The Cofactor Expansion

Chapter 3 • Section 3-1

"Unlocking the determinant, one submatrix at a time! 🗝️"

📦 The Determinant

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$:

$$ \det A = ad - bc $$

Area of the parallelogram!

We use vertical bars:

$$ \det A = |A| $$

Not absolute value!

Determinant = 0

To find the determinant of any $n \times n$ matrix:

Pick any row or column (pick the one with the most zeros!).
For each entry $a_{ij}$ in that row/col:
- Find its Sign: $(-1)^{i+j}$ (Checkerboard pattern).
- Find its Minor: Determinant of the submatrix leaving out row $i$ and col $j$.
- Multiply: Entry $\times$ Sign $\times$ Minor.
Sum them up!

$$ \det A = \sum_{j=1}^n a_{ij} C_{ij} $$

Checkerboard Pattern

+ - + -
- + - +
+ - + -
- + - +

Which row or column would you expand along for $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 0 \\ 7 & 8 & 9 \end{bmatrix}$?

🔄

Swapping two rows flips the sign: $\det B = -\det A$.

✖️

Multiplying a row by $k$ scales determinant by $k$: $\det B = k \det A$.

➕

Adding a multiple of one row to another does not change the determinant! $\det B = \det A$.

If $A$ is Upper or Lower Triangular (zeros below/above diagonal):

$$ \det A = a_{11} \times a_{22} \times \dots \times a_{nn} $$

Just multiply the diagonal!

Compute $\det A = \begin{vmatrix} 2 & 0 & -3 \\ 0 & 4 & 0 \\ 1 & 0 & -2 \end{vmatrix}$.

Strategy: Use row ops to get more zeros!

$R_3 \leftarrow 2R_3 - R_1$: $\begin{vmatrix} 2 & 0 & -3 \\ 0 & 4 & 0 \\ 0 & 0 & -1 \end{vmatrix}$ (Careful! We scaled a row, so we must divide by 2 later, or track the factor).
Actually, let's just expand along Row 2 (it already has two zeros!):

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

If $\det A = 5$, what is $\det(2A)$ for a $3 \times 3$ matrix?