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Matrix Inverse — Definition, Formula & Examples

The matrix inverse of a square matrix AA is another matrix, written A−1A^{-1}, that produces the identity matrix when multiplied by AA. In other words, A⋅A−1=A−1⋅A=IA \cdot A^{-1} = A^{-1} \cdot A = I, much like how multiplying a number by its reciprocal gives 1.

Let AA be an n×nn \times n square matrix. If there exists an n×nn \times n matrix BB such that AB=BA=InAB = BA = I_n, where InI_n is the n×nn \times n identity matrix, then AA is said to be invertible (or nonsingular) and BB is the unique inverse of AA, denoted A−1A^{-1}. A matrix is invertible if and only if its determinant is nonzero.

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:
  • a,b,c,da, b, c, d = The four entries of the 2×2 matrix A
  • ad−bcad - bc = The determinant of A; must be nonzero for the inverse to exist

How It Works

Finding a matrix inverse lets you solve systems of linear equations in one step. If you have Ax=bAx = b, then x=A−1bx = A^{-1}b. For a 2×22 \times 2 matrix, a direct formula exists using the determinant. For larger matrices, you can use row reduction on the augmented matrix [A∣I][A \mid I], reducing the left side to II so the right side becomes A−1A^{-1}. Alternatively, you can compute the inverse using the adjugate (classical adjoint) and determinant: A−1=1det⁡(A)adj⁡(A)A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A).

Worked Example

Problem: Find the inverse of A=[4726]A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}.
Step 1: Compute the determinant.
det⁡(A)=(4)(6)−(7)(2)=24−14=10\det(A) = (4)(6) - (7)(2) = 24 - 14 = 10
Step 2: Swap the diagonal entries and negate the off-diagonal entries to form the adjugate matrix.
[6−7−24]\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix}
Step 3: Divide every entry by 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 A⋅A−1=IA \cdot A^{-1} = I.
[4726][0.6−0.7−0.20.4]=[1001]✓\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} \checkmark
Answer: A−1=[0.6−0.7−0.20.4]A^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}

Another Example

Problem: Use the inverse to solve [4726][xy]=[53]\begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 3 \end{bmatrix}.
Step 1: From the previous example, the inverse is already known.
A−1=[0.6−0.7−0.20.4]A^{-1} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}
Step 2: Multiply both sides on the left by A−1A^{-1}.
[xy]=A−1b=[0.6−0.7−0.20.4][53]\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1}b = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}\begin{bmatrix} 5 \\ 3 \end{bmatrix}
Step 3: Carry out the multiplication.
[(0.6)(5)+(−0.7)(3)(−0.2)(5)+(0.4)(3)]=[0.90.2]\begin{bmatrix} (0.6)(5)+(-0.7)(3) \\ (-0.2)(5)+(0.4)(3) \end{bmatrix} = \begin{bmatrix} 0.9 \\ 0.2 \end{bmatrix}
Answer: x=0.9x = 0.9 and y=0.2y = 0.2.

Why It Matters

Matrix inverses appear throughout a linear algebra or precalculus course whenever you need to solve systems of equations, reverse transformations, or decrypt encoded messages (as in Hill cipher problems). Engineers and data scientists use matrix inverses to fit models to data through least-squares regression. Understanding when an inverse does and does not exist also builds intuition for concepts like rank and linear independence that matter in more advanced work.

Common Mistakes

Mistake: Forgetting to check whether the determinant is zero before applying the formula.
Correction: Always compute the determinant first. If det⁡(A)=0\det(A) = 0, the matrix is singular and no inverse exists.
Mistake: Swapping the wrong entries: negating the diagonal and swapping the off-diagonal for a 2×2 matrix.
Correction: For [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}, you swap aa and dd (the main-diagonal entries) and negate bb and cc (the off-diagonal entries), not the other way around.

Related Terms