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Inverse of a Matrix

Inverse of a Matrix
Matrix Inverse
Multiplicative Inverse of a Matrix

For a square matrix A, the inverse is written A-1. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses.

Note: Not all square matrices have inverses. A square matrix which has an inverse is called invertible or nonsingular, and a square matrix without an inverse is called noninvertible or singular.

 

AA-1 = A-1A = I

Example: For matrix 2x2 matrix A with values: row 1: 4, 3; row 2: 3, 2, its inverse is A inverse equals a 2x2 matrix with row 1: -2, 3 and row 2: 3, -4 since
 


AA-1 = Matrix equation showing [4,3; 3,2] multiplied by [-2,3; 3,-4] equals identity matrix [1,0; 0,1]

and A-1A = Matrix equation: [-2, 3; 3, -4] multiplied by [4, 3; 3, 2] equals identity matrix [1, 0; 0, 1].

 

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

1. Shortcut for 2x2 matrices

For 2x2 matrix A with elements: top row [a, b], bottom row [c, d], the inverse can be found using this formula:

Formula: A⁻¹ = 1/det(A) × [d, -b; -c, a] = 1/(ad-bc) × [d, -b; -c, a]

Example: Matrix inverse example: [1,2; 3,4]⁻¹ = (1/-2)[4,-2; -3,1] = [-2,1; 3/2,-1/2]

2. Augmented matrix method

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

Example: The following steps result in Matrix [1 2 / 3 4] raised to the power of -1, representing the inverse of a 2x2 matrix..

Row reduction steps finding inverse of matrix [[1,2],[3,4]] using augmented matrix, yielding inverse [[-2,1],[3/2,-1/2]]

so we see that Matrix equation showing [1 2; 3 4] inverse equals [-2 1; 3/2 -1/2].

3. Adjoint method

A-1 = The formula 1 divided by det A, representing the reciprocal of the determinant of matrix A.(adjoint of A)   or   A-1 = The formula 1 divided by det A, representing the reciprocal of the determinant of matrix A.(cofactor matrix of A)T

Example: The following steps result in A-1 for 3×3 matrix A with rows [1,2,3], [0,4,5], [1,0,6], used to illustrate matrix inverse concept..

The cofactor matrix for A is 3×3 matrix with rows [24, 5, -4], [-12, 3, 2], [-2, -5, 4], representing an example inverse matrix A⁻¹., so the adjoint is 3×3 matrix with rows [24, -12, -2], [5, 3, -5], [-4, 2, 4], representing an example matrix A for finding its inverse.. Since det A = 22, we get

A⁻¹ = 1/22 × matrix([24,-12,-2],[5,3,-5],[-4,2,4]) = matrix([12/11,-6/11,-1/11],[5/22,3/22,-5/22],[-2/11,1/11,2/11]).

 

See also

Determinant of a matrix, cofactor

Key Formula

A−1=1ad−bc[d−b−ca]for A=[abcd]A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \quad \text{for } A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
Where:
  • AA = A 2×2 square matrix
  • a,b,c,da, b, c, d = The entries of the 2×2 matrix
  • ad−bcad - bc = The determinant of A (must be nonzero for the inverse to exist)
  • A−1A^{-1} = The inverse matrix such that AA^{-1} = A^{-1}A = I

Worked Example

Problem: Find the inverse of the 2×2 matrix A = [[4, 7], [2, 6]].
Step 1: Compute the determinant of A using the formula ad − bc.
det⁡(A)=(4)(6)−(7)(2)=24−14=10\det(A) = (4)(6) - (7)(2) = 24 - 14 = 10
Step 2: Since the determinant is 10 (not zero), the inverse exists. Swap the diagonal entries a and d, and negate the off-diagonal entries b and c.
[d−b−ca]=[6−7−24]\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}
Step 3: Multiply the resulting matrix by 1 over the determinant.
A−1=110[6−7−24]=[0.6−0.7−0.20.4]A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}
Step 4: Verify by checking that AA⁻¹ equals the identity matrix.
AA−1=[4726][0.6−0.7−0.20.4]=[1001]=I✓AA^{-1} = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \checkmark
Answer: A−1=110[6−7−24]=[0.6−0.7−0.20.4]A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}

Another Example

This example uses a 3×3 matrix and the Gauss-Jordan (augmented matrix) method, which works for any size square matrix — unlike the 2×2 shortcut formula used in the first example.

