## Homogeneous Equations

A **Homogeneous** equations has a solution that satisfies the equation $A\vec{x} = \vec{0}$.

Therefore, $x_1 = 4x_3 \\ x_2 = -2x_3 \\ x_3 = x_3$ or in **Parametric Form**, $\vec{x} = x_3 \begin{bmatrix} 4\\-2\\1 \end{bmatrix}$

This is the homogeneous solution.

#### Non-Homogeneous

A solution that satisfies the equation $A\vec{x} = \vec{b}$ where A is a matrix of n x m dimensions and b is a matrix of n x 1 dimensions (is it unnecessary to say that?)

(2)To solve for the non-homogeneous $\vec{x}$, simply augment with $\vec{b}$ and solve. We get $\vec{x} = x_5 \begin{bmatrix} 4\\-2\\1 \end{bmatrix} + \begin{bmatrix} 9\\-3\\0 \end{bmatrix}$ in Parametric form. Notice that the solution here looks very much like the homogeneous solution if $\vec{b} = \vec{0}$ but there is now a point (9,-3,0) that this line is forced through. That's the difference. The Homogeneous solution is a set of all lines that satisfy the equation, but the non-homogeneous solution is that particular solution. Therefore, all the lines such $A\vec{x} = \vec{b}$ will have the same $x_5 \begin{bmatrix} 4\\-2\\1 \end{bmatrix}$ regardless of $\vec{b}$, but it only changes the point the line is forced through.