Every square matrix has a single number attached to it, its determinant. It tells you whether a system has a unique solution, whether a matrix has an inverse, and how a transformation changes area. It grows out of matrix arithmetic and elimination.
Key idea
$$\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$$ For a $3 \times 3$ matrix, expand along the first row with the signs $+, -, +$: each entry times the $2 \times 2$ determinant left when you cross out its row and column (its minor).
Why it works
Solve $ax + by = e$, $cx + dy = f$ by elimination: multiply the first equation by $d$ and the second by $b$, then subtract. You get $(ad - bc)x = ed - bf$. If $ad - bc \ne 0$ you can divide, and there is exactly one solution; if $ad - bc = 0$ you cannot. Geometrically, $|ad - bc|$ is the area of the parallelogram formed by the columns $\langle a, c \rangle$ and $\langle b, d \rangle$.
Worked examples
Worked example: Two by two
Find $\begin{vmatrix} 4 & -3 \\ 2 & 5 \end{vmatrix}$.
- $ad - bc = 4 \cdot 5 - (-3) \cdot 2 = 20 + 6$.
Answer: $26$
Worked example: Three by three
Find $\begin{vmatrix} 2 & 1 & 0 \\ 1 & 3 & -1 \\ 0 & 2 & 4 \end{vmatrix}$.
- Expand along row 1 with signs $+, -, +$: $$2\begin{vmatrix} 3 & -1 \\ 2 & 4 \end{vmatrix} - 1\begin{vmatrix} 1 & -1 \\ 0 & 4 \end{vmatrix} + 0\begin{vmatrix} 1 & 3 \\ 0 & 2 \end{vmatrix}.$$
- $$= 2(12 + 2) - 1(4 - 0) + 0 = 28 - 4.$$
Answer: $24$
Worked example: When is it zero?
For what $k$ is $\begin{vmatrix} k & 6 \\ 2 & 4 \end{vmatrix} = 0$?
- $4k - 12 = 0$, so $k = 3$.
- Then the rows $(3, 6)$ and $(2, 4)$ are proportional, so the system they describe has no unique solution.
Answer: $k = 3$
Common mistake
Computing $ad + bc$, or losing the alternating signs in a $3 \times 3$ expansion. For $2 \times 2$: main diagonal product minus the other diagonal product. For $3 \times 3$: the signs follow the checkerboard $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$. A quick check: swapping two rows should change the sign of the determinant.