Inverse of a 3×3 Matrix Calculator

Published on: October 5 2025
Final Answer: Free • Full Steps: Pro

This Inverse of a 3×3 Matrix Calculator finds the inverse of a \(3 \times 3\) matrix and shows the working line by line. Nine \(2 \times 2\) minors are worked out, an alternating sign pattern turns them into cofactors, and transposing those gives the adjugate. Dividing the adjugate by the determinant finishes it, which is why a determinant of zero leaves no inverse. Following the same pattern every time keeps a long calculation checkable.

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Work out the determinant.
  4. Find the 2 by 2 minor for each position.
  5. Apply the sign pattern to get the cofactors.
  6. Transpose the cofactors to get the adjugate.
  7. Divide every entry by the determinant.

Formula:

\(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\)
See Example 1 Hide Example 1

Problem

\(A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Work out the determinant.
  4. Find the 2 by 2 minor for each position.
  5. Apply the sign pattern to get the cofactors.
  6. Transpose the cofactors to get the adjugate.
  7. Divide every entry by the determinant.

Step 1

Step 1 - Set up the problem.

In this problem: Labelling the entries left to right and top to bottom gives \(a\) through \(i\).

\(A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: The inverse is built from the matrix of cofactors, transposed, and then scaled down by the determinant.

\(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\)

Step 3

Step 3 - Work out the determinant.

In this problem: Expanding along the top row gives \(-24\), \(-40\) and \(-15\), and \(-24 - (-40) + (-15) = 1\), which is not zero.

\(\det(A) = -24 - (-40) + (-15) = 1\)

Step 4

Step 4 - Find the 2 by 2 minor for each position.

In this problem: Nine minors are still to be worked out, so each position is taken in turn.

\(M_{ij} = \det(\text{the 2 by 2 block left when row } i \text{ and column } j \text{ are covered})\)

Step 4a

Step 4a - Find the 2 by 2 minor for each position.

In this problem: Covering row \(1\) and column \(1\) leaves \(\begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix}\), and \(1 \cdot 0 - 4 \cdot 6 = -24\).

\(M_{11} = \begin{pmatrix} 1 & 4 \\ 6 & 0 \end{pmatrix} = -24\)

Step 4b

Step 4b - Find the 2 by 2 minor for each position.

In this problem: Covering row \(1\) and column \(2\) leaves \(\begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix}\), and \(0 \cdot 0 - 4 \cdot 5 = -20\).

\(M_{12} = \begin{pmatrix} 0 & 4 \\ 5 & 0 \end{pmatrix} = -20\)

Step 4c

Step 4c - Find the 2 by 2 minor for each position.

In this problem: Covering row \(1\) and column \(3\) leaves \(\begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix}\), and \(0 \cdot 6 - 1 \cdot 5 = -5\).

\(M_{13} = \begin{pmatrix} 0 & 1 \\ 5 & 6 \end{pmatrix} = -5\)

Step 4d

Step 4d - Find the 2 by 2 minor for each position.

In this problem: Covering row \(2\) and column \(1\) leaves \(\begin{pmatrix} 2 & 3 \\ 6 & 0 \end{pmatrix}\), and \(2 \cdot 0 - 3 \cdot 6 = -18\).

\(M_{21} = \begin{pmatrix} 2 & 3 \\ 6 & 0 \end{pmatrix} = -18\)

Step 4e

Step 4e - Find the 2 by 2 minor for each position.

In this problem: Covering row \(2\) and column \(2\) leaves \(\begin{pmatrix} 1 & 3 \\ 5 & 0 \end{pmatrix}\), and \(1 \cdot 0 - 3 \cdot 5 = -15\).

\(M_{22} = \begin{pmatrix} 1 & 3 \\ 5 & 0 \end{pmatrix} = -15\)

Step 4f

Step 4f - Find the 2 by 2 minor for each position.

In this problem: Covering row \(2\) and column \(3\) leaves \(\begin{pmatrix} 1 & 2 \\ 5 & 6 \end{pmatrix}\), and \(1 \cdot 6 - 2 \cdot 5 = -4\).

\(M_{23} = \begin{pmatrix} 1 & 2 \\ 5 & 6 \end{pmatrix} = -4\)

Step 4g

Step 4g - Find the 2 by 2 minor for each position.

