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 helps you find the inverse of a 3×3 matrix and shows each step clearly. It works by finding the determinant, computing the cofactors and adjugate matrix, and then multiplying by the reciprocal of the determinant. This makes it useful for checking answers, understanding how a matrix inverse is found, and practising linear algebra step by step.

Step-by-step method

  1. Identify the entries a through i in the 3×3 matrix.
  2. Find the determinant by breaking it into the a-part, b-part, and c-part.
  3. Find each 2×2 minor to build the minor matrix M.
  4. Apply the cofactor sign pattern S to M to get the cofactor matrix C.
  5. Transpose C to get Cᵀ, which is adj( A ).
  6. Multiply adj( A ) by 1 / det( A ) and simplify.

Formulas:

\(\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\)
\(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\)

Example 1: Take the values below.

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

Step 1 - Identify the entries a through i in the 3×3 matrix.

In this problem: For a 3×3 matrix, label the entries from left to right and top to bottom as a through i.

\(\begin{gathered} a = 1,\; b = 2,\; c = 3 \\ d = 0,\; e = 1,\; f = 4 \\ g = 5,\; h = 6,\; i = 0 \end{gathered}\)

Step 2A - Find the a-part of the determinant.

In this problem: The a-part is \(a(ei - fh)\).

\(a(ei - fh) = 1(1 \cdot 0 - 4 \cdot 6) = -24\)

Step 2B - Find the b-part of the determinant.

In this problem: The b-part is \(b(di - fg)\). This part is subtracted in the determinant formula.

\(b(di - fg) = 2(0 \cdot 0 - 4 \cdot 5) = -40\)

Step 2C - Find the c-part of the determinant.

In this problem: The c-part is \(c(dh - eg)\).

\(c(dh - eg) = 3(0 \cdot 6 - 1 \cdot 5) = -15\)

Step 2D - Combine the determinant parts.

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

\(\det(A) = -24 - (-40) + (-15) = 1,\quad 1 \neq 0\)

Step 3A - Find minor M11.

In this problem: Delete row 1 and column 1, then take the determinant of the remaining 2×2 matrix.

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

Step 3B - Find minor M12.

In this problem: Delete row 1 and column 2, then take the determinant of the remaining 2×2 matrix.

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

Step 3C - Find minor M13.

In this problem: Delete row 1 and column 3, then take the determinant of the remaining 2×2 matrix.

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

Step 3D - Find minor M21.

In this problem: Delete row 2 and column 1, then take the determinant of the remaining 2×2 matrix.

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

Step 3E - Find minor M22.

In this problem: Delete row 2 and column 2, then take the determinant of the remaining 2×2 matrix.

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

Step 3F - Find minor M23.

In this problem: Delete row 2 and column 3, then take the determinant of the remaining 2×2 matrix.

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

Step 3G - Find minor M31.

In this problem: Delete row 3 and column 1, then take the determinant of the remaining 2×2 matrix.

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

Step 3H - Find minor M32.

In this problem: Delete row 3 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{32} = \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = 1 \cdot 4 - 3 \cdot 0 = 4\)

Step 3I - Find minor M33.

In this problem: Delete row 3 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 0 = 1\)

Step 4 - Apply the cofactor sign pattern S to M to get the cofactor matrix C.

In this problem: M is the minor matrix and S is the cofactor sign pattern. Apply S to M to get the cofactor matrix C.

\(M = \begin{pmatrix} -24 & -20 & -5 \\ -18 & -15 & -4 \\ 5 & 4 & 1 \end{pmatrix},\quad S = \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}\;\Rightarrow\; C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}\)

Step 5 - Transpose C to get Cᵀ, which is adj( A ).

In this problem: Transpose the cofactor matrix by switching rows and columns.

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

Step 6 - Multiply adj( A ) by 1 / det( A ) and simplify.

In this problem: Multiply each entry of \(\operatorname{adj}(A)\) by the reciprocal of the determinant and simplify.

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

Final answer:

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

Example 2: Take the values below.

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

Step 1 - Identify the entries a through i in the 3×3 matrix.

In this problem: For a 3×3 matrix, label the entries from left to right and top to bottom as a through i.

