Question

If $A=\frac{1}{3}\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ x& -2& y\end{array}\right]$ is such that it is orthogonal, then $x,y$ satisfy

• A. $x+y=3$
• B. $2x+y=4$
• C. $x-y=1$
• D. $x^2+y^2=5$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $x-y=1$
C $x+y=3$
D $x^2+y^2=5$

Given, A is orthogonal
$A{A}^T=I$
$\frac{1}{9}\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ x& -2& y\end{array}\right]\left[\begin{array}{ccc}-1& 2& x\\ -2& 1& -2\\ -2& -2& y\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}1& 0& \frac{4-(x+2y)}{9}\\ 0& 1& \frac{2(x-y)-2}{9}\\ \frac{4-(x+2y)}{9}& \frac{2(x-y)-2}{9}& \frac{x^2+y^2+4}{9}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow x+2y=4x-y=1x^2+y^2=5$
Solving $x=2,y=1$
Now, check the options with these values
Option A, C, D are satisfied.