In this problem: Covering row \(3\) and column \(1\) leaves \(\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\), and \(2 \cdot 4 - 3 \cdot 1 = 5\).

\(M_{31} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = 5\)

Step 4h

Step 4h - Find the 2 by 2 minor for each position.

In this problem: Covering row \(3\) and column \(2\) leaves \(\begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix}\), and \(1 \cdot 4 - 3 \cdot 0 = 4\).

\(M_{32} = \begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix} = 4\)

Step 4i

Step 4i - Find the 2 by 2 minor for each position.

In this problem: Covering row \(3\) and column \(3\) leaves \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\), and \(1 \cdot 1 - 2 \cdot 0 = 1\).

\(M_{33} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = 1\)

Step 5

Step 5 - Apply the sign pattern to get the cofactors.

In this problem: The sign pattern flips every second position, so \(-20 \to 20\), \(-18 \to 18\), \(-4 \to 4\), \(4 \to -4\).

\(\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} \odot \begin{pmatrix} -24 & -20 & -5 \\ -18 & -15 & -4 \\ 5 & 4 & 1 \end{pmatrix} = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}\)

Step 6

Step 6 - Transpose the cofactors to get the adjugate.

In this problem: Reflecting the cofactor matrix along its main diagonal turns its rows into columns, which gives the adjugate.

\(\operatorname{adj}(A) = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}\)

Step 7

Step 7 - Divide every entry by the determinant.

In this problem: Dividing every entry by \(1\) gives \(\begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}\).

\(A^{-1} = \frac{1}{1} \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}\)

Final Answer

\(A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}\)
See Example 2 Hide Example 2

Problem

\(A = \begin{pmatrix} 2 & 0 & 1 \\ 3 & -1 & 2 \\ 1 & 4 & 5 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Work out the determinant.
  4. Find the 2 by 2 minor for each position.
  5. Apply the sign pattern to get the cofactors.
  6. Transpose the cofactors to get the adjugate.
  7. Divide every entry by the determinant.

Step 1

Step 1 - Set up the problem.

In this problem: Labelling the entries left to right and top to bottom gives \(a\) through \(i\).

\(A = \begin{pmatrix} 2 & 0 & 1 \\ 3 & -1 & 2 \\ 1 & 4 & 5 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: The inverse is built from the matrix of cofactors, transposed, and then scaled down by the determinant.

\(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\)

Step 3

Step 3 - Work out the determinant.

In this problem: Expanding along the top row gives \(-26\), \(0\) and \(13\), and \(-26 - 0 + 13 = -13\), which is not zero.

\(\det(A) = -26 - 0 + 13 = -13\)

Step 4

Step 4 - Find the 2 by 2 minor for each position.

In this problem: Nine minors are still to be worked out, so each position is taken in turn.

\(M_{ij} = \det(\text{the 2 by 2 block left when row } i \text{ and column } j \text{ are covered})\)

Step 4a

Step 4a - Find the 2 by 2 minor for each position.

In this problem: Covering row \(1\) and column \(1\) leaves \(\begin{pmatrix} -1 & 2 \\ 4 & 5 \end{pmatrix}\), and \((-1) \cdot 5 - 2 \cdot 4 = -13\).

\(M_{11} = \begin{pmatrix} -1 & 2 \\ 4 & 5 \end{pmatrix} = -13\)

Step 4b

Step 4b - Find the 2 by 2 minor for each position.

In this problem: Covering row \(1\) and column \(2\) leaves \(\begin{pmatrix} 3 & 2 \\ 1 & 5 \end{pmatrix}\), and \(3 \cdot 5 - 2 \cdot 1 = 13\).

\(M_{12} = \begin{pmatrix} 3 & 2 \\ 1 & 5 \end{pmatrix} = 13\)

Step 4c

Step 4c - Find the 2 by 2 minor for each position.

In this problem: Covering row \(1\) and column \(3\) leaves \(\begin{pmatrix} 3 & -1 \\ 1 & 4 \end{pmatrix}\), and \(3 \cdot 4 - (-1) \cdot 1 = 13\).

\(M_{13} = \begin{pmatrix} 3 & -1 \\ 1 & 4 \end{pmatrix} = 13\)

Step 4d

Step 4d - Find the 2 by 2 minor for each position.

