Skill: Calculate the inverse of a 3 by 3 matrix using cofactors

To find the inverse of a $3\times3$ matrix, first calculate its determinant. If the determinant is nonzero, find each cofactor, arrange them in a cofactor matrix, transpose that matrix, and divide by the determinant.

Consider

$$A=\begin{pmatrix} 1&2&0\\ 0&1&1\\ 1&0&1 \end{pmatrix}.$$

1. Calculate the determinant. Expand along the first row. The signs across that row are $+,-,+$:

$$\det(A) =1\begin{vmatrix}1&1\\0&1\end{vmatrix} -2\begin{vmatrix}0&1\\1&1\end{vmatrix} +0\begin{vmatrix}0&1\\1&0\end{vmatrix} =1(1)-2(-1)+0=3.$$

Since $\det(A)=3\ne0$, the inverse exists.

2. Find all the cofactors. Each cofactor is the determinant of the $2\times2$ minor, multiplied by its sign. The signs follow the pattern

$$\begin{pmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{pmatrix}.$$

For example, the cofactor in position $(1,2)$ is

$$C_{12}=-\begin{vmatrix}0&1\\1&1\end{vmatrix}=1.$$

Calculating the other cofactors in the same way gives

$$C=\begin{pmatrix} 1&1&-1\\ -2&1&2\\ 2&-1&1 \end{pmatrix}.$$

3. Transpose the cofactor matrix and divide by the determinant.

$$A^{-1}=\frac{1}{\det(A)}C^{T} =\frac13\begin{pmatrix} 1&-2&2\\ 1&1&-1\\ -1&2&1 \end{pmatrix}.$$

The inverse undoes the transformation represented by $A$: multiplying $A$ by $A^{-1}$ gives the identity matrix $I$. If the determinant had been zero, the inverse would not exist.