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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],B=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]$
then $(A+B)^2$ equals
MathematicsMatricesJEE Main
Options:
  • A $A^2+B^2$
  • B $A^2+B^2+2 A B$
  • C $A^2+B^2+A B-B A$
  • D None of these
Solution:
2244 Upvotes Verified Answer
The correct answer is: $A^2+B^2$
$A B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]$ $=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]$ and
$B A=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$$=\left[\begin{array}{cc}-i & 0 \\ 0 & i\end{array}\right]=-A B$
$\therefore A B+B A=0$
Hence, $(A+B)^2=A^2+B^2$

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