\(\begin{gathered} a = 2,\; b = 0,\; c = 1 \\ d = 1,\; e = 1,\; f = 0 \\ g = 3,\; h = 2,\; i = 1 \end{gathered}\)

Step 2A - Find the a-part of the determinant.

In this problem: The a-part is \(a(ei - fh)\).

\(a(ei - fh) = 2(1 \cdot 1 - 0 \cdot 2) = 2\)

Step 2B - Find the b-part of the determinant.

In this problem: The b-part is \(b(di - fg)\). This part is subtracted in the determinant formula.

\(b(di - fg) = 0(1 \cdot 1 - 0 \cdot 3) = 0\)

Step 2C - Find the c-part of the determinant.

In this problem: The c-part is \(c(dh - eg)\).

\(c(dh - eg) = 1(1 \cdot 2 - 1 \cdot 3) = -1\)

Step 2D - Combine the determinant parts.

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

\(\det(A) = 2 - 0 + (-1) = 1,\quad 1 \neq 0\)

Step 3A - Find minor M11.

In this problem: Delete row 1 and column 1, then take the determinant of the remaining 2×2 matrix.

\(M_{11} = \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - 0 \cdot 2 = 1\)

Step 3B - Find minor M12.

In this problem: Delete row 1 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{12} = \begin{vmatrix} 1 & 0 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 - 0 \cdot 3 = 1\)

Step 3C - Find minor M13.

In this problem: Delete row 1 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{13} = \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot 3 = -1\)

Step 3D - Find minor M21.

In this problem: Delete row 2 and column 1, then take the determinant of the remaining 2×2 matrix.

\(M_{21} = \begin{vmatrix} 0 & 1 \\ 2 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot 2 = -2\)

Step 3E - Find minor M22.

In this problem: Delete row 2 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{22} = \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 3 = -1\)

Step 3F - Find minor M23.

In this problem: Delete row 2 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{23} = \begin{vmatrix} 2 & 0 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - 0 \cdot 3 = 4\)

Step 3G - Find minor M31.

In this problem: Delete row 3 and column 1, then take the determinant of the remaining 2×2 matrix.

\(M_{31} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = 0 \cdot 0 - 1 \cdot 1 = -1\)

Step 3H - Find minor M32.

In this problem: Delete row 3 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{32} = \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = 2 \cdot 0 - 1 \cdot 1 = -1\)

Step 3I - Find minor M33.

In this problem: Delete row 3 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{33} = \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 0 \cdot 1 = 2\)

Step 4 - Apply the cofactor sign pattern S to M to get the cofactor matrix C.

In this problem: M is the minor matrix and S is the cofactor sign pattern. Apply S to M to get the cofactor matrix C.

\(M = \begin{pmatrix} 1 & 1 & -1 \\ -2 & -1 & 4 \\ -1 & -1 & 2 \end{pmatrix},\quad S = \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}\;\Rightarrow\; C = \begin{pmatrix} 1 & -1 & -1 \\ 2 & -1 & -4 \\ -1 & 1 & 2 \end{pmatrix}\)

Step 5 - Transpose C to get Cᵀ, which is adj( A ).

In this problem: Transpose the cofactor matrix by switching rows and columns.

\(\operatorname{adj}(A) = C^{T} = \begin{pmatrix} 1 & 2 & -1 \\ -1 & -1 & 1 \\ -1 & -4 & 2 \end{pmatrix}\)

Step 6 - Multiply adj( A ) by 1 / det( A ) and simplify.

In this problem: Multiply each entry of \(\operatorname{adj}(A)\) by the reciprocal of the determinant and simplify.

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

Final answer:

\(A^{-1} = \begin{pmatrix} 1 & 2 & -1 \\ -1 & -1 & 1 \\ -1 & -4 & 2 \end{pmatrix}\)
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. Identify the entries a through i in the 3×3 matrix.
  2. Find the determinant by breaking it into the a-part, b-part, and c-part.
  3. Find each 2×2 minor to build the minor matrix M.
  4. Apply the cofactor sign pattern S to M to get the cofactor matrix C.
  5. Transpose C to get Cᵀ, which is adj( A ).
  6. Multiply adj( A ) by 1 / det( A ) and simplify.