Problem: Find the inverse of the 3×3 matrix using the augmented matrix (Gauss-Jordan) method: A = [[1, 0, 1], [0, 2, 1], [1, 1, 1]].
Step 1: Set up the augmented matrix [A | I] by placing the 3×3 identity matrix to the right of A.
[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 entry in row 3, column 1 by replacing R₃ with R₃ − R₁.
[101100021010010−101]\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: Swap R₂ and R₃ for convenience, then eliminate to get a leading 1 in row 2. After swapping: replace new R₃ with R₃ − 2·R₂.
[101100010−10100121−2]\left[\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 2 & 1 & -2 \end{array}\right]
Step 4: Eliminate the entry in row 1, column 3 by replacing R₁ with 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]
Step 5: The left side is now the identity matrix, so the right side is A⁻¹. Verify: det(A) = 1(2·1 − 1·1) − 0 + 1(0·1 − 2·1) = 1 − 2 = −1 ≠ 0, confirming invertibility.
A−1=[−1−12−10121−2]A^{-1} = \begin{bmatrix} -1 & -1 & 2 \\ -1 & 0 & 1 \\ 2 & 1 & -2 \end{bmatrix}
Answer: A−1=[−1−12−10121−2]A^{-1} = \begin{bmatrix} -1 & -1 & 2 \\ -1 & 0 & 1 \\ 2 & 1 & -2 \end{bmatrix}

Frequently Asked Questions

How do you know if a matrix has an inverse?
A matrix has an inverse if and only if it is square and its determinant is nonzero. If the determinant equals zero, the matrix is called singular and has no inverse. You can also check whether the row-reduced form of the matrix is the identity matrix; if it is, the matrix is invertible.
What is the difference between the inverse and the transpose of a matrix?
The inverse A−1A^{-1} undoes multiplication by AA, meaning AA−1=IAA^{-1} = I. The transpose ATA^T simply flips the matrix over its main diagonal, swapping rows and columns. Every matrix has a transpose, but only square matrices with nonzero determinants have inverses. These are completely different operations.
Why can't non-square matrices have inverses?
A true (two-sided) inverse requires AA−1=A−1A=IAA^{-1} = A^{-1}A = I, which demands that both products be defined and result in identity matrices. This is only possible when AA is square. Non-square matrices can have one-sided inverses (left or right), but not a full inverse.

Inverse of a Matrix vs. Transpose of a Matrix

Inverse of a MatrixTranspose of a Matrix
DefinitionMatrix A⁻¹ such that AA⁻¹ = A⁻¹A = IMatrix Aᵀ formed by swapping rows and columns of A
ExistenceOnly for square matrices with nonzero determinantExists for every matrix of any size
Effect on original"Undoes" multiplication by AReflects entries across the main diagonal
Key propertyAA⁻¹ = I(Aᵀ)ᵀ = A

Why It Matters

Matrix inverses are essential for solving systems of linear equations: if Ax=bA\mathbf{x} = \mathbf{b}, you can find the solution directly as x=A−1b\mathbf{x} = A^{-1}\mathbf{b}. You will encounter them in linear algebra, physics, computer graphics, and data science whenever transformations need to be reversed. In precalculus and algebra courses, finding a 2×2 or 3×3 inverse is a standard exam topic.

Common Mistakes

Mistake: Forgetting to check that the determinant is nonzero before applying the inverse formula.
Correction: Always compute det(A) first. If det(A) = 0, the matrix is singular and has no inverse — attempting the formula will produce division by zero.
Mistake: In the 2×2 formula, swapping the wrong entries — negating the diagonal instead of the off-diagonal.
Correction: Remember: swap a and d (the main diagonal entries), and negate b and c (the off-diagonal entries). A common mnemonic is "swap the diagonal, flip the signs of the other two."

Related Terms