Inverting a 3x3 Matrix
Inverting a 3x3 matrix
- Firstly, find the determinant of your matrix.
- Second, you form a matrix of minors
- To do this, you find the minor of each value in the matrix and overwrite the original value in your result in a new matrix.
- Next, you form a matrix of co-factors.
- A cofactor is just a signed minor.
- This means we put in pluses and minuses in a pattern.
- \(\begin{pmatrix}+ & - \\ - & +\end{pmatrix}\)
- Finally, you find \(\frac{1}{\text{det of the ORIGINAL matrix}}\) and times it by a transposed version of your co-factor matrix.
- This gives \(\mathbf{A^{-1}}\) for any matrix.
Applying this to a 2x2 matrix to start
- \(\mathbf{B}=\begin{pmatrix}2 & -3 \\ 1 & -5\end{pmatrix}\)
- \(\det \mathbf{B}=-10+3=-7\)
- \(\mathbf{M}=\begin{pmatrix}-5 & 1 \\ -3 & 2\end{pmatrix}\)
- \(\mathbf{C}=\begin{pmatrix}-5 & -1 \\ 3 & -2\end{pmatrix}\)
- \(\mathbf{C^T}=\begin{pmatrix}-5 & 3 \\ -1 & 2\end{pmatrix}\)
- \(\mathbf{B^{-1}}=-\frac{1}{7}\begin{pmatrix}-5 & 3 \\ -1 & 2\end{pmatrix}\)
Applying this to a 3x3 matrix
- \(\mathbf{A}=\begin{pmatrix}1 & 3 & 1 \\ 0 & 4 & 1 \\ 2 & -1 & 0\end{pmatrix}\)
- \(\det \mathbf{A}=(1\times1)-(3\times -2)+(1\times -8)=1+6-8=-1\)
- \(\mathbf{M}=\begin{pmatrix}1 & -2 & -8 \\ 1 & -2 & -7 \\ -1 & 1 & 4\end{pmatrix}\)
- \(\mathbf{C}=\begin{pmatrix}1 & 2 & -8 \\ -1 & -2 & 7 \\ -1 & -1 & 4\end{pmatrix}\)
- \(\mathbf{C^T}=\begin{pmatrix}1 & -1 & -1 \\ 2 & -2 & -1 \\ -8 & 7 & 4\end{pmatrix}\)
- \(\mathbf{A^{-1}}=-1\begin{pmatrix}1 & -1 & -1 \\ 2 & -2 & -1 \\ -8 & 7 & 4\end{pmatrix}=\begin{pmatrix}-1 & 1 & 1 \\ -2 & 2 & 1 \\ 8 & -7 & -4\end{pmatrix}\)