3×3 Matrix Inverse Formula

1. What is Matrix Inverse?

The inverse of a matrix A, denoted as \(A^{-1}\), is a matrix that when multiplied by the original matrix gives the identity matrix.

\[A \cdot A^{-1} = A^{-1} \cdot A = I\]

where \(I\) is the identity matrix:

\[I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

2. Main Inverse Formula

The 3×3 Matrix Inverse Formula

\[A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)\]

\[A^{-1} = \frac{\text{adj}(A)}{\det(A)}\]

where:

\(\det(A)\) = Determinant of matrix A
\(\text{adj}(A)\) = Adjugate (or Adjoint) of matrix A

Condition for Invertibility

\(A^{-1}\) exists if and only if \(\det(A) \neq 0\) (Matrix is non-singular)
\(A^{-1}\) does NOT exist if \(\det(A) = 0\) (Matrix is singular)

3. General 3×3 Matrix

Let A be a 3×3 matrix:

\[A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\]

4. Step 1: Calculate the Determinant

Determinant of 3×3 Matrix

\[\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]

Or using the expansion method:

\[\det(A) = a\begin{vmatrix} e & f \\ h & i \end{vmatrix} - b\begin{vmatrix} d & f \\ g & i \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix}\]

5. Step 2: Find the Matrix of Minors

Minor of Each Element

The minor \(M_{ij}\) of element at position (i,j) is the determinant of the 2×2 matrix that remains after removing row i and column j.

\[M = \begin{bmatrix} M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33} \end{bmatrix}\]

where:

\(M_{11} = ei - fh\)
\(M_{12} = di - fg\)
\(M_{13} = dh - eg\)
\(M_{21} = bi - ch\)
\(M_{22} = ai - cg\)
\(M_{23} = ah - bg\)
\(M_{31} = bf - ce\)
\(M_{32} = af - cd\)
\(M_{33} = ae - bd\)

6. Step 3: Apply the Cofactor Signs

Matrix of Cofactors

Apply the checkerboard pattern of signs to the matrix of minors:

Sign pattern:

\[\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\]

Cofactor matrix C:

\[C = \begin{bmatrix} +(ei-fh) & -(di-fg) & +(dh-eg) \\ -(bi-ch) & +(ai-cg) & -(ah-bg) \\ +(bf-ce) & -(af-cd) & +(ae-bd) \end{bmatrix}\]

7. Step 4: Find the Adjugate (Adjoint) Matrix

Transpose of Cofactor Matrix

The adjugate matrix is the transpose of the cofactor matrix:

\[\text{adj}(A) = C^T\] \[\text{adj}(A) = \begin{bmatrix} (ei-fh) & -(bi-ch) & (bf-ce) \\ -(di-fg) & (ai-cg) & -(af-cd) \\ (dh-eg) & -(ah-bg) & (ae-bd) \end{bmatrix}\]

8. Step 5: Calculate the Inverse

Final Inverse Formula

\[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} (ei-fh) & -(bi-ch) & (bf-ce) \\ -(di-fg) & (ai-cg) & -(af-cd) \\ (dh-eg) & -(ah-bg) & (ae-bd) \end{bmatrix}\]

Or divide each element of adj(A) by det(A)

9. Complete Formula in One Expression

\[A^{-1} = \frac{1}{a(ei-fh) - b(di-fg) + c(dh-eg)} \begin{bmatrix} (ei-fh) & -(bi-ch) & (bf-ce) \\ -(di-fg) & (ai-cg) & -(af-cd) \\ (dh-eg) & -(ah-bg) & (ae-bd) \end{bmatrix}\]

10. Worked Example

Example: Find the inverse of matrix A

Given:

\[A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{bmatrix}\]

Step 1: Calculate the determinant \[\det(A) = 1(1 \cdot 0 - 4 \cdot 6) - 2(0 \cdot 0 - 4 \cdot 5) + 3(0 \cdot 6 - 1 \cdot 5)\] \[\det(A) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5)\] \[\det(A) = -24 + 40 - 15 = 1\]

Step 2: Find matrix of minors