left(0\right)-z\left(x\right)+z\left(x\right)& z\left(2y\right)-z\left(y\right)+z(-y)& z\left(z\right)-z(-z)+z\left(z\right)\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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

$\left[\begin{array}{ccc}2x^2& 0& 0\\ 0& 6y^2& 0\\ 0& 0& 3z^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Since matrices are equal,

corresponding elements are equal,

$2x^2=1$

$x^2=\frac{1}{2}$

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

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

$6y^2=1$

$y^2=\frac{1}{6}$

$y=\pm \sqrt{\frac{1}{6}}$

$y=\pm \frac{1}{\sqrt{6}}$

$3z^2=1$

$z^2=\frac{1}{3}$

$z=\pm \sqrt{\frac{1}{3}}$

$z=\pm \frac{1}{\sqrt{3}}$

Thus, $x=\pm \frac{1}{\sqrt{2}},y=\pm \frac{1}{\sqrt{6}},z=\pm \frac{1}{\sqrt{3}}$