# MATH 323 Lecture 3

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## Homogeneous System

A system of equations is said to be homogeneous if all constants on the right-hand side of the equations are 0.

Homogeneous systems are always consistent usually with the trivial solution ${\displaystyle (x_{1},x_{2},\ldots ,x_{n})=(0,0,\ldots ,0)}$

Theorem. An ${\displaystyle m\times n}$ homogeneous system of linear equations has a nontrivial solution iff ${\displaystyle n>m}$.

Proof. With fewer equations than unknowns, there will be at least ${\displaystyle n-m}$ free variables. Any free variable has an infinite number of possible values, therefore the system has infinitely many solutions.

Q.E.D.

## Matrix Notation

Matrices use CAPITAL LETTERS (${\displaystyle A}$, ${\displaystyle B}$, etc.)

Entries use lowercase letters (${\displaystyle a}$, ${\displaystyle b}$, etc.)

With ${\displaystyle m}$ rows and ${\displaystyle n}$ columns, a matrix is said to be of size ${\displaystyle m\times n}$

${\displaystyle A={\begin{bmatrix}a_{11}&\dots &a_{1n}\\\vdots &\ddots &\vdots \\a_{m1}&\dots &a_{mn}\end{bmatrix}}=\left(a_{ij}\right)}$

A single element may be referenced with subscript row ${\displaystyle i}$, column ${\displaystyle j}$: ${\displaystyle a_{ij}}$

A row ${\displaystyle i}$ may be extracted with the following notation:

${\displaystyle {\vec {a}}(i,:)=(a_{i1},\ldots ,a_{in})}$

A column ${\displaystyle j}$ may be extracted with the following notation:

${\displaystyle {\vec {a}}(:,j)={\vec {a}}_{j}={\begin{bmatrix}a_{1j}\\\vdots \\a_{mj}\end{bmatrix}}}$

Matrix constructor form:

${\displaystyle M_{m\times n}(\mathbb {R} )={\mbox{set (ring) of }}m\times n{\mbox{ matrices over }}\mathbb {R} }$

### Vectors

2 forms:

1. Row vector (${\displaystyle 1\times n}$): ${\displaystyle {\vec {v}}=(a_{11},\ldots ,a_{1n})}$
2. Column vector (${\displaystyle n\times 1}$): ${\displaystyle {\begin{bmatrix}a_{11}\\\vdots \\a_{n1}\end{bmatrix}}}$

Vectors of either type constitute ${\displaystyle m}$-dimensional vector space (represented by ${\displaystyle \mathbb {R} ^{m}}$) in Euclidean space

If ${\displaystyle {\vec {a}}_{1},\ldots ,{\vec {a}}_{n}}$ are vectors in ${\displaystyle \mathbb {R} ^{m}}$ and ${\displaystyle c_{1},\ldots ,c_{n}}$ are scalars, then their sum is of the form ${\displaystyle c_{1}{\vec {a}}_{1}+\dots +c_{n}{\vec {a}}_{n}}$ and is said to be a linear combination of the vectors ${\displaystyle a_{1},\ldots ,a_{n}}$

Theorem 1.3.1. A system of equations is consistent iff ${\displaystyle {\vec {b}}}$ is a linear combination of column vectors ${\displaystyle {\vec {a}}_{i}}$ of the coefficient matrix ${\displaystyle A}$.

${\displaystyle x_{1}{\vec {a}}_{1}+\dots +x_{n}{\vec {a}}_{n}={\vec {b}}}$

### Operations

Two matrices may be added only if they have the same size:

{\displaystyle {\begin{aligned}A+B&=C\\c_{ij}=a_{ij}+b_{ij}\end{aligned}}}

Just add the corresponding cells of each matrix

##### Example

${\displaystyle {\begin{bmatrix}2&-1&3\\-1&-2&1\end{bmatrix}}+{\begin{bmatrix}-1&0&1\\0&-2&-2\end{bmatrix}}={\begin{bmatrix}1&-1&4\\-1&-4&3\end{bmatrix}}}$

#### Multiplication

Matrix ${\displaystyle A}$ may multiply into matrix ${\displaystyle B}$ only if ${\displaystyle A}$ has size ${\displaystyle m\times n}$ and ${\displaystyle B}$ has size ${\displaystyle n\times k}$. The resulting matrix will have size ${\displaystyle m\times k}$

Take row ${\displaystyle i}$, multiply each of its ${\displaystyle n}$ cells by the corresponding cells of column ${\displaystyle j}$. The resulting matrix cell will be the sum of the components' products:

${\displaystyle c_{ij}=\sum _{t=1}^{n}a_{it}b_{tj}}$
where ${\displaystyle 1\leq i\leq m}$ and ${\displaystyle 1\leq j\leq k}$

It's just a dot product!

Warning: ${\displaystyle AB\neq BA}$
##### Application

Consider the following equation: ${\displaystyle 2x_{1}-3x_{2}+x_{3}=1}$

Write the coefficient matrix:

${\displaystyle {\begin{bmatrix}2&-3&1\end{bmatrix}}}$

... and vector of unknowns:

${\displaystyle {\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}}$

Multiplying the coefficients by the unknowns yields the same equation:

${\displaystyle {\begin{bmatrix}2&-3&1\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}2x_{1}-3x_{2}+x_{3}\end{bmatrix}}}$
##### Example

[itex]\begin{bmatrix}2 & -1 & 1 \\ -1 & 0 & 1\end{bmatrix} \, \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 + (-1) \cdot 0 + 1 \cdot 1 & 2 \cdot (-1) + (-1) \cdot 2 + 1 \cdot (-1) \\ (-1) \cdot 1 + 0 \cdot 0 + 1 \cdot 1 & (-1) \cdot (-1) + 0 \cdot 2 + 1 \cdot (-1) \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ 0 & 0 \end{bmatrix}