Step 1

Step 1 - Identify the entries a through i in the 3×3 matrix.

In this problem: For a 3×3 matrix, label the entries from left to right and top to bottom as a through i.

\(\begin{gathered} a = 1,\; b = 2,\; c = 3 \\ d = 0,\; e = 1,\; f = 4 \\ g = 5,\; h = 6,\; i = 0 \end{gathered}\)

Step 2A

Step 2A - Find the a-part of the determinant.

In this problem: The a-part is \(a(ei - fh)\).

\(a(ei - fh) = 1(1 \cdot 0 - 4 \cdot 6) = -24\)

Step 2B

Step 2B - Find the b-part of the determinant.

In this problem: The b-part is \(b(di - fg)\). This part is subtracted in the determinant formula.

\(b(di - fg) = 2(0 \cdot 0 - 4 \cdot 5) = -40\)

Step 2C

Step 2C - Find the c-part of the determinant.

In this problem: The c-part is \(c(dh - eg)\).

\(c(dh - eg) = 3(0 \cdot 6 - 1 \cdot 5) = -15\)

Step 2D

Step 2D - Combine the determinant parts.

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

\(\det(A) = -24 - (-40) + (-15) = 1,\quad 1 \neq 0\)

Step 3A

Step 3A - Find minor M11.

In this problem: Delete row 1 and column 1, then take the determinant of the remaining 2×2 matrix.

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

Step 3B

Step 3B - Find minor M12.

In this problem: Delete row 1 and column 2, then take the determinant of the remaining 2×2 matrix.

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

Step 3C

Step 3C - Find minor M13.

In this problem: Delete row 1 and column 3, then take the determinant of the remaining 2×2 matrix.

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

Step 3D

Step 3D - Find minor M21.

In this problem: Delete row 2 and column 1, then take the determinant of the remaining 2×2 matrix.

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

Step 3E

Step 3E - Find minor M22.

In this problem: Delete row 2 and column 2, then take the determinant of the remaining 2×2 matrix.

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

Step 3F

Step 3F - Find minor M23.

In this problem: Delete row 2 and column 3, then take the determinant of the remaining 2×2 matrix.

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

Step 3G

Step 3G - Find minor M31.

In this problem: Delete row 3 and column 1, then take the determinant of the remaining 2×2 matrix.

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

Step 3H

Step 3H - Find minor M32.

In this problem: Delete row 3 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{32} = \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = 1 \cdot 4 - 3 \cdot 0 = 4\)

Step 3I

Step 3I - Find minor M33.

In this problem: Delete row 3 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{33} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \cdot 1 - 2 \cdot 0 = 1\)

Step 4

Step 4 - Apply the cofactor sign pattern S to M to get the cofactor matrix C.

In this problem: M is the minor matrix and S is the cofactor sign pattern. Apply S to M to get the cofactor matrix C.

\(M = \begin{pmatrix} -24 & -20 & -5 \\ -18 & -15 & -4 \\ 5 & 4 & 1 \end{pmatrix},\quad S = \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}\;\Rightarrow\; C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}\)

Step 5

Step 5 - Transpose C to get Cᵀ, which is adj( A ).

In this problem: Transpose the cofactor matrix by switching rows and columns.

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

Step 6

Step 6 - Multiply adj( A ) by 1 / det( A ) and simplify.

In this problem: Multiply each entry of \(\operatorname{adj}(A)\) by the reciprocal of the determinant and simplify.

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

Final Answer

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

Problem

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

Approach

Step-by-step method

  1. Identify the entries a through i in the 3×3 matrix.
  2. Find the determinant by breaking it into the a-part, b-part, and c-part.
  3. Find each 2×2 minor to build the minor matrix M.
  4. Apply the cofactor sign pattern S to M to get the cofactor matrix C.
  5. Transpose C to get Cᵀ, which is adj( A ).
  6. Multiply adj( A ) by 1 / det( A ) and simplify.

Step 1

Step 1 - Identify the entries a through i in the 3×3 matrix.

In this problem: For a 3×3 matrix, label the entries from left to right and top to bottom as a through i.

