Section 5 Definitions

## Homogeneous Equations

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

(1)
\begin{align} A = \begin{bmatrix} 2&4&0&|&0\\2&5&2&|&0\\1&3&2&|&0 \end{bmatrix} = \begin{bmatrix} 1&0&-4&|&0\\0&1&2&|&0\\0&0&0&|&0 \end{bmatrix}. \end{align}

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)
\begin{align} A = \begin{bmatrix} 2&4&0\\2&5&2\\1&3&2 \end{bmatrix} \text{ and } \vec{b} = \begin{bmatrix} 6\\3\\0 \end {bmatrix} \end{align}

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.

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