A natural question to ask here is whether or not there is generalized version of the determinant for $3 \times 3$ matrices that can tell us if a $3 \times 3$ matrix has an inverse. While there are many quantities one could compute that will provide such information, there is exactly one way to do it that will satisfy the following properties
Therefore this quantity is a natural choice for the generalized notion of determinant.
There are several methods for computing the determinant of a $3 \times 3$ matrix. One straightforward method is the method of “cofactor expansion” which computes the $3 \times 3$ determinant as a weighted sum of three $2 \times 2$ determinant.
For a $3 \times 3$ matrix $A$, the $(i,j)$-minor of $A$ is the determinant of the submatrix formed by deleting the $i$-th row and $j$-th column of $A$.
Geometrically, we can consider a minor to be a measurement that can determine if the projection of the columns to certain coordinate planes are collinear. The algebraic meaning is more complicated, it has connections to the concept of “multi-linear forms”. This leads to the concept of cofactors.
For a $3 \times 3$ matrix $A$, the $(i,j)$-cofactor of $A$ is the $(i,j)$-minor times $(-1)^{i+j}$.
In most practical situations, we only need to consider cofactors along the first row or column (i.e., $i=1$ or $j=1$ in most situations this semester).
This formula, also known as formula computes the determinant as a weighted sum of cofactors. For a $3 \times 3$ matrix $A$, with the first row being
\[\begin{bmatrix} a_{11} & a_{12} & a_{13} \end{bmatrix},\]the determinant of $A$ can be computed via cofactor expansion formula along the first row given by
\[\det A = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}\]where $C_{ij}$ is the $(i,j)$-cofactor of $A$. In general, the cofactor expansion formula along the $i$-th row is
\[\det A = a_{i1} C_{i1} + a_{i2} C_{i2} + a_{i3} C_{i3}.\]Similarly, the cofactor expansion formula along the $j$-th column is given by
\[\det A = a_{1j} C_{1j} + a_{2j} C_{2j} + a_{3j} C_{3j}\]One important fact is that regardless of how we perform the cofactor expansion (along any row or column), the result is always the same (see Leibniz formula below).
Exercise. Using the cofactor expansion formula, compute the determinant of the following matrices
\[\begin{aligned} A &= \begin{bmatrix} 2 & 3 & 5 \\ 4 & 1 & 2 \\ 1 & 2 & 6 \\ \end{bmatrix} & B &= \begin{bmatrix} 0 & 2 & 4 \\ 5 & 1 & 2 \\ 0 & 4 & 6 \\ \end{bmatrix} & C &= \begin{bmatrix} 1 & 0 & 2 \\ 4 & 2 & 2 \\ 5 & 0 & 6 \\ \end{bmatrix} \end{aligned}\]The most crucial fact, is that
\[\det A \ne 0 \quad\Longleftrightarrow\quad A \text{ has an inverse}.\]This is not easy to prove. We will study this in the context of adjugate matrix.
Another formula for the $3 \times 3$ determinant is defined in terms permutations. This is the Leibniz formula:
\[\det \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} = aei + bfg + cdh - ceg - bdi - afh\]Here, the determinant is expressed as a sum of 6 terms. Each term has 3 factors, one from each row, and the column indices corresponding to the 6 permutations. It is easy to check that this formula is equivalent to the cofactor expansion formula.