How can x1 be free if its coefficient is zero?

[dietcookie]

Homework Statement

Find the Eigenvector of the matrix.

Homework Equations

Ax=λx

A=

[2 0 1]
[0 3 4]
[0 0 1]

The Attempt at a Solution

Ok I'm just having a major brain fart here, been doing this all day. For λ=2, I solved for x and get this solution,

[0 1 0][x1] [0]
[0 0 1][x2]=[0]
[0 0 0][x3] [0]

The associated eigenvector from the book for λ=2 is (1,0,0) which I know is correct but did I forget how to read a matrix but the augmented matrix I RREFed is as follows,

[0 1 0 0]
[0 0 1 0]
[0 0 0 0]

So x1 and x2 is zero, how is x1=1?

 

[dietcookie]

Ok maybe I just figured it out, since x2 and x3 are equal to zero, does that mean that x1 is free? That seems to make sense, but how can it be free when it's coefficient is zero?

 

[thegreenlaser]

Ok maybe I just figured it out, since x2 and x3 are equal to zero, does that mean that x1 is free? That seems to make sense, but how can it be free when it's coefficient is zero?

As you wrote,
$$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right)$$
This gives you three equations:
$$\begin{eqnarray} 0 \cdot x_1 &=& 0 \\ x_2 &=& 0 \\ x_3 &=& 0 \end{eqnarray}$$
However, the first equation tells you nothing about x1. That equation is true regardless of what you set x1 equal to. To 'parameterize' (although it's a little silly in this case), you can let x1 = t, where t is an arbitrary real number. Then, the vector
$$\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} t \\ 0 \\ 0 \end{array} \right)$$
is a solution for any t. Set t=1 for simplicity, and your eigenvector is
$$\left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right)$$