Skill: Calculate minors of a 3 by 3 matrix

To find the minor $M_{ij}$ of an entry in row $i$, column $j$:

1. Delete that entry’s row and column.
2. Find the determinant of the remaining $2\times2$ matrix. For $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, calculate $ad-bc$.

Minors are unsigned in the sense that you do not apply the alternating signs used for cofactors. A minor itself can still be negative.

For example, find all the minors of

$$A=\begin{pmatrix} 1&2&3\\ 0&4&5\\ 1&0&6 \end{pmatrix}.$$

For the entry in row 1, column 1, delete row 1 and column 1:

$$M_{11}=\begin{vmatrix}4&5\\0&6\end{vmatrix} =(4)(6)-(5)(0)=24.$$

Repeat this process for each entry. The resulting minors are

$$\begin{aligned} M_{11}&=\begin{vmatrix}4&5\\0&6\end{vmatrix}=24, & M_{12}&=\begin{vmatrix}0&5\\1&6\end{vmatrix}=-5, & M_{13}&=\begin{vmatrix}0&4\\1&0\end{vmatrix}=-4,\\[4pt] M_{21}&=\begin{vmatrix}2&3\\0&6\end{vmatrix}=12, & M_{22}&=\begin{vmatrix}1&3\\1&6\end{vmatrix}=3, & M_{23}&=\begin{vmatrix}1&2\\1&0\end{vmatrix}=-2,\\[4pt] M_{31}&=\begin{vmatrix}2&3\\4&5\end{vmatrix}=-2, & M_{32}&=\begin{vmatrix}1&3\\0&5\end{vmatrix}=5, & M_{33}&=\begin{vmatrix}1&2\\0&4\end{vmatrix}=4. \end{aligned}$$

The minors, arranged in the same positions as the original entries, are

$$\begin{pmatrix} 24&-5&-4\\ 12&3&-2\\ -2&5&4 \end{pmatrix}.$$

Each minor is just the $2\times2$ determinant left after deleting the corresponding row and column.