# Solution to this system of equations

I'm currently reading the Michael Artin book "Algebra" and at page 4 I've encountered a detail, that's not totally clear.

Here's the relevant excerpt from the book.

Thus the matrix equation $$\begin{bmatrix}0 & -1 & 2 \\ 3&4&-6\\\end{bmatrix}\begin{bmatrix}x_1 \\x_2\\x_3\end{bmatrix}=\begin{bmatrix}2 \\ 1\end{bmatrix}$$ represents the following system of two equations in three unknowns: \begin{align}-x_2+2x_3&=2\\3x_1+4x_2-6x_3&=1.\end{align} Equation $(1.8)$ exhibits one solution: $x_1=1,\,x_2=4,\,x_3=3.$

What bugs me is the statement "one solution", because the way I see it, this system has a solution space with one degree of freedom and the answer vector could be formulated as.

$(3-\frac{2}{3}x_3, \; 2x_3 - 2, \; x_3)$

And $(1, 4, 3)$ is only one of many possible solutions.

What am I missing here? Or am I reading the text wrong?

• Read the next words. "There are others." – Ted Shifrin May 29 '14 at 19:28
• Ah, I see those next words are in the second edition but not in the first. – Ted Shifrin May 29 '14 at 19:34
• I guess they were put in the second ed. for good reason then :) – Morgan Wilde May 29 '14 at 19:34
• Yup. Indeed. You win :) – Ted Shifrin May 29 '14 at 19:38