TITLE: How to find 3 x 3 matrix inverses
QUESTION [6 upvotes]: Is there a way of finding the inverse of a $3 \times 3$ matrix without forming an augmented matrix with the identity matrix? Also, is there a quick way of checking that a  $3 \times 3$ matrix's inverse exists, without trying to compute it? Thanks.

REPLY [9 votes]: Starting with a $3\times 3$ matrix
$$
A=\left(\begin{matrix}
a & b & c \\
d & e & f \\
g & h & i
\end{matrix}\right)
$$
1. Check that the matrix is invertible. First we need to check that we can actually invert the matrix. This amounts to showing that its determinant is non-zero.
$$
\det A=a\det\left(\begin{matrix}e & f \\ h & i\end{matrix}\right)-b\det\left(\begin{matrix}d & f \\ g & i\end{matrix}\right)+c\det\left(\begin{matrix}d & e \\ g & h\end{matrix}\right)
$$
2. Find the cofactors of the matrix. For each of the $(i,j)$-th entries of the matrix $A$, we denote the cofactor $C_{ij}$ by $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the determinant of the $2\times 2$ matrix obtained from $A$ by removing the $i$-th row and the $j$-th column. For example
$$
M_{23}=\det\left(\begin{matrix}
a & b & - \\
- & - & - \\
g & h & -
\end{matrix}\right)
=\det\left(\begin{matrix}a & b \\ g & h\end{matrix}\right)=ah-bg
$$
So $C_{23}=(-1)^{2+3}(ah-bg)=bg-ah$.
3. Find the adjugate matrix. After finding all the cofactors, the adjugate of $A$ is
$$
\text{adj }A=\left(\begin{matrix}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{matrix}\right)^T=\left(\begin{matrix}
C_{11} & C_{21} & C_{31} \\
C_{12} & C_{22} & C_{32} \\
C_{13} & C_{23} & C_{33}
\end{matrix}\right)
$$
That is you make a matrix out of the cofactors, and then you transpose the matrix.
4. We're done! Finally, we can can write the inverse as
$$
A^{-1}=\frac{1}{\det A}\text{adj }A
$$
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