The Gauss-Jordan method

The 2 × 2 formula does not generalise simply. Beyond that, the inverse is computed by Gaussian elimination carried all the way: the Gauss-Jordan method.

The principle

Write A and the identity matrix side by side, then apply elementary row operations to both halves at once, until the left part becomes the identity:

   [ A | I ]     -----------------> [ I | A⁻¹ ]
                 row operations

Why it works: each elementary operation amounts to multiplying on the left by an invertible matrix. If the sequence of operations turns A into I, then that same sequence is A⁻¹ — and applying it to I makes it appear.

A worked example

Let us invert A = [[1, 0, 2], [2, -1, 3], [4, 1, 8]].

[ 1   0   2 | 1  0  0 ]
[ 2  -1   3 | 0  1  0 ]      R2 <- R2 - 2 R1
[ 4   1   8 | 0  0  1 ]      R3 <- R3 - 4 R1

[ 1   0   2 |  1  0  0 ]
[ 0  -1  -1 | -2  1  0 ]     R2 <- -R2
[ 0   1   0 | -4  0  1 ]

[ 1   0   2 |  1  0  0 ]
[ 0   1   1 |  2 -1  0 ]     R3 <- R3 - R2
[ 0   1   0 | -4  0  1 ]

[ 1   0   2 |  1  0  0 ]
[ 0   1   1 |  2 -1  0 ]     R3 <- -R3
[ 0   0  -1 | -6  1  1 ]

The left part is triangular; it remains to go back up to clear what sits above the pivots:

[ 1   0   2 |  1  0  0 ]     R1 <- R1 - 2 R3
[ 0   1   1 |  2 -1  0 ]     R2 <- R2 - R3
[ 0   0   1 |  6 -1 -1 ]

[ 1   0   0 | -11   2   2 ]
[ 0   1   0 |  -4   0   1 ]        <-  A⁻¹ has appeared on the right
[ 0   0   1 |   6  -1  -1 ]

The check, which is essential: multiply A by the result and you must land on the identity.

[ 1  0  2 ]   [ -11   2   2 ]   [ 1  0  0 ]
[ 2 -1  3 ] × [  -4   0   1 ] = [ 0  1  0 ]   ✔
[ 4  1  8 ]   [   6  -1  -1 ]   [ 0  0  1 ]

When the method fails

If, during elimination, a row of the left part becomes entirely zero, it is over: A is not invertible.

[ 1   2 | 1  0 ]     R2 <- R2 - 2 R1     [ 1   2 |  1  0 ]
[ 2   4 | 0  1 ]                         [ 0   0 | -2  1 ]
                                           ^^^^^
                                 no pivot can be obtained here

So the algorithm does not merely compute the inverse: it also decides whether it exists. That is an advantage over the 2 × 2 formula, which presupposes the determinant is known.

The cost

Gauss-Jordan on an n × n matrix   ->   about n³ operations

That is acceptable, but it is roughly three times the cost of simply solving a system. Hence the practical rule of the next lesson: invert a matrix only when you genuinely need the inverse itself.

Summary

• Gauss-Jordan: start from [ A | I ] and end at [ I | A⁻¹ ].
• Elementary operations apply to both halves simultaneously.
• Go down to reach echelon form, then back up to clear above the pivots.
• An entirely zero row on the left signals a non-invertible matrix.
• Always check by computing A × A⁻¹ = I.
• Cost: about n³, roughly three system solves.