Question

If $A=\frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]$ satisfies $A^{T}A=I$, then $x+y$

• A. $3$
• B. $0$
• C. $-3$
• D. $1$

Answer 1

The correct option is C $-3$

$A=\frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]A^T=\frac{1}{3}\left[\begin{array}{ccc}1& 2& x\\ 2& 1& 2\\ 2& -2& y\end{array}\right]$

$A^{T}A=\left[\begin{array}{ccc}9& 0& x+2y+4\\ 0& 9& 2x+2-2y\\ x+2y+4& 2x+2-2y& x^2+y^2+4\end{array}\right]\times \frac{1}{9}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Two equation are $x+2y+4=0$

$2x-2y=2=0$.

On solving\rightarrow $(-2,-1)$

$x=-2,y=-1$

$x+y=-3$.

$C$ is correct.