Linear Algebra Determinants: Basic Techniques and Properties Part 1

Introduction to Determinants: Basic Techniques and Properties

These lecture notes, authored by Mayada Shahada for MATH 211 (Linear Algebra with Applications), cover the definition, calculation, and properties of determinants for square matrices of size $n \times n$.

Determinant of a $2 \times 2$ Matrix

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is defined as: $\det(A) = ad - bc$

Notation

To denote the determinant, vertical bars are often used instead of square brackets: $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$

Numerical Example

Given matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, find $\det(A)$. $\det(A) = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = (1)(4) - (2)(3) = 4 - 6 = -2$

Determinant of an $n \times n$ Matrix: Preliminary Definitions

The determinant of an $n \times n$ matrix is a single real number related to geometric properties such as area or volume. It is defined recursively using determinants of smaller $(n-1) \times (n-1)$ submatrices.

1. Sign of a Position

Let $A = [a_{ij}]$ be an $n \times n$ matrix. The sign assigned to the $(i,j)$ position is determined by the formula: $\text{Sign} = (-1)^{i+j}$

• If $(i+j)$ is even, the sign is $+1$ (or simply $+$).
• If $(i+j)$ is odd, the sign is $-1$ (or simply $-$).

For a $4 \times 4$ matrix, the signs follow a checkerboard pattern: $\begin{pmatrix} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{pmatrix}$

2. The Minor of a Matrix

Let $A = [a_{ij}]$ be an $n \times n$ matrix. The $ij^{th}$ minor of $A$, denoted as $minor(A)_{ij}$, is the determinant of the $(n-1) \times (n-1)$ matrix resulting from deleting the $i^{th}$ row and the $j^{th}$ column of $A$.

Example calculation of a minor: Let $A = \begin{pmatrix} 1 & 1 & 3 \\ 2 & 4 & 1 \\ 5 & 2 & 6 \end{pmatrix}$

• To find $minor(A)_{12}$, remove Row 1 and Column 2: $\begin{vmatrix} 2 & 1 \\ 5 & 6 \end{vmatrix} = (2)(6) - (1)(5) = 12 - 5 = 7$
• To find $minor(A)_{23}$, remove Row 2 and Column 3: $\begin{vmatrix} 1 & 1 \\ 5 & 2 \end{vmatrix} = (1)(2) - (1)(5) = 2 - 5 = -3$

3. The Cofactor of a Matrix

The $ij^{th}$ cofactor of $A$, denoted $cof(A)_{ij}$, incorporates the positional sign with the minor: $cof(A)_{ij} = (-1)^{i+j} \times minor(A)_{ij}$

Example calculation of a cofactor: Using the previous matrix $A$, find $cof(A)_{12}$.

• $i+j = 1+2 = 3$ (odd, so sign is negative).
• $minor(A)_{12} = 7$.
• $cof(A)_{12} = (-1)^{1+2} \times 7 = -1 \times 7 = -7$

Cofactor Expansion (Laplace Expansion)

The determinant of an $n \times n$ matrix can be computed by summing the products of the elements of any single row or column by their corresponding cofactors.

Formulas

1. Expanding along the $i^{th}$ row: $\det(A) = a_{i1}cof(A)_{i1} + a_{i2}cof(A)_{i2} + \dots + a_{in}cof(A)_{in} = \sum_{j=1}^{n} a_{ij} cof(A)_{ij}$

2. Expanding along the $j^{th}$ column: $\det(A) = a_{1j}cof(A)_{1j} + a_{2j}cof(A)_{2j} + \dots + a_{nj}cof(A)_{nj} = \sum_{i=1}^{n} a_{ij} cof(A)_{ij}$

Comprehensive Example

Let $A = \begin{pmatrix} 1 & 1 & 3 \\ 2 & 4 & 1 \\ 5 & 2 & 6 \end{pmatrix}$. Find $\det(A)$.

