- Home
- Standard 12
- Mathematics
જો $A = \left[ {\begin{array}{*{20}{c}}1&2\\{ - 3}&0\end{array}} \right]$ અને $B = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\2&3\end{array}} \right]$, તો
${A^2} = A$
${B^2} = B$
$AB \ne BA$
$AB = BA$
Solution
(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, $AB \ne BA$.
