Inverse via Gauss-Jordan Elimination - Linear Algebra Step by Step

Find A⁻¹ by augmenting A with the identity matrix to form [A|I], then row-reducing until the left block becomes I. The right block becomes A⁻¹. Each row operation is shown step by step.

Gauss-Jordan method Augment: [A|I]. Row-reduce the left block to I using elimination and back-substitution. The right block transforms into A⁻¹ in lockstep.

Example

Reduce [A|I] to [I|A inverse].

highlighted = computed this step

Step 1 — Set up augmented matrix

Set up the augmented matrix for row reduction.

R=\left[\begin{array}{ccc|c}1&2&1&0\\3&7&0&1\end{array}\right]

Step 2 — Update row 2

Row-reduce with R2 <- R2 - 3*R1.

R=\left[\begin{array}{ccc|c}1&2&1&0\\\hl{0}&\hl{1}&\hl{-3}&\hl{1}\end{array}\right]

Step 3 — Update row 1

Row-reduce with R1 <- R1 - 2*R2.

R=\left[\begin{array}{ccc|c}\hl{1}&\hl{0}&\hl{7}&\hl{-2}\\0&1&-3&1\end{array}\right]

Step 4 — Read inverse

Read the inverse matrix A inverse.

Ainv=\begin{bmatrix}7&-2\\-3&1\end{bmatrix},\quad \text{A inverse complete}