In this problem: Covering row \(2\) and column \(1\) leaves \(\begin{pmatrix} 0 & 1 \\ 4 & 5 \end{pmatrix}\), and \(0 \cdot 5 - 1 \cdot 4 = -4\).

\(M_{21} = \begin{pmatrix} 0 & 1 \\ 4 & 5 \end{pmatrix} = -4\)

Step 4e

Step 4e - Find the 2 by 2 minor for each position.

In this problem: Covering row \(2\) and column \(2\) leaves \(\begin{pmatrix} 2 & 1 \\ 1 & 5 \end{pmatrix}\), and \(2 \cdot 5 - 1 \cdot 1 = 9\).

\(M_{22} = \begin{pmatrix} 2 & 1 \\ 1 & 5 \end{pmatrix} = 9\)

Step 4f

Step 4f - Find the 2 by 2 minor for each position.

In this problem: Covering row \(2\) and column \(3\) leaves \(\begin{pmatrix} 2 & 0 \\ 1 & 4 \end{pmatrix}\), and \(2 \cdot 4 - 0 \cdot 1 = 8\).

\(M_{23} = \begin{pmatrix} 2 & 0 \\ 1 & 4 \end{pmatrix} = 8\)

Step 4g

Step 4g - Find the 2 by 2 minor for each position.

In this problem: Covering row \(3\) and column \(1\) leaves \(\begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}\), and \(0 \cdot 2 - 1 \cdot (-1) = 1\).

\(M_{31} = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} = 1\)

Step 4h

Step 4h - Find the 2 by 2 minor for each position.

In this problem: Covering row \(3\) and column \(2\) leaves \(\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\), and \(2 \cdot 2 - 1 \cdot 3 = 1\).

\(M_{32} = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} = 1\)

Step 4i

Step 4i - Find the 2 by 2 minor for each position.

In this problem: Covering row \(3\) and column \(3\) leaves \(\begin{pmatrix} 2 & 0 \\ 3 & -1 \end{pmatrix}\), and \(2 \cdot (-1) - 0 \cdot 3 = -2\).

\(M_{33} = \begin{pmatrix} 2 & 0 \\ 3 & -1 \end{pmatrix} = -2\)

Step 5

Step 5 - Apply the sign pattern to get the cofactors.

In this problem: The sign pattern flips every second position, so \(13 \to -13\), \(-4 \to 4\), \(8 \to -8\), \(1 \to -1\).

\(\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} \odot \begin{pmatrix} -13 & 13 & 13 \\ -4 & 9 & 8 \\ 1 & 1 & -2 \end{pmatrix} = \begin{pmatrix} -13 & -13 & 13 \\ 4 & 9 & -8 \\ 1 & -1 & -2 \end{pmatrix}\)

Step 6

Step 6 - Transpose the cofactors to get the adjugate.

In this problem: Reflecting the cofactor matrix along its main diagonal turns its rows into columns, which gives the adjugate.

\(\operatorname{adj}(A) = \begin{pmatrix} -13 & 4 & 1 \\ -13 & 9 & -1 \\ 13 & -8 & -2 \end{pmatrix}\)

Step 7

Step 7 - Divide every entry by the determinant.

In this problem: Dividing every entry by \(-13\) gives \(\begin{pmatrix} 1 & -\frac{4}{13} & -\frac{1}{13} \\ 1 & -\frac{9}{13} & \frac{1}{13} \\ -1 & \frac{8}{13} & \frac{2}{13} \end{pmatrix}\).

\(A^{-1} = \frac{1}{-13} \begin{pmatrix} -13 & 4 & 1 \\ -13 & 9 & -1 \\ 13 & -8 & -2 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{4}{13} & -\frac{1}{13} \\ 1 & -\frac{9}{13} & \frac{1}{13} \\ -1 & \frac{8}{13} & \frac{2}{13} \end{pmatrix}\)

Final Answer

\(A^{-1} = \begin{pmatrix} 1 & -\frac{4}{13} & -\frac{1}{13} \\ 1 & -\frac{9}{13} & \frac{1}{13} \\ -1 & \frac{8}{13} & \frac{2}{13} \end{pmatrix}\)
Matrix A

Example Problems

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