Linear algebra: nonhomogeneous system

[z-component]

Homework Statement

Let A = $$\left( \begin{array}{l} \begin{array}{*{20}c} 0 & 1 & { - 1} \\ \end{array} \\ \begin{array}{*{20}c} 2 & 1 & 1 \\ \end{array} \\ <br /> \end{array} \right)$$. Suppose that for some b in $$\mathbb{R}^2$$, $$p = \left( {\begin{array}{*{20}c} 1 \\ { - 1} \\ 1 \\ \end{array} } \right)$$ is one particular solution to the nonhomogeneous equation Ax = b. What is the general form of the solution to this equation (Ax = b)?

Homework Equations

A relevant definition: A nonhomogeneous system is one with Ax = b, b is not equal to zero. A is a matrix and x and b are vectors.

The Attempt at a Solution

Since a homogeneous equation is Ax = 0, b = 0 in this particular question. Given p, I can presume that the general form of the solution will include p, but I'm not sure how to proceed.

 

[z-component]

The solution is:

$$\left( {\begin{array}{*{20}c}<br /> 1 \\<br /> { - 1} \\<br /> 1 \\<br /> <br /> \end{array} } \right) + x_3 \left( {\begin{array}{*{20}c}<br /> { - 1} \\<br /> 1 \\<br /> 1 \\<br /> <br /> \end{array} } \right),{\text{ }}x & _3 \in \mathbb{R}$$

But I can't make sense out of it.

 

[Hurkyl]

Well, first off you can verify that actually is a family of solutions, right?

You seem to think the homogeneous equation Ax=0 has some bearing; well, what is the solution space to Ax=0?

 

[z-component]

I'm guessing since it's in the solution, the solution space is contained in R, but I'm not sure how to verify that it is actually a family of solutions. I'm trying to understand the presentation of the solution in terms of two vectors, one being multiplied by x3 for some reason.

 

[Hurkyl]

I'm not sure how to verify that it is actually a family of solutions.

Multiply it (on the left) by A. What do you get?

 

[daniel_i_l]

Given one solution and all the solutions to the respective homogeneous equation how do you find the solutions to the non-homogeneous equation?
HINT: what happens if you add the two equations together?