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Question

# Find the inverse of each of the following matrices and verify that ${A}^{-1}A={I}_{3}$. (i) $\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$ (ii) $\left[\begin{array}{ccc}2& 3& 1\\ 3& 4& 1\\ 3& 7& 2\end{array}\right]$

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Solution

## $\left(\mathrm{i}\right)A=\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{C}_{11}=\left|\begin{array}{cc}4& 3\\ 3& 4\end{array}\right|=7,{C}_{12}=-\left|\begin{array}{cc}1& 3\\ 1& 4\end{array}\right|=-1\mathrm{and}{C}_{13}=\left|\begin{array}{cc}1& 4\\ 1& 3\end{array}\right|=-1\phantom{\rule{0ex}{0ex}}{C}_{21}=-\left|\begin{array}{cc}3& 3\\ 3& 4\end{array}\right|=-3,{C}_{22}=\left|\begin{array}{cc}1& 3\\ 1& 4\end{array}\right|=1\mathrm{and}{C}_{23}=-\left|\begin{array}{cc}1& 3\\ 1& 3\end{array}\right|=0\phantom{\rule{0ex}{0ex}}{C}_{31}=\left|\begin{array}{cc}3& 3\\ 4& 3\end{array}\right|=-3,{C}_{32}=-\left|\begin{array}{cc}1& 3\\ 1& 3\end{array}\right|=0\mathrm{and}{C}_{33}=\left|\begin{array}{cc}1& 3\\ 1& 4\end{array}\right|=1\phantom{\rule{0ex}{0ex}}\mathrm{adj}A={\left[\begin{array}{ccc}7& -1& -1\\ -3& 1& 0\\ -3& 0& 1\end{array}\right]}^{T}=\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{and}\left|A\right|=1\phantom{\rule{0ex}{0ex}}{A}^{-1}=\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Now},{A}^{-1}A=\left[\begin{array}{ccc}7& -3& -3\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]={I}_{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)B=\left[\begin{array}{ccc}2& 3& 1\\ 3& 4& 1\\ 3& 7& 2\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{C}_{11}=\left|\begin{array}{cc}4& 1\\ 7& 2\end{array}\right|=1,{C}_{12}=-\left|\begin{array}{cc}3& 1\\ 3& 2\end{array}\right|=-3\mathrm{and}{C}_{13}=\left|\begin{array}{cc}3& 4\\ 3& 7\end{array}\right|=9\phantom{\rule{0ex}{0ex}}{C}_{21}=-\left|\begin{array}{cc}3& 1\\ 7& 2\end{array}\right|=1,{C}_{22}=\left|\begin{array}{cc}2& 1\\ 3& 2\end{array}\right|=1\mathrm{and}{C}_{23}=-\left|\begin{array}{cc}2& 3\\ 3& 7\end{array}\right|=-5\phantom{\rule{0ex}{0ex}}{C}_{31}=\left|\begin{array}{cc}3& 1\\ 4& 1\end{array}\right|=-1,{C}_{32}=-\left|\begin{array}{cc}2& 1\\ 3& 1\end{array}\right|=1\mathrm{and}{C}_{33}=\left|\begin{array}{cc}2& 3\\ 3& 4\end{array}\right|=-1\phantom{\rule{0ex}{0ex}}\mathrm{adj}B={\left[\begin{array}{ccc}1& -3& 9\\ 1& 1& -5\\ -1& 1& -1\end{array}\right]}^{T}=\left[\begin{array}{ccc}1& 1& -1\\ -3& 1& 1\\ 9& -5& -1\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{and}\left|B\right|=2\phantom{\rule{0ex}{0ex}}{B}^{-1}=\frac{1}{2}\left[\begin{array}{ccc}1& 1& -1\\ -3& 1& 1\\ 9& -5& -1\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{Now},{B}^{-1}B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]={I}_{3}$

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