A) \[\frac{1}{10}\left( \begin{matrix} 4 & 2 \\ -3 & 1 \\ \end{matrix} \right)\]
B) \[\frac{1}{10}\left( \begin{matrix} 1 & -2 \\ 3 & 4 \\ \end{matrix} \right)\]
C) \[\frac{1}{10}\left( \begin{matrix} 4 & 2 \\ -3 & 1 \\ \end{matrix} \right)\]
D) \[\left( \begin{matrix} 4 & 2 \\ -3 & 1 \\ \end{matrix} \right)\]
Correct Answer: A
Solution :
Let \[A=\left[ \begin{matrix} 1 & -2 \\ 3 & 4 \\ \end{matrix} \right]\] \[|A|\,\,=4+6=10\ne 0\] Now, \[{{A}_{11}}=4\], \[a=-6,\,A=\left| \,\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & -2 & -5 \\ \end{matrix}\, \right|\], \[{{A}_{21}}=-(-2)=2\], \[{{A}_{22}}=1\] \[\therefore \]\[adj\,(A)=\left[ \begin{matrix} 4 & 2 \\ -3 & 1 \\ \end{matrix} \right]\] \[\therefore \] \[{{A}^{-1}}=\frac{adj\,(A)}{|A|}=\frac{1}{10}\left[ \begin{matrix} 4 & 2 \\ -3 & 1 \\ \end{matrix} \right]\]. Trick: Check from the options \[A{{A}^{-1}}=I\,\] Þ \[A{{A}^{-1}}\]=\[\left[ \begin{matrix} 1 & -2 \\ 3 & 4 \\ \end{matrix} \right]\,\left[ \begin{matrix} \frac{4}{10} & \frac{2}{10} \\ \frac{-3}{10} & \frac{1}{10} \\ \end{matrix} \right]\]= \[\left[ \begin{matrix} \frac{10}{10} & 0 \\ 0 & \frac{10}{10} \\ \end{matrix} \right]\] = \[A{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=I\].
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