Lesson 3.2: Calculation and Properties of the Determinant

The machinery for computing and reasoning about the scaling factor of space.

In our last lesson, we developed a deep geometric intuition for the determinant. Now, it's time to build the machinery to compute this magical number for any square matrix.

The 3x3 Case: Cofactor Expansion
This method breaks down the determinant of an `n x n` matrix into a combination of determinants of smaller `(n-1) x (n-1)` matrices.

For a general 3x3 matrix expanded along the first row, the formula is:

det(A)=aefhibdfgi+cdegh\det(A) = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}
Key Properties of the Determinant

1. Identity Matrix: det(I)=1\det(I) = 1.

2. Row Swaps: Swapping two rows flips the sign of the determinant.

3. Row Replacement: Adding a multiple of one row to another does not change the determinant.

4. Triangular Matrix: The determinant is the product of the diagonal entries.

5. Invertibility: AA is invertible if and only if det(A)0\det(A) \neq 0.

6. Multiplicative Property: det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B).

7. Inverse: det(A1)=1/det(A)\det(A^{-1}) = 1 / \det(A).

8. Transpose: det(AT)=det(A)\det(A^T) = \det(A).

A Smarter Way to Calculate
Brute-force cofactor expansion is very slow for large matrices. The most efficient way to calculate a determinant is to use row operations to get to REF, then multiply the pivots.

Step 1: Use Row Operations to get to REF

Given A=[246135124]A = \begin{bmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 1 & 2 & 4 \end{bmatrix}, factor 2 from R1, then perform elimination.

det(A)=2det([123012001])\det(A) = 2 \cdot \det(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix})

Step 2: Multiply the Diagonals of U and the factored scalars

The REF matrix `U` has a determinant of 1 (1x1x1).

The final determinant is det(A)=2×1=2\det(A) = 2 \times 1 = 2. This is far faster than a full cofactor expansion.