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 $A^2+B^2$

Given, $A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],B=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$

$A+B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]+\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]=\left[\begin{array}{cc}0& 1-i\\ i+1& 0\end{array}\right]$

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

$A^2=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$B^2=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$A^2+B^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$