Editing Gaussian elimination (section)

=== Finding the inverse of a matrix ===
{{See also|Invertible matrix}}
A variant of Gaussian elimination called '''Gauss–Jordan elimination''' can be used for finding the inverse of a matrix, if it exists. If {{math|''A''}} is an {{math|''n'' × ''n''}} square matrix, then one can use row reduction to compute its [[invertible matrix|inverse matrix]], if it exists. First, the {{math|''n'' × ''n''}} [[identity matrix]] is augmented to the right of {{math|''A''}}, forming an {{math|''n'' × 2''n''}} [[block matrix]] {{math|[''A'' {{!}} ''I'']}}. Now through application of elementary row operations, find the reduced echelon form of this {{math|''n'' × 2''n''}} matrix. The matrix {{math|''A''}} is invertible if and only if the left block can be reduced to the identity matrix {{math|''I''}}; in this case the right block of the final matrix is {{math|''A''<sup>−1</sup>}}. If the algorithm is unable to reduce the left block to {{math|''I''}}, then {{math|''A''}} is not invertible.

For example, consider the following matrix:
<math display="block">A =
 \begin{bmatrix}
  2 & -1 &  0 \\
 -1 &  2 & -1 \\
  0 & -1 &  2
 \end{bmatrix}.
</math>

To find the inverse of this matrix, one takes the following matrix augmented by the identity and row-reduces it as a 3&nbsp;×&nbsp;6 matrix:
<math display="block">[ A | I ] = 
 \left[\begin{array}{ccc|ccc}
   2 & -1 &  0 & 1 & 0 & 0 \\
  -1 &  2 & -1 & 0 & 1 & 0 \\
   0 & -1 &  2 & 0 & 0 & 1
 \end{array}\right].
</math>

By performing row operations, one can check that the reduced row echelon form of this augmented matrix is
<math display="block">[ I | B ] = 
 \left[\begin{array}{rrr|rrr}
  1 & 0 & 0 & \frac34 & \frac12 & \frac14 \\
  0 & 1 & 0 & \frac12 &      1  & \frac12 \\
  0 & 0 & 1 & \frac14 & \frac12 & \frac34
 \end{array}\right].
</math>

One can think of each row operation as the left product by an [[elementary matrix]]. Denoting by {{math|''B''}} the product of these elementary matrices, we showed, on the left, that {{math|1=''BA'' = ''I''}}, and therefore, {{math|1=''B'' = ''A''<sup>−1</sup>}}. On the right, we kept a record of {{math|1=''BI'' = ''B''}}, which we know is the inverse desired. This procedure for finding the inverse works for square matrices of any size.