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Module 2/Systems & Gaussian Elimination

Computing the Inverse by Row Reduction

Once you can row-reduce a matrix, you can invert one. This lesson turns Gauss–Jordan elimination into a machine for computing A−1A^{-1}: augment AA with the identity, reduce the left block to InI_n, and read the inverse off the right block. You will also learn to spot, mid-reduction, exactly when a matrix has no inverse at all.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

Compute the inverse of A=[120230001]A=\begin{bmatrix}1&2&0\\2&3&0\\0&0&1\end{bmatrix} by row reduction.

True or false: a 4×44\times4 matrix whose reduced row echelon form has exactly three leading 1's (pivots) is invertible.

What you’ll be able to do

  • State and use the criterion that a square matrix AA is invertible if and only if rref⁡(A)=In\operatorname{rref}(A)=I_n, equivalently rank⁡(A)=n\operatorname{rank}(A)=n.
  • Explain why applying the row operations that carry AA to InI_n to the identity matrix produces A−1A^{-1} (Theorem 5.3).
  • Execute the augmented-matrix algorithm [ A∣In ]→[ In∣A−1 ][\,A\mid I_n\,]\to[\,I_n\mid A^{-1}\,] on 2×22\times2 and 3×33\times3 matrices.
  • Detect non-invertibility during elimination by recognising a missing pivot (a zero row), i.e. rank⁡(A)<n\operatorname{rank}(A)<n.
  • Verify a computed inverse with the check AA−1=InAA^{-1}=I_n.

In your course

· MATH2015 · Linear Algebra & Probability
§5.3 Calculating the inverse of a matrix
  • Theorem 5.3Computing the inverse by row operations
    If an invertible n×nn\times n matrix AA is reduced to InI_n by elementary row operations, applying the same operations in the same order to InI_n yields A−1A^{-1}.
  • Theorem 5.2Elementary operations as matrix multiplication
    An elementary row operation on AA equals left-multiplication by the corresponding elementary matrix.
  • Theorem 5.1Rank is preserved by row operations
    Elementary row transformations do not change the row rank of a matrix.
  • Example 5.5Inverse of [1022−13418]\begin{bmatrix}1&0&2\\2&-1&3\\4&1&8\end{bmatrix}
    A−1=[−1122−4016−1−1]A^{-1}=\begin{bmatrix}-11&2&2\\-4&0&1\\6&-1&-1\end{bmatrix}.
  • Example 5.6A singular matrix
    [123456789]\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix} has rank 22, so it is not invertible.
The invertibility criterion A invertible  ⟺  rref⁡(A)=In  ⟺  rank⁡(A)=nA\text{ invertible}\iff\operatorname{rref}(A)=I_n\iff\operatorname{rank}(A)=n opens §5.3, where the method and both worked examples are stated. Elementary row operations are Definition 5.1 (§5.1); reduced row echelon form is Definition 5.4 (§5.2).
1

Invertible means full rank

A square n×nn\times n matrix AA is invertible exactly when it has full rank, rank⁡(A)=n\operatorname{rank}(A)=n. Because row operations never change the rank (Theorem 5.1), this is the same as saying the reduced row echelon form of AA is the identity: A is invertible  ⟺  rref⁡(A)=In.A\text{ is invertible}\iff\operatorname{rref}(A)=I_n. When this holds, every column carries a pivot (a leading 11), so there is no free variable and no zero row.

2

The augmented-matrix algorithm

To invert AA, write the n×2nn\times 2n augmented matrix [ A∣In ][\,A\mid I_n\,] and run Gauss–Jordan elimination until the left block becomes InI_n. The right block is then A−1A^{-1}: [ A∣In ] → row reduce  [ In∣A−1 ].[\,A\mid I_n\,]\ \xrightarrow{\ \text{row reduce}\ }\ [\,I_n\mid A^{-1}\,]. Every operation you apply to the left is applied simultaneously to the right, so the right block records exactly the sequence of operations that turned AA into InI_n.

