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
    $AB = BA$

Answer

Correct option: C.
$AB \ne BA$
c
(c) Since ${A^2} = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 4}&2\\{ - 3}&{ - 6}\end{array}} \right] \ne A$

${B^2} = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\4&9\end{array}} \right] \ne B$

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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