If  $5A=\left[\begin{array}{cc}3& -4\\ 4& x\end{array}\right]$ and  $A{A}^T=A^{T}A=I$, then  $x=$

$A{A}^T,A=\frac{1}{5}\left[\begin{array}{cc}3& -4\\ 4& x\end{array}\right]$

$\Rightarrow \frac{1}{5}\left[\begin{array}{cc}3& -4\\ 4& x\end{array}\right]\frac{1}{5}\left[\begin{array}{cc}3& 4\\ -4& x\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \frac{1}{25}\left[\begin{array}{cc}9+16& 12-4x\\ 12-4x& 16+x^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}25& 12-4x\\ 12-4x& 16+x^2\end{array}\right]=\left[\begin{array}{cc}25& 0\\ 0& 25\end{array}\right]$

$12-4x=0\Rightarrow 4x=12\Rightarrow x=3$

$16+x^2=25\Rightarrow x^2=9\Rightarrow x=\pm 3\Rightarrow x=3$