3

Reading off non-invertibility

If at any stage a row of the left block becomes all zeros, that block can never become InI_n: a pivot is missing and rank⁡(A)<n\operatorname{rank}(A)<n. The matrix is not invertible, and you may stop immediately—you do not even need to reach reduced row echelon form, since the gap already shows at the row-echelon stage.

4

Always check your answer

Row reduction has many arithmetic steps, so verify the result by multiplying: a correct inverse satisfies AA−1=InAA^{-1}=I_n (and then A−1A=InA^{-1}A=I_n automatically). This single matrix product catches almost every slip, and the course notes recommend doing it every time.

Invertibility criterion (§5.3)

A square n×nn\times n matrix AA is invertible if and only if rank⁡(A)=n\operatorname{rank}(A)=n; equivalently, if and only if rref⁡(A)=In\operatorname{rref}(A)=I_n.

Intuition. Full rank means the nn columns are linearly independent, so AA maps Rn\mathbb{R}^n bijectively onto Rn\mathbb{R}^n. Reduction can reach InI_n only when nothing collapses—a zero row would signal a lost dimension.
Theorem 5.3 (computing the inverse)

If an invertible n×nn\times n matrix AA is transformed into the identity matrix InI_n by a sequence of elementary row operations, then applying those same operations, in the same order, to InI_n yields A−1A^{-1}.

Intuition. Each row operation is left-multiplication by an elementary matrix EiE_i (Theorem 5.2). If Ek⋯E1A=InE_k\cdots E_1 A=I_n, then Ek⋯E1=A−1E_k\cdots E_1=A^{-1}, and that product is exactly what you obtain by applying the same operations to InI_n. The augmented matrix [ A∣I ][\,A\mid I\,] simply carries both computations side by side.
Theorem 5.2 (elementary matrices)

Each elementary row operation on an m×nm\times n matrix AA equals left-multiplication by the corresponding m×mm\times m elementary matrix, obtained by applying that operation to ImI_m.

Intuition. This is the engine behind Theorem 5.3: it rewrites a whole sequence of row operations as a single matrix product, which can then be inverted and rearranged with ordinary algebra.
Theorem 5.1 (rank is preserved)

Elementary row transformations do not change the (row) rank of a matrix.

Intuition. This is why we may replace AA by its reduced row echelon form when deciding invertibility: the rank we read off at the end is the rank we started with.

Worked examples

Example 1

Find the inverse of A=[1213]A=\begin{bmatrix}1&2\\1&3\end{bmatrix} by row reduction.

  1. 1

    Augment with the identity: [12101301]\left[\begin{array}{cc|cc}1&2&1&0\\1&3&0&1\end{array}\right].

  2. 2

    Clear below the first pivot, R2→R2−R1R_2\to R_2-R_1: [121001−11]\left[\begin{array}{cc|cc}1&2&1&0\\0&1&-1&1\end{array}\right].

  3. 3

    Clear above the second pivot, R1→R1−2R2R_1\to R_1-2R_2: [103−201−11]\left[\begin{array}{cc|cc}1&0&3&-2\\0&1&-1&1\end{array}\right].

  4. 4

    The left block is I2I_2, so the right block is the inverse.

  5. 5

    Check: [1213][3−2−11]=[1001]\begin{bmatrix}1&2\\1&3\end{bmatrix}\begin{bmatrix}3&-2\\-1&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}. ✓

Answer. A−1=[3−2−11]A^{-1}=\begin{bmatrix}3&-2\\-1&1\end{bmatrix}
Example 2

(Example 5.5) Find the inverse of A=[1022−13418]A=\begin{bmatrix}1&0&2\\2&-1&3\\4&1&8\end{bmatrix}.

Example 3

(Example 5.6) Determine whether A=[123456789]A=\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix} is invertible.