Inverting a 3x3 Matrix | Generated by AI
Yes, you’re on the right track! You’re describing the standard method for finding the inverse of a 3×3 matrix \( A \) (often denoted \( A^{-1} \)) using the cofactor matrix (or more precisely, the adjugate matrix, which is the transpose of the cofactor matrix). The “divider 1/value * matrix” part refers to scaling by \( 1/\det(A) \), where \( \det(A) \) is the determinant of \( A \). And “rorate” might mean “transpose” (rotating rows to columns)—that’s a key step too.
I’ll explain the method step by step, why it works, and how to compute it for a general 3×3 matrix. This comes from the general formula for matrix inverses: \( A^{-1} = \frac{1}{\det(A)} \adj(A) \), where \( \adj(A) \) is the adjugate.
Step-by-Step Method for a 3×3 Matrix
Let’s say your matrix is:
\[
A = \begin{pmatrix}
a & b & c
d & e & f
g & h & i
\end{pmatrix}
\]
-
Compute the Determinant \( \det(A) \) (the “value” you divide by): \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] If \( \det(A) = 0 \), the matrix has no inverse (it’s singular). Otherwise, proceed.
- Find the Cofactor Matrix \( C \):
- The cofactor \( C_{ij} \) for element \( a_{ij} \) is \( (-1)^{i+j} \) times the determinant of the 2×2 submatrix obtained by deleting row \( i \) and column \( j \).
- You don’t need to compute all cofactors from scratch—you can compute them along one row (or column) and use properties to fill the rest, but for 3×3, it’s quick to do all nine.
The cofactors are: \[ C_{11} = (-1)^{2} \det\begin{pmatrix} e & f \ h & i \end{pmatrix} = ei - fh \] \[ C_{12} = (-1)^{3} \det\begin{pmatrix} d & f \ g & i \end{pmatrix} = -(di - fg) \] \[ C_{13} = (-1)^{4} \det\begin{pmatrix} d & e \ g & h \end{pmatrix} = dh - eg \] \[ C_{21} = (-1)^{3} \det\begin{pmatrix} b & c \ h & i \end{pmatrix} = -(bi - ch) \] \[ C_{22} = (-1)^{4} \det\begin{pmatrix} a & c \ g & i \end{pmatrix} = ai - cg \] \[ C_{23} = (-1)^{5} \det\begin{pmatrix} a & b \ g & h \end{pmatrix} = -(ah - bg) \] \[ C_{31} = (-1)^{4} \det\begin{pmatrix} b & c \ e & f \end{pmatrix} = bf - ce \] \[ C_{32} = (-1)^{5} \det\begin{pmatrix} a & c \ d & f \end{pmatrix} = -(af - cd) \] \[ C_{33} = (-1)^{6} \det\begin{pmatrix} a & b \ d & e \end{pmatrix} = ae - bd \]
So the cofactor matrix is: \[ C = \begin{pmatrix} ei-fh & -(di-fg) & dh-eg
-(bi-ch) & ai-cg & -(ah-bg)
bf-ce & -(af-cd) & ae-bd \end{pmatrix} \] -
Transpose to Get the Adjugate \( \adj(A) = C^T \) (this is the “rorate” step—swap rows and columns): \[ \adj(A) = \begin{pmatrix} ei-fh & -(bi-ch) & bf-ce
-(di-fg) & ai-cg & -(af-cd)
dh-eg & -(ah-bg) & ae-bd \end{pmatrix} \] - Scale by \( 1/\det(A) \) to Get the Inverse: \[ A^{-1} = \frac{1}{\det(A)} \adj(A) \] Each entry of \( \adj(A) \) gets multiplied by \( 1/\det(A) \).
Example
Let \( A = \begin{pmatrix} 1 & 2 & 3 \ 0 & 1 & 4 \ 5 & 6 & 0 \end{pmatrix} \).
- \( \det(A) = 1(1\cdot0 - 4\cdot6) - 2(0\cdot0 - 4\cdot5) + 3(0\cdot6 - 1\cdot5) = 1(-24) - 2(-20) + 3(-5) = -24 + 40 - 15 = 1 \).
- Cofactors (quickly): \( C = \begin{pmatrix} -24 & 20 & -5 \ -15 & -15 & 5 \ -14 & -10 & 1 \end{pmatrix} \) (I’ll skip full calc here, but you can verify).
- \( \adj(A) = C^T = \begin{pmatrix} -24 & -15 & -14 \ 20 & -15 & -10 \ -5 & 5 & 1 \end{pmatrix} \).
- \( A^{-1} = \frac{1}{1} \adj(A) = \begin{pmatrix} -24 & -15 & -14 \ 20 & -15 & -10 \ -5 & 5 & 1 \end{pmatrix} \).
You can check: \( A \cdot A^{-1} = I \) (the identity matrix).
Why Does This Work?
The key reason is a property of determinants and cofactors: for any square matrix \( A \), \[ A \cdot \adj(A) = \det(A) \cdot I \] (where \( I \) is the identity matrix). This comes from the Laplace expansion of the determinant along rows/columns. Each entry in \( A \cdot \adj(A) \) expands to a sum that equals \( \det(A) \) on the diagonal and 0 elsewhere (because off-diagonal terms cancel via cofactor signs).
To get the inverse, divide both sides by \( \det(A) \) (assuming it’s nonzero): \[ A \cdot \left( \frac{1}{\det(A)} \adj(A) \right) = I \implies A^{-1} = \frac{1}{\det(A)} \adj(A) \] Similarly, \( \adj(A) \cdot A = \det(A) \cdot I \), so it works from both sides.
This method generalizes to any \( n \times n \) matrix, but for 3×3, it’s efficient. For larger matrices, Gaussian elimination is often faster computationally.
If you have a specific matrix you’d like me to invert or clarify a step, just share it!