Row operations not only can be used for solving any system of linear equations, but can also help us find the inverse of a square matrix, especially when the square matrix is larger than 2 x 2. We illustrate the method through the following example:[br][br][u]Example[/u]: Let [math]A=\begin{pmatrix}2 & 7 & 1 \\ 1 & 4 & -1 \\ 1 & 3 & 0\end{pmatrix}[/math]. Find [math]A^{-1}[/math] if exists.[br][br]We write down the augmented matrix of the form [math]\left( \ A \ | \ I \right)[/math], where [math]I[/math] is the 3 x 3 identity matrix:[br][br][math]\left(\begin{array}{ccc|ccc}2 & 7 & 1 & 1 & 0 & 0\\1 & 4 & -1 & 0 & 1 & 0 \\ 1 & 3 & 0 & 0 & 0 & 1\end{array}\right)[/math][br][br]Then use the row reduction algorithm to transform the matrix [math]A[/math] into the one in [u]reduced echelon form[/u]. It is very important that when you do each row operation on the augmented matrix, you must do it for the [u]entire row[/u] i.e. including the entries on the right side of the vertical line. The following are the computation steps:[br][br][math]\left(\begin{array}{ccc|ccc}2 & 7 & 1 & 1 & 0 & 0\\1 & 4 & -1 & 0 & 1 & 0 \\ 1 & 3 & 0 & 0 & 0 & 1\end{array}\right)\rightarrow \left(\begin{array}{ccc|ccc}1 & 4 & -1 & 0 & 1 & 0 \\2 & 7 & 1 & 1 & 0 & 0\\ 1 & 3 & 0 & 0 & 0 & 1\end{array}\right)[/math][br][math]\rightarrow \left(\begin{array}{ccc|ccc}1 & 4 & -1 & 0 & 1 & 0 \\0 & -1 & 3 & 1 & -2 & 0\\ 0 & -1 & 1 & 0 & -1 & 1\end{array}\right)\rightarrow \left(\begin{array}{ccc|ccc}1 & 0 & 11 & 4 & -7 & 0 \\0 & 1 & -3 & -1 & 2 & 0\\ 0 & 0 & -2 & -1 & 1 & 1\end{array}\right)[/math][br][math]\rightarrow \left(\begin{array}{ccc|ccc}1 & 0 & 0 & -\frac32 & -\frac32 & -\frac{11}2 \\0 & 1 & 0 & \frac12 & \frac12 & -\frac32 \\ 0 & 0 & 1 & \frac12 & -\frac12 & \frac12\end{array}\right)[/math][br][br]The 3 x 3 matrix on the right side of the vertical line is the inverse of [math]A[/math]. That is, [br][br][math]A^{-1}=\begin{pmatrix}-\frac32 & -\frac32 & -\frac{11}2 \\\frac12 & \frac12 & -\frac32 \\\frac12 & -\frac12 & \frac12\end{pmatrix}[/math][br][br]For any invertible matrix, the matrix in reduced echelon form that you will obtain by the above method must be an identity matrix. If the matrix in reduced echelon form is not an identity matrix, it means the given matrix is [u]NOT invertible[/u].[br][br]

As we know, we can express the row reduction steps in terms of elementary matrices. Given an invertible matrix [math]A[/math]. Suppose we transform the augmented matrix [math]\left( A \ | \ I \right)[/math] to [math]\left( I \ | \ B \right)[/math] after r steps of row operations, where [math]B[/math] is the matrix appeared on the right side of the vertical line in the last step. There exists a sequence of elementary matrices [math]E_1, E_2, \ldots, E_r[/math] such that [br][br][math]E_r\cdots E_1\left( A \ | \ I \right)=\left(E_r\cdots E_1A \ | \ E_r\cdots E_1I \right)=\left( I \ | \ B \right)[/math][br][br]Hence, we have [math]E_r\cdots E_1A=I[/math] and [math]E_r\cdots E_1=E_r\cdots E_1I=B[/math]. In other words, [math]BA=I[/math] i.e. [math]B=A^{-1}[/math].