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

Expanding along row 1:

$\det(A) = 2 \cdot \det\begin{pmatrix} -1 & 4 \\ 2 & 1 \end{pmatrix} - 1 \cdot \det\begin{pmatrix} 0 & 4 \\ 5 & 1 \end{pmatrix} + 3 \cdot \det\begin{pmatrix} 0 & -1 \\ 5 & 2 \end{pmatrix}$

$= 2(-1 - 8) - 1(0 - 20) + 3(0 + 5) = 2(-9) - 1(-20) + 3(5) = -18 + 20 + 15 = 17.$

Column 1 has a zero entry at position $(2, 1)$, so one term vanishes:

$\det(A) = 2 \cdot C_{11} + 0 \cdot C_{21} + 5 \cdot C_{31}$

$= 2 \cdot \det\begin{pmatrix} -1 & 4 \\ 2 & 1 \end{pmatrix} + 0 + 5 \cdot (-1)^{3+1}\det\begin{pmatrix} 1 & 3 \\ -1 & 4 \end{pmatrix}$

$= 2(-9) + 5(4 + 3) = -18 + 35 = 17.$

Same answer. Choosing a row/column with zeros saves work.

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

Expand along column 2 (two zeros): only rows 3 contributes.

$\det(A) = (-1)^{3+2} \cdot 1 \cdot \det\begin{pmatrix} 1 & 0 & 3 \\ 2 & 0 & 1 \\ 0 & 3 & 4 \end{pmatrix} = -\det\begin{pmatrix} 1 & 0 & 3 \\ 2 & 0 & 1 \\ 0 & 3 & 4 \end{pmatrix}.$

Expand this $3 \times 3$ along column 2: only row 3 contributes.

$= -(-1)^{3+2} \cdot 3 \cdot \det\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = -(- 3)(1 - 6) = -(-3)(-5) = -15.$

For an upper triangular $3 \times 3$ matrix, expanding along column 1 repeatedly:

$\det\begin{pmatrix} a & * & * \\ 0 & b & * \\ 0 & 0 & c \end{pmatrix} = a \cdot \det\begin{pmatrix} b & * \\ 0 & c \end{pmatrix} = a \cdot bc = abc.$

Generalizing: the determinant of a triangular matrix is the product of diagonal entries.

For a block upper triangular matrix $\begin{pmatrix} A & B \\ 0 & D \end{pmatrix}$ where $A$ is $k \times k$ and $D$ is $\ell \times \ell$:

$\det\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} = \det(A) \cdot \det(D).$

This generalizes the triangular matrix result.

The Vandermonde matrix with nodes $x_1, \ldots, x_n$:

$V = \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{pmatrix}$

has $\det(V) = \prod_{1 \leq i < j \leq n} (x_j - x_i)$. So $V$ is invertible iff all $x_i$ are distinct.

For $n = 3$: $\det(V) = (x_2 - x_1)(x_3 - x_1)(x_3 - x_2)$.

$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$, so $A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

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

$\det(A) = 1(0-24) - 2(0-20) + 3(0-5) = -24 + 40 - 15 = 1$.

Computing all nine cofactors and transposing gives $\text{adj}(A) = A^{-1}$ (since $\det A = 1$).

What happens if we expand using entries of row $i$ with cofactors of row $k$ ($k \neq i$)?

$\sum_{j=1}^n a_{ij} C_{kj} = 0 \quad (i \neq k).$

This is because the sum equals the determinant of a matrix with two identical rows ($i$ and $k$), which is zero. This fact is used to prove $A \cdot \text{adj}(A) = \det(A) \cdot I$.

If $\text{rank}(A) = n - 1$, then $\text{adj}(A)$ has rank 1. If $\text{rank}(A) \leq n - 2$, then $\text{adj}(A) = 0$. The adjugate captures the "next level" of singularity.

For a $3 \times 3$ matrix, each cofactor is a $2 \times 2$ determinant. Computing all nine cofactors involves nine $2 \times 2$ determinants. This is more work than row reduction for large matrices, but the adjugate formula is theoretically important.

For a $3 \times 3$ skew-symmetric matrix $A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$, expanding along row 1:

$\det(A) = 0 \cdot M_{11} - a \cdot \det\begin{pmatrix} -a & c \\ -b & 0 \end{pmatrix} + b \cdot \det\begin{pmatrix} -a & 0 \\ -b & -c \end{pmatrix} = -a(bc) + b(ac) = 0.$

Every odd-dimensional skew-symmetric matrix has determinant $0$.