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# Find the values of $x, y, z$ if the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfy the equation $A'A = I$

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Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• If $A_{i,j}$ be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
Step 1: Given
$A=\begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$
$A'=\begin{bmatrix} 0 &x & x\\ 2y & y & -y \\ z & -z & z \end{bmatrix}$
Interchange the rows and column to obtain the transpose
$A'A=\begin{bmatrix} 0 &x & x\\ 2y & y & -y \\ z & -z & z \end{bmatrix}\begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 0+x^2+x^2 &0+xy-xy & 0-xz+xz\\ 0+yx-yx & 4y^2+y^2+y^2 & 2yz-yz-yz \\ 0-xz+xz& 2yz-zy-zy & z^2+z^2+z^2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 2x^2 &0 & 0\\ 0 & 6y^2 & 0\\ 0& 0 &3z^2 \end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$
Step 2: Given A'A=$I_3$
Hence the given matrices are equal
Hence the corresponding elements should also be equal
$\Rightarrow 2x^2=1$
$x^2=\frac{1}{2}$
$x=\pm \sqrt{\frac{1}{2}}$
$3z^2=1$
$z^2=\frac{1}{3}$
$z=\pm \sqrt{\frac{1}{3}}$
$\Rightarrow 6y^2=1$
$y^2=\frac{1}{6}$
$y=\pm \sqrt{\frac{1}{6}}$
Hence $x=\pm \sqrt{\frac{1}{2}}$
$y=\pm \sqrt{\frac{1}{6}}$
$z=\pm \sqrt{\frac{1}{3}}$