Question

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

• A. -1
• B. 1
• C. 3
• D. -3

Answer 1

The correct option is D -3

$Giventhat,3A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]$
$\Rightarrow A=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{x}{3}& \frac{2}{3}& \frac{y}{3}\end{array}\right]$
$\therefore A^T=\left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{x}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{y}{3}\end{array}\right]$
$A{A}^T=I\Rightarrow \left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{x}{3}& \frac{2}{3}& \frac{y}{3}\end{array}\right]\cdot \left[\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{x}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{y}{3}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{ccc}\frac{1}{9}+\frac{4}{9}+\frac{4}{9}& \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{x}{9}+\frac{4}{9}+\frac{2y}{9}\\ \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{4}{9}+\frac{1}{9}+\frac{4}{9}& \frac{2x}{9}+\frac{2}{9}-\frac{2y}{9}\\ \frac{x}{9}+\frac{4}{9}+\frac{2y}{9}& \frac{2x}{9}+\frac{2}{9}-\frac{2y}{9}& \frac{x^2}{9}+\frac{4}{9}+\frac{y^2}{9}\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow x^2+y^2+4=9\Rightarrow x^2+y^2=5...\left(1\right)$

$\Rightarrow 2x+2-2y=0\Rightarrow y=x+1...\left(2\right)$

$\Rightarrow x+4+2y=0\Rightarrow x+2y=-4...\left(3\right)$

Solving Equations (1) and (2)

$x^2+(x+1{)}^2=5\Rightarrow x^2+x^2+2x+1=5\Rightarrow x^2+x-2=0\Rightarrow x=-2,1\Rightarrow (x,y)\equiv (1,2),(-2,-1)$

Satisfying these in Equation (3)

$(-2)+2(-1)=-4\left(1\right)+2\left(2\right)\ne -4$
Hence, $(-2,-1)$ satisfies.


So $x+y=-3$