Theorem

A homogenous system is always consistent

Proof:

Let $A\vec{x} = \vec{0}$ be a homogeneous system, where $A = [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n]$. The augmented matrix for this system will always be in the form: $$[\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n \ \vec{0}]$$ Since the solution must always be $\vec{0}$, there can never be a pivot in the augmented column. Consequently, the system can never be inconsistent because there will never be a row in the form: $$[0 \ 0 \ \cdots \ 0 \ b]$$ where $b$ is a non-zero entry in the augmented column. For a homogeneous system, any value in the $b$ column will always be zero. Therefore, a homogeneous system is always consistent.

Theorem

If the system is consistent, then the solution to the non-homogenous system $A\vec{x}=\vec{b}$ is the sum of the particular solution, $\vec{p}$, to the non-homogeneous system and a linear combination of the solutions to the related homogenous linear system $A\vec{x}=\vec{0}$

Proof

Theorem

The homogenous equation $A\vec{x}=\vec{0}$ has a non-trivial solution if and only if the equation has at least one free variable.

Proof:

Let $A=[a_1 \ a_2 \ \cdots \ a_n]$ and let $\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$.

The homogenous equation $A\vec{x}=\vec{0}$ has a non-trivial solution.

$\leftrightarrow$ $x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_n\vec{a}_n = \vec{0}$ has a non-trivial solution.

$\leftrightarrow$ There exists some column in $A$ say $a_i$ where $a_i = \vec{0}$. There also exists some component in $\vec{x}$ say $x_i$ where $x \ne 0$.

$\leftrightarrow$ $A$ has a column of zeroes, so there must not be a pivot value in the entire column.

$\leftrightarrow$ There is a free variable in our system.