Question

If $A=\left[\begin{array}{cc}0& x\\ y& 0\end{array}\right]$ and $A^3+A=O,$ then which of the following is correct

• A. $xy=-1$
• B. $xy=0$
• C. $xy=\frac{1}{2}$
• D. $xy=1$

Answer 1

The correct option is A $xy=-1$

$A^2=A\cdot A=\left[\begin{array}{cc}0& x\\ y& 0\end{array}\right]\left[\begin{array}{cc}0& x\\ y& 0\end{array}\right]=\left[\begin{array}{cc}xy& 0\\ 0& xy\end{array}\right]$
$\Rightarrow A^3=A^2\cdot A=\left[\begin{array}{cc}xy& 0\\ 0& xy\end{array}\right]\left[\begin{array}{cc}0& x\\ y& 0\end{array}\right]=\left[\begin{array}{cc}0& x^{2}y\\ x{y}^2& 0\end{array}\right]$
$\Rightarrow A^3+A=O,\left(\text{Given}\right)\Rightarrow \left[\begin{array}{cc}0& x^{2}y\\ x{y}^2& 0\end{array}\right]+\left[\begin{array}{cc}0& x\\ y& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& x+x^{2}y\\ y+x{y}^2& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow x+x^{2}y=0x(1+xy)=0\Rightarrow x=0,xy=-1\cdots \left(i\right)$ and
$y+y^{2}x=0\Rightarrow y(1+xy)=0\Rightarrow y=0,xy=-1\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow xy=-1$