Solving linear system using "reduced row echelon form" when augmented columns is "zero"

Question

I just learnt about Reduced Row Echelon Form and solved few rather large matrices. However I have no idea how to approach this rather simple system:

$$x+2y=0\\ 3x+6y=0$$

Here is my 2x2 augmented matrix: $$ A = \left[ \begin{array}{cc|c} 1&2&0\\ 3&6&0 \end{array} \right]$$

Then I calculated Reduced Row Echelon Form by formula $R2 = 3R1 - R2$: $$ RREF(A) = \left[ \begin{array}{cc|c} 1&2&0\\ 0&0&0 \end{array} \right]$$

Now what? I can guess that I have infinite answers because I have zero at last column. However I am not sure about it.

P.S: I am actually trying to find eigenvector of $A=\begin{bmatrix}3&2\\3&8 \end{bmatrix}$. I found $\lambda = 2,9$. Here is my solution to find eigenvector for $\lambda=2$:

$$ \begin{align*} (A - \lambda I)\vec{V} &= 0\\ (\begin{bmatrix}3&2\\3&8\end{bmatrix} - \begin{bmatrix}2&0\\0&2\end{bmatrix})\vec{V} &= 0\\ \begin{bmatrix}1&2\\3&6\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\ \begin{bmatrix}1x+2y\\3x+6y\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\ \end{align*} $$

Does this means that my matrix have infinite number of eigenvectors?

Answer 1

First thing: you do not have infinite answers: this means $x=\infty$, $y=\infty$ which is nonsense. What you have is an infinite number of answers: it is the number of answers that is infinite, not the answers themselves. (Analogy: "I have a large number of friends on Facebook" is not the same as "I have large friends on Facebook".)

Second thing: whether or not there are an infinite number of answers has nothing to do with the right hand side. It is because the left hand side includes a non-leading (non-pivot) column. So you would let the variable corresponding to this column be a parameter, and then solve for the rest in terms of this parameter.

If the zeros on the RHS worry you, start off by solving the following system: $$\left\{\eqalign{x+2y&=3\cr 3x+6y&=9\ .\cr}\right.$$ Then go back to your question - it is really exactly the same.

See if you can finish the problem.

Re: your PS. If the matrix $A$ has an eigenvalue $\lambda$, then there are always an infinite number of eigenvectors corresponding to $\lambda$.

Answer 2

It is correct that you have an infinite number of solutions. The reason for this is because the equations are multiples of each other. E2 = 3E1

$$3(x+2y=0)$$ $$3x+6y=0$$

This is true for all systems of equations that are multiples of each other.

For example:

$$2x+4y=8$$ $$6x+12y=24$$

will also have an infinite number of solutions because they're multiples of each other.

Answer 3

it means that you have only one independent variable and $t(2,-1)$ where $t\in \mathbb R$ is your solution space.