Recipe

Find a matrix inverse by row reduction

Use when: you have a square matrix $A$ and need $A^{-1}$ , for a size where no small shortcut formula applies

Goal: either the actual matrix $A^{-1}$ , or a confirmed conclusion that $A$ is singular and has none

Taught in Computing an inverse by row reduction

Quick recipe

Bolt the identity onto $A$ 's right side
Row-reduce all the way to reduced row echelon form
Verify by multiplying back

1
Form the augmented matrix $[\,A\mid I_n\,]$ , with $A$ on the left and the same-size identity on the right.

Example

Take $M = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ . Bolting $I_2$ onto its right side gives $\left[\begin{array}{cc|cc} 2 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right]$

Why: applying every row operation to this single object is what makes the identity's block track, step by step, exactly what those operations do to $A$ .
2
Use the same three row operations as Gaussian elimination, but continue past row echelon form until every pivot reads $1$ and every entry above and below each pivot reads $0$ .

Example

Continuing with $[\,M \mid I_2\,]$ from Step 1, swap rows so the top-left entry is already a $1$ ( $R_1 \leftrightarrow R_2$ ): $\left[\begin{array}{cc|cc} 1 & 1 & 0 & 1 \\ 2 & 1 & 1 & 0 \end{array}\right]$ Clear below the first pivot with $R_2 \to R_2 - 2R_1$ : $\left[\begin{array}{cc|cc} 1 & 1 & 0 & 1 \\ 0 & -1 & 1 & -2 \end{array}\right]$ Scale the second row so its pivot reads $1$ : $R_2 \to -R_2$ : $\left[\begin{array}{cc|cc} 1 & 1 & 0 & 1 \\ 0 & 1 & -1 & 2 \end{array}\right]$ Clear above the second pivot with $R_1 \to R_1 - R_2$ : $\left[\begin{array}{cc|cc} 1 & 0 & 1 & -1 \\ 0 & 1 & -1 & 2 \end{array}\right]$ The left block has reached $I_2$ , so read $M^{-1}$ straight off the right block: $M^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$
- the left block reaches $I_n$ — the right block has become $A^{-1}$ ; read it off.
- the left block develops an all-zero row before reaching $I_n$ — stop; $A$ is singular and has no inverse.
Why: carrying $A$ to $I_n$ through a sequence of elementary row operations is the same thing, by definition, as multiplying $A$ by the matrix those operations build, and that product equaling $I_n$ is exactly what it means to be $A^{-1}$ .
3
Compute $AA^{-1}$ (or $A^{-1}A$ ) and confirm it equals $I_n$ .

Example

$MM^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2(1)+1(-1) & 2(-1)+1(2) \\ 1(1)+1(-1) & 1(-1)+1(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
That's $I_2$ exactly, so $M^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$ checks out.

Why: this is the definition of an inverse being checked directly, not assumed from the row-reduction bookkeeping alone.

Common mistakes

Applying a row operation to only the left block and forgetting to apply the exact same operation to the right block.
Stopping at row echelon form instead of continuing to REDUCED row echelon form (clearing above each pivot too, not just below).
Trying to fix an all-zero row by scaling it, which can never manufacture a nonzero pivot from nothing.
Reporting the row-reduced left block as the answer instead of reading $A^{-1}$ off the right block.