# MATH 323 Lecture 1

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## Course Information

2 in-class exams and a final (all in C E 222)

1. 7th Week
2. 12th Week
3. TBA

Basic Scientific Calculator is allowed on exam (TI-34)

## Chapter 1: Matrices and Systems of Equations

A system is a set of unknown variables ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$ and functions ${\displaystyle f_{1}(x_{1},x_{2},\ldots ,x_{n})=0,f_{2}(\ldots ),\ldots }$.

• Linear functions are function for which the exponent on all unknown variables (x's) is 1. Can be plotted as a line/plane in 2D/3D space
• ${\displaystyle f(x_{1},\ldots ,x_{n})=a_{1}x_{1}+a_{2}x_{2}+\dots +a_{n}x_{n}+b}$
• Quadratic functons are non-linear functions for which the largest exponent on all x's is 2

Consider systems of linear equations, each function is often written in form ${\displaystyle a_{1}x_{1}+a_{2}x_{2}+\dots +a_{n}x_{n}=b}$

${\displaystyle {\begin{cases}a_{11}x_{1}+a_{12}x_{2}+\dots +a_{1n}x_{n}=b_{1}\\a_{21}x_{1}+a_{22}x_{2}+\dots +a_{2n}x_{n}=b_{2}\\\dots \\a_{m1}x_{1}+a_{m2}x_{2}+\dots +a_{mn}x_{n}=b_{m}\end{cases}}}$

There are ${\displaystyle m}$ equations and ${\displaystyle n}$ unknowns. We will refer to this system as a ${\displaystyle m\times n}$ system.

### Rules

1. The order in which any two equations are written may be interchanged
2. Both sides of an equation may be multiplied by the same non-zero real number
3. A multiple of one equation may be added to / subtracted from another equation

These are more applicable to lesser systems of equations, and leads to a system that is equivalent to the original.

### Examples

1. ${\displaystyle {\begin{cases}x_{1}+2x_{2}=5\\2x_{1}+3x_{2}=8\end{cases}}}$ (2 × 2, the unique solution to which is ${\displaystyle x=(1,2)}$)
2. ${\displaystyle {\begin{cases}x_{1}-x_{2}+x_{3}=2\\2x_{1}+x_{2}-x_{3}=4\end{cases}}}$ (2 × 3, solutions are ${\displaystyle \left\{(2,\beta ,\beta )\mid \forall \beta \in \mathbb {R} \right\}}$
3. ${\displaystyle {\begin{cases}x_{1}+x_{2}=2\\x_{1}-x_{2}=1\\x_{1}=4\end{cases}}}$ (3 × 2, no solution that satisfies answers)

In examples 1 & 2, the systems were consistent, but since the solution set for 3 was empty, it is inconsistent

### 2 × 2 Systems

In general, ${\displaystyle {\begin{cases}a_{11}x_{1}+a_{12}x_{2}=b_{1}\\a_{21}x_{1}+a_{22}x_{2}=b_{2}\end{cases}}}$

Plotting each equation yields a line.

1. If a unique solution exists, it will be at the coordinates of the lines' intersection ({x_1}^0, {x_2}^0
2. If empty solutions (inconsistent), both lines will be parallel
3. If infinite solutions, both lines are the same line

### Equivalent Systems

Given 2 systems I and II, with potentially different number of equations but same number of unknowns (${\displaystyle m\times n}$ and ${\displaystyle k\times n}$), they are called equivalent if they share the same solution set.

(Prof's notation: ${\displaystyle I\sim II}$)

Consider the following systems:

${\displaystyle {\begin{cases}3x_{1}+2x_{2}-x_{3}=-2\\x_{2}=3\\2x_{3}=4\end{cases}}}$

${\displaystyle {\begin{cases}3x_{1}+2x_{2}-x_{3}=-2\\-3x_{1}-x_{2}+x_{3}=5\\3x_{1}+2x_{2}+x_{3}=2\end{cases}}}$

The two systems are equivalent because they both have unique solutions (-2, 3, 2)

Suppose we take two equations ${\displaystyle f_{i},f_{j},0\leq i,j\leq m}$ from a ${\displaystyle m\times n}$ system and replace ${\displaystyle f_{j}}$ by ${\displaystyle f_{j}+\alpha f_{i},\alpha \in \mathbb {R} }$, the original system and the new system are equivalent. The opposite direction also applies: ${\displaystyle f_{j}=f_{j}-\alpha f_{i}}$

### Strict Triangular Form

Given a ${\displaystyle m\times n}$ system of equations where ${\displaystyle m=n}$, in ${\displaystyle k}$th equation, the first ${\displaystyle k-1}$ variables/coefficient are all zero and the coefficient of ${\displaystyle x_{k}}$ is not 0

{\displaystyle {\begin{aligned}a_{11}\,x_{1}+\dots +a_{1n}\,x_{n}&=b_{1}\\0x_{1}+a_{22}\,x_{2}+\dots +a_{2n}\,x_{n}&=b_{2}\\0x_{1}+0x_{2}+a_{33}\,x_{3}+\dots +a_{3n}\,x_{n}&=b_{3}\\\dots \\0x_{1}+0x_{2}+\dots +0x_{n-1}+a_{mn}\,x_{n}&=b_{m-1}\end{aligned}}}

There is always a unique solution to a system of this form.

Solve the bottom 1-term equation, then plug that back into the previous equation, and continue back-substitution until all equations are satisfied.

Goal of rules is to reduce a system into strict triangular form:

${\displaystyle {\begin{cases}x_{1}+2x_{2}+x_{3}=3\\3x_{1}-x_{2}-3x_{3}=-1\\2x_{1}+3x_{2}+x_{3}=4\\\end{cases}}}$

${\displaystyle (2)-3(1)\leftarrow -7x_{2}-6x_{3}=-10}$ ${\displaystyle (3)-2(1)\leftarrow -x_{1}-x_{3}=-2}$ ${\displaystyle (3)-{\frac {1}{7}}(2)\leftarrow -{\frac {1}{7}}x_{3}=-{\frac {4}{7}}}$

Solution is (3,-2,4)

## Homework

Section 1.1: 1c, 2, 3, 5cd, 6ceh, 7, 8

Section 1.2: 3, 5egi, 8, 9