Determinants

By the end of this lesson, you will be able to compute determinants of 2×2 and 3×3 matrices using cofactor expansion, interpret the determinant geometrically, and use determinants to assess invertibility.

Determinant of a 2×2 Matrix

$$\det\begin{pmatrix}a&b\c&d\end{pmatrix} = ad - bc$$

The determinant is a scalar that encodes properties of the matrix.

Invertibility: $A$ is invertible iff $\det(A) \neq 0$.

Determinant of a 3×3 Matrix

Cofactor expansion along the first row:

$$\det\begin{pmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}\end{pmatrix}$$

$= a_{11}\det\begin{pmatrix}a_{22}&a_{23}\a_{32}&a_{33}\end{pmatrix} - a_{12}\det\begin{pmatrix}a_{21}&a_{23}\a_{31}&a_{33}\end{pmatrix} + a_{13}\det\begin{pmatrix}a_{21}&a_{22}\a_{31}&a_{32}\end{pmatrix}$

Sign pattern for cofactors:

$$\begin{pmatrix}+&-&+\-&+&-\+&-&+\end{pmatrix}$$

Example 1 — 3×3 Determinant

$$\det\begin{pmatrix}2&-1&3\0&4&-2\5&1&0\end{pmatrix}$$

Expand along row 1:

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

$= 2(0+2) + 1(0+10) + 3(0-20) = 4 + 10 - 60 = -46$

Properties of Determinants

1. $\det(AB) = \det(A)\det(B)$
2. $\det(A^T) = \det(A)$
3. $\det(kA) = k^n\det(A)$ for $n\times n$ matrix
4. If any row/column is all zeros: $\det(A) = 0$
5. Swapping two rows changes the sign of the determinant
6. If two rows are identical: $\det(A) = 0$

Geometric Interpretation

For a 2×2 matrix with column vectors $\vec{u}$ and $\vec{v}$:

$|\det(A)|$ = area of the parallelogram formed by $\vec{u}$ and $\vec{v}$.

For a 3×3 matrix: $|\det(A)|$ = volume of the parallelepiped.

$\det(A) = 0$ means the vectors are linearly dependent (coplanar in 3D) and the matrix is singular (not invertible).

Example 2 — Area Using Determinant

Find the area of the parallelogram with vertices at $O(0,0)$, $A(3,1)$, $B(1,4)$, $C(4,5)$.

$\text{Area} = \left|\det\begin{pmatrix}3&1\1&4\end{pmatrix}\right| = |12 - 1| = 11$

Cramer's Rule Preview

For a 2×2 system $AX = B$ with $\det(A) \neq 0$:

$$x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}$$

where $A_x$ replaces the $x$-column with $B$, and $A_y$ replaces the $y$-column.

Summary

2×2: $\det\begin{pmatrix}a&b\c&d\end{pmatrix} = ad - bc$