🎓 Chapter 3: Determinants

A focused summary for linking determinant definitions, computation techniques, algebraic properties, and geometric meaning—empowering confident problem solving in linear algebra.

🔰 Introduction to Determinants

Determinants assign a scalar to each square matrix. This scalar captures the scaling and orientation effect of the linear transformation $(x \mapsto Ax)$ and answers crucial questions: is $A$ invertible? How does $A$ transform area or volume? Can we obtain explicit solutions to $Ax = b$?

Why determinants matter

  • $\det A \neq 0 \iff A$ is invertible.
  • $|\det A|$ measures signed area/volume scaling.
  • Determinants justify Cramer's rule and the adjugate formula for $A^{-1}$.
  • They bridge algebraic manipulations with geometric intuition.

Recursive definition

For $1 \times 1$ matrices, $\det[a] = a$. For $n \geq 2$, determinants are defined via cofactor expansions along any row or column using minors and cofactors $C_{ij}$.

Prerequisite: minors & cofactors

✍️ Notation & First Computation

Notation

If $A = [a_{ij}]$ is $n \times n$, the determinant can be denoted $\det A$ or by vertical bars $|A|$:

$$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, \qquad |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}$$

For a $2 \times 2$ matrix, the determinant formula is:

$$\det A = a_{11}a_{22} - a_{12}a_{21}$$
Example · Compute $\det A$

Let

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

Expanding along the first row:

$$\begin{align} \det A &= 1 \cdot \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} + 3 \cdot \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} \\[0.5em] &= 1(4 \cdot 6 - 5 \cdot 0) - 2(0 \cdot 6 - 5 \cdot 1) + 3(0 \cdot 0 - 4 \cdot 1) \\[0.5em] &= 1(24) - 2(-5) + 3(-4) \\[0.5em] &= 24 + 10 - 12 = 22 \end{align}$$

Result: $\det A = 22$.

🧩 Cofactor Expansion

Theorem (Cofactor expansion) For any row $i$ or column $j$:
$$\det A = \sum_{k=1}^{n} a_{ik} C_{ik} \quad \text{(expansion along row } i\text{)}$$ $$\det A = \sum_{k=1}^{n} a_{kj} C_{kj} \quad \text{(expansion along column } j\text{)}$$
where $C_{ij} = (-1)^{i+j} \det M_{ij}$ and $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$.
Example · Expand along row 3

For

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

expand along the third row:

$$\det A = 4 C_{31} + (-2) C_{32} + 5 C_{33}$$

Computing the cofactors:

$$\begin{align} C_{31} &= (-1)^{3+1} \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = +(1 \cdot 2 - (-1) \cdot 3) = +5 \\[0.8em] C_{32} &= (-1)^{3+2} \begin{vmatrix} 2 & -1 \\ 0 & 2 \end{vmatrix} = -(2 \cdot 2 - (-1) \cdot 0) = -4 \\[0.8em] C_{33} &= (-1)^{3+3} \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix} = +(2 \cdot 3 - 1 \cdot 0) = +6 \end{align}$$

Therefore:

$$\det A = 4(5) + (-2)(-4) + 5(6) = 20 + 8 + 30 = 58$$
Theorem (Triangular matrices) If $A$ is upper triangular, lower triangular, or diagonal, then:
$$\det A = a_{11} \cdot a_{22} \cdot a_{33} \cdots a_{nn} = \prod_{i=1}^{n} a_{ii}$$

⚙️ Row Operation Properties

Theorem (Effect of elementary row operations) Let $A$ be square and $B$ the result of applying a single row operation to $A$:
  • Row replacement: adding a multiple of one row to another leaves the determinant unchanged, $\det B = \det A$.
  • Row swap: swapping two rows changes the sign, $\det B = -\det A$.
  • Row scaling: multiplying a row by scalar $k$ scales the determinant, $\det B = k \det A$.

Combine these rules with reduction to triangular form to compute determinants efficiently while recording necessary adjustments.

Example · Using row operations

To compute $\det A$ where:

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

Factor out 2 from the first row: $\det A = 2 \det \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 0 & 2 & 4 \end{bmatrix}$

Then use row operations to create zeros and reduce to triangular form.

