Question

Find the inverse of the following matrix, using elementary transformations: 

${\displaystyle A=\left[\begin{array}{ccc}2& 3& 1\\ 2& 4& 1\\ 3& 7& 2\end{array}\right]}$

Answer 1

${\displaystyle A=\left[\begin{array}{ccc}2& 3& 1\\ 2& 4& 1\\ 3& 7& 2\end{array}\right]}$ 

AA^-1 =I 

${\displaystyle \left[\begin{array}{ccc}2& 3& 1\\ 2& 4& 1\\ 3& 7& 2\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}$ 

R₂ → R₂ - R₁

R₃ → R₃ - R₁

${\displaystyle \left[\begin{array}{ccc}2& 3& 1\\ 0& 1& 0\\ 1& 4& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ -1& 0& 1\end{array}\right]}$ 

R₁ ↔ R₃ 

${\displaystyle \left[\begin{array}{ccc}1& 4& 1\\ 0& 1& 0\\ 2& 3& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}-1& 0& 1\\ -1& 1& 0\\ 1& 0& 0\end{array}\right]}$ 

R₃ → R₃ - 2R₁

${\displaystyle \left[\begin{array}{ccc}1& 4& 1\\ 0& 1& 0\\ 0& -5& -1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}-1& 0& 1\\ -1& 1& 0\\ 3& 0& -2\end{array}\right]}$ 

R₁ → R₁ - 4R₂

R₃ → R₃ - 5R₂

${\displaystyle \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}3& -4& 1\\ -1& 1& 0\\ -2& 5& -2\end{array}\right]}$ 

${\displaystyle R_1\to R_1+R_3}$

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1& 1& -1\\ -1& 1& 0\\ -2& 5& -2\end{array}\right]}$ 

${\displaystyle R_3\to -R_3}$

${\displaystyle \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1& 1& -1\\ -1& 1& 0\\ 2& -5& 2\end{array}\right]}$ 

${\displaystyle I.A^1=\left[\begin{array}{ccc}1& 1& -1\\ -1& 1& 0\\ 2& -5& 2\end{array}\right]}$ 

${\displaystyle \Rightarrow A^1=\left[\begin{array}{ccc}1& 1& -1\\ -1& 1& 0\\ 2& -5& 2\end{array}\right]}$