Calculating the inverse of a matrix through row reduction

Another way to calculate the inverse of a matrix is to use row reduction (also called Gaussian elimination). The rules of row reduction are

• multiply any row by a nonzero constant \(\lambda \neq 0\);

• interchange rows;

• add a multiple of one row to another.

To find the inverse of \(A\), we set up a new matrix made up of \(A\) and \(I\) next to each other with a line between them just to remind us where the separation is between the orignal matrices i.e. we write a rectangular matrix of the form \((A|I)\). Then we apply rules of row reduction until we obtain the form \((I|B)\). Then \(B = A^{-1}\). An example will make this method clearer. Suppose we have a matrix \(A\) with

\[\begin{split} \begin{aligned} A = \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix}\, . \end{aligned} \end{split}\]

Row reduction proceeds as follows: first we remove twice the first row from the second row and then we remove three times the second row from the first row i.e.

\[\begin{split} \begin{aligned} \left( \begin{matrix} 1 & 3 \\ 2 & 7 \end{matrix} \left\vert \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right. \right) \xrightarrow{r_2 \to r_2 - 2 r_1} \left( \begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix} \left\vert \begin{matrix} 1 & 0 \\ -2 & 1 \end{matrix} \right. \right) \xrightarrow{r_1 \to r_1 - 3 r_2} \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \left\vert \begin{matrix} 7 & -3 \\ -2 & 1 \end{matrix} \right. \right) \end{aligned} \end{split}\]

where \(r_j\) denotes the \(j^{\rm th}\) row. The inverse of \(A\) is therefore

\[\begin{split} \begin{aligned} A^{-1} = \begin{pmatrix} 7 & -3 \\ -2 & 1 \end{pmatrix}\, . \end{aligned} \end{split}\]

You can check that indeed \(A^{-1} A = I\).