\(\begin{gathered} a = 2,\; b = 0,\; c = 1 \\ d = 1,\; e = 1,\; f = 0 \\ g = 3,\; h = 2,\; i = 1 \end{gathered}\)

Step 2A

Step 2A - Find the a-part of the determinant.

In this problem: The a-part is \(a(ei - fh)\).

\(a(ei - fh) = 2(1 \cdot 1 - 0 \cdot 2) = 2\)

Step 2B

Step 2B - Find the b-part of the determinant.

In this problem: The b-part is \(b(di - fg)\). This part is subtracted in the determinant formula.

\(b(di - fg) = 0(1 \cdot 1 - 0 \cdot 3) = 0\)

Step 2C

Step 2C - Find the c-part of the determinant.

In this problem: The c-part is \(c(dh - eg)\).

\(c(dh - eg) = 1(1 \cdot 2 - 1 \cdot 3) = -1\)

Step 2D

Step 2D - Combine the determinant parts.

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

\(\det(A) = 2 - 0 + (-1) = 1,\quad 1 \neq 0\)

Step 3A

Step 3A - Find minor M11.

In this problem: Delete row 1 and column 1, then take the determinant of the remaining 2×2 matrix.

\(M_{11} = \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = 1 \cdot 1 - 0 \cdot 2 = 1\)

Step 3B

Step 3B - Find minor M12.

In this problem: Delete row 1 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{12} = \begin{vmatrix} 1 & 0 \\ 3 & 1 \end{vmatrix} = 1 \cdot 1 - 0 \cdot 3 = 1\)

Step 3C

Step 3C - Find minor M13.

In this problem: Delete row 1 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{13} = \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot 3 = -1\)

Step 3D

Step 3D - Find minor M21.

In this problem: Delete row 2 and column 1, then take the determinant of the remaining 2×2 matrix.

\(M_{21} = \begin{vmatrix} 0 & 1 \\ 2 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot 2 = -2\)

Step 3E

Step 3E - Find minor M22.

In this problem: Delete row 2 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{22} = \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 3 = -1\)

Step 3F

Step 3F - Find minor M23.

In this problem: Delete row 2 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{23} = \begin{vmatrix} 2 & 0 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - 0 \cdot 3 = 4\)

Step 3G

Step 3G - Find minor M31.

In this problem: Delete row 3 and column 1, then take the determinant of the remaining 2×2 matrix.

\(M_{31} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = 0 \cdot 0 - 1 \cdot 1 = -1\)

Step 3H

Step 3H - Find minor M32.

In this problem: Delete row 3 and column 2, then take the determinant of the remaining 2×2 matrix.

\(M_{32} = \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = 2 \cdot 0 - 1 \cdot 1 = -1\)

Step 3I

Step 3I - Find minor M33.

In this problem: Delete row 3 and column 3, then take the determinant of the remaining 2×2 matrix.

\(M_{33} = \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 0 \cdot 1 = 2\)

Step 4

Step 4 - Apply the cofactor sign pattern S to M to get the cofactor matrix C.

In this problem: M is the minor matrix and S is the cofactor sign pattern. Apply S to M to get the cofactor matrix C.

\(M = \begin{pmatrix} 1 & 1 & -1 \\ -2 & -1 & 4 \\ -1 & -1 & 2 \end{pmatrix},\quad S = \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}\;\Rightarrow\; C = \begin{pmatrix} 1 & -1 & -1 \\ 2 & -1 & -4 \\ -1 & 1 & 2 \end{pmatrix}\)

Step 5

Step 5 - Transpose C to get Cᵀ, which is adj( A ).

In this problem: Transpose the cofactor matrix by switching rows and columns.

\(\operatorname{adj}(A) = C^{T} = \begin{pmatrix} 1 & 2 & -1 \\ -1 & -1 & 1 \\ -1 & -4 & 2 \end{pmatrix}\)

Step 6

Step 6 - Multiply adj( A ) by 1 / det( A ) and simplify.

In this problem: Multiply each entry of \(\operatorname{adj}(A)\) by the reciprocal of the determinant and simplify.

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

Final Answer

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