Question

If $A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],B=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$, then $(A+B{)}^2$ equals

• A. $A^2+B^2$
• B. $A^2+B^2+2AB$
• C. $A^2+B^2+AB-BA$
• D. None of these

Answer 1

The correct option is A $A^2+B^2$

$A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],B=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$
$A^2=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$B^2=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$AB=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$
$AB=\left[\begin{array}{cc}i& 0\\ 0& -i\end{array}\right]$
Now, $BA=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
$\Rightarrow BA=\left[\begin{array}{cc}-i& 0\\ 0& i\end{array}\right]$
$\Rightarrow BA=-AB$
Now, $(A+B{)}^2=A^2+B^2+AB+BA$
$=A^2+B^2$
Hence, $(A+B{)}^2=A^2+B^2$