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 finds the inverse of a \(2 \times 2\) matrix and shows the working line by line. The determinant comes first, because a determinant of zero leaves no inverse at all. The main diagonal entries swap places, the other two change sign, and every entry is then divided by the determinant. Following the same pattern every time makes the inverse easy to check and repeat.

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Work out the determinant.
  4. Swap the main diagonal and negate the other two entries.
  5. Multiply every entry by one over the determinant.

Formula:

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

Problem

\(A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Work out the determinant.
  4. Swap the main diagonal and negate the other two entries.
  5. Multiply every entry by one over the determinant.

Step 1

Step 1 - Set up the problem.

In this problem: Reading the matrix left to right and top to bottom gives \(a = 4\), \(b = 7\), \(c = 2\) and \(d = 6\).

\(A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: The inverse rearranges the four entries and then scales them down by the determinant, which is why a determinant of zero leaves no inverse.

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

Step 3

Step 3 - Work out the determinant.

In this problem: The main diagonal gives \(4 \cdot 6 = 24\) and the other gives \(7 \cdot 2 = 14\), so \(24 - 14 = 10\), which is not zero.

\(\det(A) = 24 - 14 = 10\)

Step 4

Step 4 - Swap the main diagonal and negate the other two entries.

In this problem: Swapping \(4\) and \(6\) and changing the signs of \(7\) and \(2\) gives \(\begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix}\).

\(\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix}\)

Step 5

Step 5 - Multiply every entry by one over the determinant.

In this problem: Dividing every entry by \(10\) gives \(\begin{pmatrix} \frac{3}{5} & -\frac{7}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}\).

\(A^{-1} = \frac{1}{10} \begin{pmatrix} 6 & -7 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} & -\frac{7}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}\)

Final Answer

\(A^{-1} = \begin{pmatrix} \frac{3}{5} & -\frac{7}{10} \\ -\frac{1}{5} & \frac{2}{5} \end{pmatrix}\)
See Example 2 Hide Example 2

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Work out the determinant.
  4. Swap the main diagonal and negate the other two entries.
  5. Multiply every entry by one over the determinant.

Step 1

Step 1 - Set up the problem.

In this problem: Reading the matrix left to right and top to bottom gives \(a = 3\), \(b = -2\), \(c = 5\) and \(d = 1\).

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

Step 2

Step 2 - Write the formula.

In this problem: The inverse rearranges the four entries and then scales them down by the determinant, which is why a determinant of zero leaves no inverse.

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

Step 3

Step 3 - Work out the determinant.

In this problem: The main diagonal gives \(3 \cdot 1 = 3\) and the other gives \((-2) \cdot 5 = -10\), so \(3 - (-10) = 13\), which is not zero.

\(\det(A) = 3 - (-10) = 13\)

Step 4

Step 4 - Swap the main diagonal and negate the other two entries.

In this problem: Swapping \(3\) and \(1\) and changing the signs of \(-2\) and \(5\) gives \(\begin{pmatrix} 1 & 2 \\ -5 & 3 \end{pmatrix}\).

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

Step 5

Step 5 - Multiply every entry by one over the determinant.

In this problem: Dividing every entry by \(13\) gives \(\begin{pmatrix} \frac{1}{13} & \frac{2}{13} \\ -\frac{5}{13} & \frac{3}{13} \end{pmatrix}\).

\(A^{-1} = \frac{1}{13} \begin{pmatrix} 1 & 2 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{13} & \frac{2}{13} \\ -\frac{5}{13} & \frac{3}{13} \end{pmatrix}\)

Final Answer

\(A^{-1} = \begin{pmatrix} \frac{1}{13} & \frac{2}{13} \\ -\frac{5}{13} & \frac{3}{13} \end{pmatrix}\)
Matrix A

Example Problems

Click one to load it into the box and solve it.