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# Find the inverse matrix using Gauss-Jordan method

So far our knowledge about inverse matrices is restricted to $2×2$$2×2$2xx22\times 2 matrices. The following method will allow us to find the inverse of any dimensional matrix - if it exists - using Gauss-Jordan method.
We are given a matrix and the identity of the same dimentions:
$A=\left(\begin{array}{ccc}{a}_{11}& \cdots & {a}_{1n}\\ ⋮& \ddots & ⋮\\ {a}_{n1}& \cdots & {a}_{nn}\end{array}\right)\text{and}{I}_{n}=\left(\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & 1\end{array}\right)$A=([a_(11),cdots,a_(1n)],[vdots,ddots,vdots],[a_(n1),cdots,a_(nn)])" and "I_(n)=([1,cdots,0],[vdots,ddots,vdots],[0,cdots,1])A= \left(\begin{array}{ccc}a_{11}&\cdots & a_{1n}\\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn}\end{array}\right) \text{ and } I_n= \left(\begin{array}{ccc}1&\cdots & 0\\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1\end{array}\right) If we define the matrix $\left(\begin{array}{cccccc}{a}_{11}& \cdots & {a}_{1n}& 1& \cdots & 0\\ ⋮& \ddots & ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& \cdots & {a}_{nn}& 0& \cdots & 1\end{array}\right)=\left(A\mid I\right)$$\left(\begin{array}{cccccc}{a}_{11}& \cdots & {a}_{1n}& 1& \cdots & 0\\ ⋮& \ddots & ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}& \cdots & {a}_{nn}& 0& \cdots & 1\end{array}\right)=\left(A\mid I\right)$([a_(11),cdots,a_(1n),1,cdots,0],[vdots,ddots,vdots,vdots,ddots,vdots],[a_(n1),cdots,a_(nn),0,cdots,1])=(A∣I)\left(\begin{array}{ccc | ccc}a_{11}&\cdots & a_{1n} & 1&\cdots & 0\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ a_{n1} & \cdots & a_{nn}& 0 & \cdots & 1\end{array}\right)= \left(A\mid I\right)
And we apply now Gauss-Jordan so that matrix $\mathbf{A}$$\mathbf{A}$A\mathbf{A} is transformed into $\mathbf{I}$$\mathbf{I}$I\mathbf{I}, then our initial right part of $\left(A\mid I\right)$$\left(A\mid I\right)$(A∣I)\left(A\mid I\right) will be transformed into ${\mathbf{A}}^{-1}$${\mathbf{A}}^{-1}$A^(-1)\mathbf{A}^{-1} finalising the process as
$\left(\begin{array}{cccccc}1& \cdots & 0& {a}_{11}^{-1}& \cdots & {a}_{1n}^{-1}\\ ⋮& \ddots & ⋮& ⋮& \ddots & ⋮\\ 0& \cdots & 1& {a}_{n1}^{-1}& \cdots & {a}_{nn}^{-1}\end{array}\right)=\left(I\mid {A}^{-1}\right)$$\left(\begin{array}{cccccc}1& \cdots & 0& {a}_{11}^{-1}& \cdots & {a}_{1n}^{-1}\\ ⋮& \ddots & ⋮& ⋮& \ddots & ⋮\\ 0& \cdots & 1& {a}_{n1}^{-1}& \cdots & {a}_{nn}^{-1}\end{array}\right)=\left(I\mid {A}^{-1}\right)$([1,cdots,0,a_(11)^(-1),cdots,a_(1n)^(-1)],[vdots,ddots,vdots,vdots,ddots,vdots],[0,cdots,1,a_(n1)^(-1),cdots,a_(nn)^(-1)])=(I∣A^(-1))\left(\begin{array}{ccc | ccc}1&\cdots & 0 & a_{11}^{-1}&\cdots & a_{1n}^{-1}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ 0 & \cdots & 1& a_{n1}^{-1} & \cdots & a_{nn}^{-1}\end{array}\right)= \left(I\mid A^{-1}\right)
Find the inverse of
$A=\left(\begin{array}{rrr}8& 4& 4\\ -3& -9& -6\\ 1& -7& -5\end{array}\right)$$A=\left(\begin{array}{r}8 4 4\\ -3 -9 -6\\ 1 -7 -5\end{array}\right)$A=([8,4,4],[-3,-9,-6],[1,-7,-5])A= \left(\begin{array}{rrr}8 & 4 & 4 \\ -3 & -9 & -6 \\ 1 & -7 & -5 \end{array}\right)