Consider the matrix defined in Example MI as, \begin{equation*} A= \begin{bmatrix} 1 & 2 & 1 & 2 & 1 \\ -2 & -3 & 0 & -5 & -1 \\ 1 & 1 & 0 & 2 & 1 \\ -2 & -3 & -1 & -3 & -2 \\ -1 & -3 & -1 & -3 & 1 \end{bmatrix} \end{equation*}
For its inverse, we desire a matrix $B$ so that $AB=I_5$. Emphasizing the structure of the columns and employing the definition of matrix multiplication Definition MM, \begin{align*} AB&=I_5\\ A\lbrack \vect{B}_1|\vect{B}_2|\vect{B}_3|\vect{B}_4|\vect{B}_5]&=[\vect{e}_1|\vect{e}_2|\vect{e}_3|\vect{e}_4|\vect{e}_5\rbrack\\ \lbrack A\vect{B}_1|A\vect{B}_2|A\vect{B}_3|A\vect{B}_4|A\vect{B}_5]&=[\vect{e}_1|\vect{e}_2|\vect{e}_3|\vect{e}_4|\vect{e}_5\rbrack \end{align*}
Equating the matrices column-by-column we have \begin{align*} A\vect{B}_1=\vect{e}_1&& A\vect{B}_2=\vect{e}_2&& A\vect{B}_3=\vect{e}_3&& A\vect{B}_4=\vect{e}_4&& A\vect{B}_5=\vect{e}_5. \end{align*}
Since the matrix $B$ is what we are trying to compute, we can view each column, $\vect{B}_i$, as a column vector of unknowns. Then we have five systems of equations to solve, each with 5 equations in 5 variables. Notice that all 5 of these systems have the same coefficient matrix. We will now solve each system in turn, \begin{align*} \ & \end{align*} Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_1}$, \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 2 & 1 & 1\\ -2 & -3 & 0 & -5 & -1 & 0\\ 1 & 1 & 0 & 2 & 1 & 0\\ -2 & -3 & -1 & -3 & -2 & 0\\ -1 & -3 & -1 & -3 & 1 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & -3\\ 0 & \leading{1} & 0 & 0 & 0 & 0\\ 0 & 0 & \leading{1} & 0 & 0 & 1\\ 0 & 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & 0 & \leading{1} & 1 \end{bmatrix} ; \vect{B}_1=\colvector{-3\\0\\1\\1\\1}\\ \end{align*} Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_2}$, \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 2 & 1 & 0\\ -2 & -3 & 0 & -5 & -1 & 1\\ 1 & 1 & 0 & 2 & 1 & 0\\ -2 & -3 & -1 & -3 & -2 & 0\\ -1 & -3 & -1 & -3 & 1 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & 3\\ 0 & \leading{1} & 0 & 0 & 0 & -2\\ 0 & 0 & \leading{1} & 0 & 0 & 2\\ 0 & 0 & 0 & \leading{1} & 0 & 0\\ 0 & 0 & 0 & 0 & \leading{1} & -1 \end{bmatrix} ; \vect{B}_2=\colvector{3\\-2\\2\\0\\-1}\\ \end{align*} Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_3}$, \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 2 & 1 & 0\\ -2 & -3 & 0 & -5 & -1 & 0\\ 1 & 1 & 0 & 2 & 1 & 1\\ -2 & -3 & -1 & -3 & -2 & 0\\ -1 & -3 & -1 & -3 & 1 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & 6\\ 0 & \leading{1} & 0 & 0 & 0 & -5\\ 0 & 0 & \leading{1} & 0 & 0 & 4\\ 0 & 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & 0 & \leading{1} & -2 \end{bmatrix} ; \vect{B}_3=\colvector{6\\-5\\4\\1\\-2}\\ \end{align*} Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_4}$, \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 2 & 1 & 0\\ -2 & -3 & 0 & -5 & -1 & 0\\ 1 & 1 & 0 & 2 & 1 & 0\\ -2 & -3 & -1 & -3 & -2 & 1\\ -1 & -3 & -1 & -3 & 1 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & -1\\ 0 & \leading{1} & 0 & 0 & 0 & -1\\ 0 & 0 & \leading{1} & 0 & 0 & 1\\ 0 & 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & 0 & \leading{1} & 0 \end{bmatrix} ; \vect{B}_4=\colvector{-1\\-1\\1\\1\\0}\\ \end{align*} Row-reduce the augmented matrix of the linear system $\linearsystem{A}{\vect{e}_5}$, \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 2 & 1 & 0\\ -2 & -3 & 0 & -5 & -1 & 0\\ 1 & 1 & 0 & 2 & 1 & 0\\ -2 & -3 & -1 & -3 & -2 & 0\\ -1 & -3 & -1 & -3 & 1 & 1 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 0 & -2\\ 0 & \leading{1} & 0 & 0 & 0 & 1\\ 0 & 0 & \leading{1} & 0 & 0 & -1\\ 0 & 0 & 0 & \leading{1} & 0 & 0\\ 0 & 0 & 0 & 0 & \leading{1} & 1 \end{bmatrix} ; \vect{B}_5=\colvector{-2\\1\\-1\\0\\1}\\ \end{align*} We can now collect our 5 solution vectors into the matrix $B$, \begin{align*} B= &[\vect{B}_1|\vect{B}_2|\vect{B}_3|\vect{B}_4|\vect{B}_5]\\ =& \left[\colvector{-3\\0\\1\\1\\1} \left\lvert\colvector{3\\-2\\2\\0\\-1}\right. \left\lvert\colvector{6\\-5\\4\\1\\-2}\right. \left\lvert\colvector{-1\\-1\\1\\1\\0}\right. \left\lvert\colvector{-2\\1\\-1\\0\\1}\right. \right]\\ &= \begin{bmatrix} -3 & 3 & 6 & -1 & -2 \\ 0 & -2 & -5 & -1 & 1 \\ 1 & 2 & 4 & 1 & -1 \\ 1 & 0 & 1 & 1 & 0 \\ 1 & -1 & -2 & 0 & 1 \end{bmatrix} \end{align*}
By this method, we know that $AB=I_5$. Check that $BA=I_5$, and then we will know that we have the inverse of $A$.