Question

Inverse of A[1 2 3]
2 3 1
2 1 2]

Answer 1

We have to find the inverse of,

$A=\left[\begin{tabular}{ccc}1&2&3\\&&&&&\\2&3&1\\&&&&&\\2&1&2\end{tabular}\right]$

Let me have an augmented matrix,

$(A\ |\ I)=\left[\begin{tabular}{ccc|ccc}1&2&3&1&0&0\\&&&&&\\2&3&1&0&1&0\\&&&&&\\2&1&2&0&0&1\end{tabular}\right]$

So let's begin!

$\left[\begin{tabular}{ccc|ccc}1&2&3&1&0&0\\&&&&&\\2&3&1&0&1&0\\&&&&&\\2&1&2&0&0&1\end{tabular}\right]\xrightarrow{R_2-R_3\to R_2}\left[\begin{tabular}{ccc|ccc}1&2&3&1&0&0\\&&&&&\\0&2&-1&0&1&-1\\&&&&&\\2&1&2&0&0&1\end{tabular}\right]\\\\\\\\\left[\begin{tabular}{ccc|ccc}1&2&3&1&0&0\\&&&&&\\0&2&-1&0&1&-1\\&&&&&\\2&1&2&0&0&1\end{tabular}\right]\xrightarrow{R_3-2R_1\to R_3}\left[\begin{tabular}{ccc|ccc}1&2&3&1&0&0\\&&&&&\\0&2&-1&0&1&-1\\&&&&&\\0&-3&-4&-2&0&1\end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&2&3&1&0&0\\&&&&&\\0&2&-1&0&1&-1\\&&&&&\\0&-3&-4&-2&0&1\end{tabular}\right]\xrightarrow{R_1-R_2\to R_1}\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&2&-1&0&1&-1\\&&&&&\\0&-3&-4&-2&0&1\end{tabular}\right]\\\\\\\\\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&2&-1&0&1&-1\\&&&&&\\0&-3&-4&-2&0&1\end{tabular}\right]\xrightarrow{2R_2+R_3\to R_2}\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&-6&-2&2&-1\\&&&&&\\0&-3&-4&-2&0&1\end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&-6&-2&2&-1\\&&&&&\\0&-3&-4&-2&0&1\end{tabular}\right]\xrightarrow{3R_2+R_3\to R_3}\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&-6&-2&2&-1\\&&&&&\\0&0&-22&-8&6&-2\end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&-6&-2&2&-1\\&&&&&\\0&0&-22&-8&6&-2\end{tabular}\right]\xrightarrow{-\frac{1}{2}R_3\to R_3}\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&-6&-2&2&-1\\&&&&&\\0&0&11&4&-3&1\end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&-6&-2&2&-1\\&&&&&\\0&0&11&4&-3&1\end{tabular}\right]\xrightarrow{R_2+R_3\to R_2}\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&5&2&-1&0\\&&&&&\\0&0&11&4&-3&1\end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&5&2&-1&0\\&&&&&\\0&0&11&4&-3&1\end{tabular}\right]\xrightarrow{\frac{1}{11}R_3\to R_3}\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&5&2&-1&0\\&&&&&\\0&0&1& \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&4&1&-1&1\\&&&&&\\0&1&5&2&-1&0\\&&&&&\\0&0&1& \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]\xrightarrow{R_1-4R_3\to R_1}\left[\begin{tabular}{ccc|ccc}1&0&0& -\dfrac{5}{11} & \dfrac{1}{11} & \dfrac{7}{11} \\&&&&&\\0&1&5&2&-1&0\\&&&&&\\0&0&1& \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&0& -\dfrac{5}{11} & \dfrac{1}{11} & \dfrac{7}{11} \\&&&&&\\0&1&5&2&-1&0\\&&&&&\\0&0&1& \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]\xrightarrow{R_2-5R_3\to R_2}\left[\begin{tabular}{ccc|ccc}1&0&0& -\dfrac{5}{11} & \dfrac{1}{11} & \dfrac{7}{11} \\&&&&&\\0&1&0& \dfrac{2}{11} & \dfrac{4}{11} & -\dfrac{5}{11} \\&&&&&\\0&0&1& \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]$

$\left[\begin{tabular}{ccc|ccc}1&0&0& -\dfrac{5}{11} & \dfrac{1}{11} & \dfrac{7}{11} \\&&&&&\\0&1&0& \dfrac{2}{11} & \dfrac{4}{11} & -\dfrac{5}{11} \\&&&&&\\0&0&1& \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]=(I\ |\ A^{-1})$

Finally,

$A^{-1}=\left[\begin{tabular}{ccc} -\dfrac{5}{11} & \dfrac{1}{11} & \dfrac{7}{11} \\\\ \dfrac{2}{11} & \dfrac{4}{11} & -\dfrac{5}{11} \\\\ \dfrac{4}{11} & -\dfrac{3}{11} & \dfrac{1}{11} \end{tabular}\right]$

Or,

$11A^{-1}=\left[\begin{tabular}{ccc} -5 & 1 & 7 \\\\ 2 & 4 & -5 \\\\ 4 & -3 & 1 \end{tabular}\right]$