Inverse of a 2×2 Matrix Calculator

Published on: September 28 2025

Final Answer: Free • Full Steps: Plus

This Inverse of a 2×2 Matrix Calculator helps you find the inverse of a 2×2 matrix and shows each step clearly. In Algebra 2, inverse matrices are often used when studying matrices, determinants, and systems of equations. The calculator first finds the determinant, then swaps the main diagonal entries, changes the signs of the off-diagonal entries, and multiplies by the reciprocal of the determinant.

Step-by-step method

Identify the entries a, b, c, and d in the 2×2 matrix.
Find the determinant using det( A ) = ad − bc, then check that it is not 0.
Use the inverse matrix pattern shown in the formula.
Multiply the new matrix by 1 / det( A ) and simplify.

Formulas:

\(\det(A) = ad - bc\)

\(A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\)

Example 1: Take the values below.

\(A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}\)

Step 1 - Identify the entries a, b, c, and d in the 2×2 matrix.

In this problem: For a 2×2 matrix, a is top-left, b is top-right, c is bottom-left, and d is bottom-right.

\(A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix},\quad a = 2,\; b = 1,\; c = 5,\; d = 3\)

Step 2 - Find the determinant using det( A ) = ad − bc, then check that it is not 0.

In this problem: The determinant is \(1\). Since \(1 \neq 0\), the matrix has an inverse.

\(\det(A) = 2 \cdot 3 - 1 \cdot 5 = 6 - 5 = 1,\quad 1 \neq 0\)

Step 3 - Use the inverse matrix pattern shown in the formula.

In this problem: Use the inverse matrix pattern from the formula, then substitute this problem’s values.

\(\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\)

Step 4 - Multiply the new matrix by 1 / det( A ) and simplify.

In this problem: Multiply each entry by the reciprocal of the determinant and simplify.

\(A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\)

Final answer:

\(A^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\)

Example 2: Take the values below.

\(A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}\)

Step 1 - Identify the entries a, b, c, and d in the 2×2 matrix.

In this problem: For a 2×2 matrix, a is top-left, b is top-right, c is bottom-left, and d is bottom-right.

\(A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix},\quad a = 4,\; b = -2,\; c = 1,\; d = 1\)

Step 2 - Find the determinant using det( A ) = ad − bc, then check that it is not 0.

In this problem: The determinant is \(6\). Since \(6 \neq 0\), the matrix has an inverse.

\(\det(A) = 4 \cdot 1 - (-2) \cdot 1 = 4 - (-2) = 6,\quad 6 \neq 0\)

Step 3 - Use the inverse matrix pattern shown in the formula.

In this problem: Use the inverse matrix pattern from the formula, then substitute this problem’s values.

\(\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}\)

Step 4 - Multiply the new matrix by 1 / det( A ) and simplify.

In this problem: Multiply each entry by the reciprocal of the determinant and simplify.

\(A^{-1} = \frac{1}{6} \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} \frac{1}{6} & \frac{1}{3} \\ -\frac{1}{6} & \frac{2}{3} \end{pmatrix}\)

Final answer:

\(A^{-1} = \begin{pmatrix} \frac{1}{6} & \frac{1}{3} \\ -\frac{1}{6} & \frac{2}{3} \end{pmatrix}\)

See Example 1 Hide Example 1

Problem

\(A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}\)

Approach

Step-by-step method

Identify the entries a, b, c, and d in the 2×2 matrix.
Find the determinant using det( A ) = ad − bc, then check that it is not 0.
Use the inverse matrix pattern shown in the formula.
Multiply the new matrix by 1 / det( A ) and simplify.

Step 1

Step 1 - Identify the entries a, b, c, and d in the 2×2 matrix.

In this problem: For a 2×2 matrix, a is top-left, b is top-right, c is bottom-left, and d is bottom-right.

\(A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix},\quad a = 2,\; b = 1,\; c = 5,\; d = 3\)

Step 2

Step 2 - Find the determinant using det( A ) = ad − bc, then check that it is not 0.

In this problem: The determinant is \(1\). Since \(1 \neq 0\), the matrix has an inverse.

\(\det(A) = 2 \cdot 3 - 1 \cdot 5 = 6 - 5 = 1,\quad 1 \neq 0\)

Step 3

Step 3 - Use the inverse matrix pattern shown in the formula.

In this problem: Use the inverse matrix pattern from the formula, then substitute this problem’s values.

\(\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\)

Step 4

Step 4 - Multiply the new matrix by 1 / det( A ) and simplify.

In this problem: Multiply each entry by the reciprocal of the determinant and simplify.

\(A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\)

Final Answer

\(\)

See Example 2 Hide Example 2

Problem

\(A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}\)

Approach

Step-by-step method

Identify the entries a, b, c, and d in the 2×2 matrix.
Find the determinant using det( A ) = ad − bc, then check that it is not 0.
Use the inverse matrix pattern shown in the formula.
Multiply the new matrix by 1 / det( A ) and simplify.

Step 1

Step 1 - Identify the entries a, b, c, and d in the 2×2 matrix.

In this problem: For a 2×2 matrix, a is top-left, b is top-right, c is bottom-left, and d is bottom-right.

\(A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix},\quad a = 4,\; b = -2,\; c = 1,\; d = 1\)

Step 2

Step 2 - Find the determinant using det( A ) = ad − bc, then check that it is not 0.

In this problem: The determinant is \(6\). Since \(6 \neq 0\), the matrix has an inverse.

\(\det(A) = 4 \cdot 1 - (-2) \cdot 1 = 4 - (-2) = 6,\quad 6 \neq 0\)

Step 3

Step 3 - Use the inverse matrix pattern shown in the formula.

In this problem: Use the inverse matrix pattern from the formula, then substitute this problem’s values.

\(\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}\)

Step 4

Step 4 - Multiply the new matrix by 1 / det( A ) and simplify.

In this problem: Multiply each entry by the reciprocal of the determinant and simplify.

\(A^{-1} = \frac{1}{6} \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} \frac{1}{6} & \frac{1}{3} \\ -\frac{1}{6} & \frac{2}{3} \end{pmatrix}\)

Final Answer

\(\)

Matrix A