Problem Study : Definition of Inverse of a Square Matrix

 
Show that $ \displaystyle A=\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)$ satisfies the equation $ \displaystyle {{A}^{2}}-3A-7I=O$ where $ \displaystyle I$ is a unit matrix of order 2. Hence using this result, find $ \displaystyle {{A}^{{-1}}}$.

Solution

      $ \displaystyle \ \ \ \ A\ \ =\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)$

      $ \displaystyle \begin{array}{l}\therefore {{A}^{2}}=\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {25-3} \\ {-5+2} \end{array}\ \ \ \begin{array}{*{20}{c}} {15-6} \\ {-3+4} \end{array}} \right)\\\\\ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {22} & 9 \\ {-3} & 1 \end{array}} \right)\end{array}$

      $ \displaystyle \begin{array}{l}\therefore {{A}^{2}}-3A-7I\\\\=\left( {\begin{array}{*{20}{c}} {22} & 9 \\ {-3} & 1 \end{array}} \right)-3\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)-7\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} {22} & 9 \\ {-3} & 1 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-15} \\ 3 \end{array}\ \ \ \begin{array}{*{20}{c}} {-9} \\ 6 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-7} & 0 \\ 0 & {-7} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} {22-15-7} & {9-9} \\ {-3+3} & {1+6-7} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\=O\end{array}$

      $ \displaystyle \therefore {{A}^{2}}-3A-7I=O$

      Multiplying both sides with $ \displaystyle {{A}^{{-1}}}$,

      $ \displaystyle \begin{array}{l}{{A}^{2}}-3A-7I=O\\\\AA{{A}^{{-1}}}-3A{{A}^{{-1}}}-7I{{A}^{{-1}}}=O{{A}^{{-1}}}\\\\AI-3I-7{{A}^{{-1}}}=O\\\\AI-3I=7{{A}^{{-1}}}\end{array}$

      $ \displaystyle \begin{array}{l}\therefore 7{{A}^{{-1}}}=AI-3I\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-2} \end{array}} \right)-3\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {5-3} \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-2-3} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 2 \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-5} \end{array}} \right)\\\\\therefore {{A}^{{-1}}}\ \ =\frac{1}{7}\left( {\begin{array}{*{20}{c}} 2 \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-5} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {\frac{2}{7}} & {\frac{3}{7}} \\ {-\frac{1}{7}} & {-\frac{5}{7}} \end{array}} \right)\end{array}$