MCQ
If $A = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]$, then
- A${A^2} = A$
- B${B^2} = B$
- ✓$AB \ne BA$
- D$AB = BA$
${B^2} = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\4&9\end{array}} \right] \ne B$
Now $AB = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&6\\3&0\end{array}} \right]$
and $BA = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 7}&4\end{array}} \right]$
Obviously, $AB \ne BA$.
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