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Theorem

A homogenous system is always consistent

Proof:

Let Ax=0 A\vec{x} = \vec{0} be a homogeneous system, where A=[a1 a2  an] A = [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n] . The augmented matrix for this system will always be in the form: [a1 a2  an 0] [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n \ \vec{0}] Since the solution must always be 0 \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  0 b] [0 \ 0 \ \cdots \ 0 \ b] where b b is a non-zero entry in the augmented column. For a homogeneous system, any value in the b 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 Ax=bA\vec{x}=\vec{b} is the sum of the particular solution, p\vec{p}, to the non-homogeneous system and a linear combination of the solutions to the related homogenous linear system Ax=0A\vec{x}=\vec{0}
Proof

Theorem

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

Proof:

Let A=[a1 a2  an]A=[a_1 \ a_2 \ \cdots \ a_n] and let x=[x1x2xn]\vec{x} = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}.

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

\leftrightarrow x1a1+x2a2++xnan=0x_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 AA say aia_i where ai=0a_i = \vec{0}. There also exists some component in x\vec{x} say xix_i where x0x \ne 0.

\leftrightarrow AA 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.