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Inverse of a Matrix (Method 2)


Two panels: general 2×2 formula A⁻¹ = (1/(ad−bc)) · [[d,−b],[−c,a]], and a worked example with A = [[4,3],[3,2]], det A = −1, A⁻¹ = [[−2,3],[3,−4]], verified A·A⁻¹ = I.

Here are three ways to find the inverse of a matrix:

1. Shortcut for 2x2 matrices

For , the inverse can be found using this formula:

Example:

2. Augmented matrix method

Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].

Example: The following steps result in .

so we see that .

3. Adjoint method

A-1 = (adjoint of A)   or   A-1 = (cofactor matrix of A)T

Example: The following steps result in A-1 for .

The cofactor matrix for A is , so the adjoint is . Since det A = 22, we get

.

See also

Determinant of a matrix, cofactor

Key Formula

[ A  ∣  I ]→row operations[ I  ∣  A−1 ][\, A \;|\; I \,] \xrightarrow{\text{row operations}} [\, I \;|\; A^{-1} \,]
Where:
  • AA = The square matrix whose inverse you want to find
  • II = The identity matrix of the same size as A
  • A−1A^{-1} = The inverse matrix, which appears on the right side after row reduction

Worked Example

Problem: Find the inverse of the matrix A using the augmented matrix method (Gauss-Jordan elimination), where A = [[2, 1], [5, 3]].
Step 1: Set up the augmented matrix by placing the identity matrix to the right of A.
[ A  ∣  I ]=[21105301][\, A \;|\; I \,] = \left[\begin{array}{cc|cc} 2 & 1 & 1 & 0 \\ 5 & 3 & 0 & 1 \end{array}\right]
Step 2: Make the pivot in row 1, column 1 equal to 1 by dividing row 1 by 2.
R1→12R1:[1121205301]R_1 \to \tfrac{1}{2}R_1 : \left[\begin{array}{cc|cc} 1 & \tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 5 & 3 & 0 & 1 \end{array}\right]
Step 3: Eliminate the 5 below the pivot by replacing row 2 with R₂ − 5R₁.
R2→R2−5R1:[112120012−521]R_2 \to R_2 - 5R_1 : \left[\begin{array}{cc|cc} 1 & \tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{5}{2} & 1 \end{array}\right]
Step 4: Make the pivot in row 2, column 2 equal to 1 by multiplying row 2 by 2.
R2→2R2:[11212001−52]R_2 \to 2R_2 : \left[\begin{array}{cc|cc} 1 & \tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & 1 & -5 & 2 \end{array}\right]
Step 5: Eliminate the ½ above the second pivot by replacing row 1 with R₁ − ½R₂.
R1→R1−12R2:[103−101−52]R_1 \to R_1 - \tfrac{1}{2}R_2 : \left[\begin{array}{cc|cc} 1 & 0 & 3 & -1 \\ 0 & 1 & -5 & 2 \end{array}\right]
Step 6: The left side is now the identity matrix, so the right side is A⁻¹.
A−1=[3−1−52]A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}
Answer: A⁻¹ = [[3, −1], [−5, 2]]. You can verify: A · A⁻¹ = [[2·3+1·(−5), 2·(−1)+1·2], [5·3+3·(−5), 5·(−1)+3·2]] = [[1, 0], [0, 1]] = I.

Another Example

Problem: Find the inverse of the 3×3 matrix B = [[1, 0, 1], [0, 2, 1], [1, 1, 1]] using the augmented matrix method.
Step 1: Write the augmented matrix [B | I].
[101100021010111001]\left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right]
Step 2: Eliminate the 1 in row 3, column 1 by replacing R₃ with R₃ − R₁.
R3→R3−R1:[101100021010010−101]R_3 \to R_3 - R_1 : \left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 & 0 & 1 \end{array}\right]
Step 3: Scale R₂ by ½ to make the pivot 1, then eliminate below and above.
R2→12R2:[10110001120120010−101]R_2 \to \tfrac{1}{2}R_2 : \left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 \\ 0 & 1 & 0 & -1 & 0 & 1 \end{array}\right]
Step 4: Eliminate the 1 in row 3, column 2: R₃ → R₃ − R₂.
R3→R3−R2:[1011000112012000−12−1−121]R_3 \to R_3 - R_2 : \left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 \\ 0 & 0 & -\tfrac{1}{2} & -1 & -\tfrac{1}{2} & 1 \end{array}\right]
Step 5: Scale R₃ by −2, then back-substitute upward to clear column 3.
R3→−2R3:[1011000112012000121−2]R_3 \to -2R_3 : \left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & \tfrac{1}{2} & 0 & \tfrac{1}{2} & 0 \\ 0 & 0 & 1 & 2 & 1 & -2 \end{array}\right]
Step 6: Eliminate column 3 entries above: R₂ → R₂ − ½R₃ and R₁ → R₁ − R₃.
[100−1−12010−10100121−2]\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & -1 & 2 \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 2 & 1 & -2 \end{array}\right]
Answer: B⁻¹ = [[−1, −1, 2], [−1, 0, 1], [2, 1, −2]].

Frequently Asked Questions

How do you know if a matrix has no inverse when using the augmented matrix method?
If at any point during row reduction you get an entire row of zeros on the left side of the augmented matrix, the matrix is singular (non-invertible). This means the determinant is zero and no inverse exists.
Why does the augmented matrix method work for finding the inverse?
Every row operation you perform is equivalent to multiplying on the left by an elementary matrix. When you transform A into I, the combined effect of all those operations equals A⁻¹. Applying those same operations to the identity matrix on the right accumulates A⁻¹ step by step. In effect, you are solving the equation AX = I for the unknown matrix X.

Augmented matrix method (Method 2) vs. 2×2 shortcut formula (Method 1)

The 2×2 shortcut uses the formula A⁻¹ = (1/det A) · [[d, −b], [−c, a]] and works only for 2×2 matrices. The augmented matrix method uses Gauss-Jordan elimination on [A | I] and works for any size square matrix. For 2×2 matrices, the shortcut is faster. For 3×3 or larger matrices, the augmented matrix method is generally the most practical hand-calculation approach.

Why It Matters

The augmented matrix method is the most versatile technique for finding a matrix inverse by hand. It scales to any size square matrix, unlike the 2×2 shortcut formula. In applications such as solving systems of linear equations, computer graphics transformations, and engineering models, computing the inverse via row reduction is a foundational skill.

Common Mistakes

Mistake: Performing a row operation on only one side of the augmented matrix.
Correction: Every row operation must be applied to the entire augmented row — both the left (A) side and the right (I) side simultaneously. Otherwise the relationship between the two sides is destroyed.
Mistake: Stopping at row echelon form instead of continuing to reduced row echelon form.
Correction: You need the left side to become the full identity matrix (all pivots equal to 1 and all other entries zero). If you only get an upper triangular form, you must continue eliminating entries above each pivot until you reach reduced row echelon form.

Related Terms