Option 1: Expansion along Row 1$\det(A) = +(1)\begin{vmatrix} 4 & 1 \\ 2 & 6 \end{vmatrix} - (1)\begin{vmatrix} 2 & 1 \\ 5 & 6 \end{vmatrix} + (3)\begin{vmatrix} 2 & 4 \\ 5 & 2 \end{vmatrix}$$\det(A) = 1(24 - 2) - 1(12 - 5) + 3(4 - 20)$$\det(A) = 22 - 7 + 3(-16) = 22 - 7 - 48 = -33$

Option 2: Expansion along Column 2$\det(A) = -(1)\begin{vmatrix} 2 & 1 \\ 5 & 6 \end{vmatrix} + (4)\begin{vmatrix} 1 & 3 \\ 5 & 6 \end{vmatrix} - (2)\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}$$\det(A) = -1(12 - 5) + 4(6 - 15) - 2(1 - 6)$$\det(A) = -7 + 4(-9) - 2(-5) = -7 - 36 + 10 = -33$

Theorems and Efficiency Shortcuts

Theorem: Determinant is Well Defined

The determinant of an $n \times n$ matrix can be computed using cofactor expansion along any row or column. Choosing lines with many zeros drastically simplifies the arithmetic.

Determinants of Special Matrices

1. Zero Row/Column: If $A$ contains a row or column of zeros, then $\det(A) = 0$.
2. Triangular Matrices:
   • Upper Triangular: All entries below the main diagonal are zero.
   • Lower Triangular: All entries above the main diagonal are zero.
   • Diagonal: Both upper and lower triangular.
   • The Product Rule: If $A$ is triangular or diagonal, $\det(A)$ is the product of the diagonal entries:      $\det(A) = a_{11} \times a_{22} \times \dots \times a_{nn}$

Numerical Example (Triangular):$\det\begin{pmatrix} 1 & 2 & 3 \\ 0 & 5 & 6 \\ 0 & 0 & 9 \end{pmatrix} = (1)(5)(9) = 45$

Elementary Row Operations and Determinants

Standard matrix operations affect the determinant value in specific ways. Let $A$ be an $n \times n$ matrix.

Type I: Row Interchanges

If matrix $B$ is obtained by switching two rows of $A$, then: $\det(B) = -\det(A)$

Type II: Scalar Multiplication

If matrix $B$ is obtained by multiplying one row of $A$ by a scalar $k$, then: $\det(B) = k \times \det(A)$

Type III: Row Addition

If matrix $B$ is obtained by adding a multiple of one row to another row, then: $\det(B) = \det(A)$(The determinant remains unchanged.)

Proportional Rows

If $A$ contains a row that is a multiple of another row, then $\det(A) = 0$.

Complex Examples and Problems

Problem 1: $4 \times 4$ Determinant using Column Expansion

Find $\det(A)$ for $A = \begin{pmatrix} 0 & 1 & -2 & 1 \\ 5 & 0 & 0 & 7 \\ 0 & 1 & -1 & 0 \\ 3 & 0 & 0 & 2 \end{pmatrix}$. Choosing Column 2 (since it contains two zeros): $\det(A) = -(1)\begin{vmatrix} 5 & 0 & 7 \\ 0 & -1 & 0 \\ 3 & 0 & 2 \end{vmatrix} - (1)\begin{vmatrix} 0 & -2 & 1 \\ 5 & 0 & 7 \\ 3 & 0 & 2 \end{vmatrix}$ Expansion of the first sub-matrix along its second row: $(-1)\begin{vmatrix} 5 & 7 \\ 3 & 2 \end{vmatrix} = (-1)(10 - 21) = 11$ Expansion of the second sub-matrix along its second column: $-(-2)\begin{vmatrix} 5 & 7 \\ 3 & 2 \end{vmatrix} = 2(10 - 21) = -22$ Final Calculation: $\det(A) = -1(11) - 1(-22) = -11 + 22 = 11$

Practice Problem 2: Row Equality

Let $A = \begin{pmatrix} 2 & 3 & 5 \\ 3 & 5 & 9 \\ 1 & 2 & 4 \end{pmatrix}$. Applying $-1 \times R_3 + R_1 \rightarrow R_1$ and $-1 \times R_3 + R_2 \rightarrow R_2$ results in: $A' = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}$ Since Row 1 equals Row 2, $\det(A) = 0$.