🔗 Determinants, Invertibility & Transpose

Theorem (Invertibility test) A square matrix $A$ is invertible if and only if $\det A \neq 0$.
Theorem (Transpose invariance) For any square matrix $A$:
$$\det(A^T) = \det(A)$$

Thus $A$ and $A^T$ always share determinant, rank, and invertibility. This means cofactor expansion can be performed along rows or columns with equal validity.

🧮 Determinants & Matrix Products

Theorem (Multiplicativity) For $n \times n$ matrices $A$ and $B$:
$$\det(AB) = (\det A)(\det B)$$

Consequently, if $A$ is invertible:

$$\det(A^{-1}) = \frac{1}{\det A} = (\det A)^{-1}$$

This follows from $\det(AA^{-1}) = \det(I) = 1$.

🗝️ Cramer's Rule & Adjugate Matrix

Cramer's rule

Theorem (Cramer's Rule) Let $A$ be an invertible $n \times n$ matrix and consider the system $Ax = b$. The unique solution has components:
$$x_i = \frac{\det A_i(b)}{\det A}, \quad i = 1, 2, \ldots, n$$
where $A_i(b)$ is the matrix formed by replacing column $i$ of $A$ with vector $b$.
Example · Solve via Cramer's rule

Solve the system:

$$\begin{cases} 3x_1 - 2x_2 = 6 \\ -5x_1 + 4x_2 = 8 \end{cases}$$

Matrix form: $Ax = b$ where

$$A = \begin{bmatrix} 3 & -2 \\ -5 & 4 \end{bmatrix}, \quad b = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$$

First, compute $\det A$:

$$\det A = (3)(4) - (-2)(-5) = 12 - 10 = 2$$

For $x_1$, replace column 1 with $b$:

$$A_1(b) = \begin{bmatrix} 6 & -2 \\ 8 & 4 \end{bmatrix} \Rightarrow \det A_1(b) = (6)(4) - (-2)(8) = 24 + 16 = 40$$ $$x_1 = \frac{40}{2} = 20$$

For $x_2$, replace column 2 with $b$:

$$A_2(b) = \begin{bmatrix} 3 & 6 \\ -5 & 8 \end{bmatrix} \Rightarrow \det A_2(b) = (3)(8) - (6)(-5) = 24 + 30 = 54$$ $$x_2 = \frac{54}{2} = 27$$

Solution: $x = \begin{bmatrix} 20 \\ 27 \end{bmatrix}$

Adjugate and inverse

The adjugate matrix $\text{adj}(A)$ is the transpose of the cofactor matrix.

Theorem (Inverse via adjugate) If $A$ is invertible, then:
$$A^{-1} = \frac{1}{\det A} \cdot \text{adj}(A)$$

Although rarely used for large-scale computation, this formula underpins theoretical derivations and proofs.

🌐 Determinants & Geometry

Area in $\mathbb{R}^2$

If $A$ is $2 \times 2$, the area of the parallelogram spanned by the columns of $A$ is $|\det A|$.

$$\text{Area} = \left|\det\begin{bmatrix} a & c \\ b & d \end{bmatrix}\right| = |ad - bc|$$

Volume in $\mathbb{R}^3$

If $A$ is $3 \times 3$, the volume of the parallelepiped determined by the columns of $A$ is $|\det A|$.

$$\text{Volume} = |\det A|$$

More generally, $|\det A|$ measures the factor by which $A$ scales oriented $n$-dimensional volume, and the sign of $\det A$ indicates whether orientation is preserved ($\det A > 0$) or reversed ($\det A < 0$).

Example · Geometric interpretation

Consider vectors $u = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$ and $v = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$:

$$\text{Area of parallelogram} = \left|\det\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}\right| = |3 \cdot 4 - 1 \cdot 1| = |11| = 11$$

📋 Quick Recap Checklist

  • Recognize and use determinant notation $\det A$ and $|A|$.
  • Compute determinants via cofactor expansions; select rows or columns tactically.
  • Leverage triangular form and row-operation properties to accelerate calculations.
  • Apply $\det A \neq 0 \iff A$ invertible and $\det(A^T) = \det A$.
  • Use $\det(AB) = (\det A)(\det B)$ and $\det(A^{-1}) = (\det A)^{-1}$.
  • Implement Cramer's rule and relate $A^{-1}$ to $\text{adj}(A)$.
  • Interpret $|\det A|$ as area/volume scaling with orientation awareness.