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?

  • 1
    $\begingroup$ Read the next words. "There are others." $\endgroup$ – Ted Shifrin May 29 '14 at 19:28
  • $\begingroup$ Ah, I see those next words are in the second edition but not in the first. $\endgroup$ – Ted Shifrin May 29 '14 at 19:34
  • $\begingroup$ I guess they were put in the second ed. for good reason then :) $\endgroup$ – Morgan Wilde May 29 '14 at 19:34
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    $\begingroup$ Yup. Indeed. You win :) $\endgroup$ – Ted Shifrin May 29 '14 at 19:38

In mathematical text, saying "there is one solution" means "there is at least one solution". It does not exclude the possibility that there are many solutions. If there is exactly one solution, one would write "there is exactly one solution" or "there is only one solution".

  • $\begingroup$ I was expecting an explicit statement if it were the case. Thanks for explaining me this. $\endgroup$ – Morgan Wilde May 29 '14 at 19:31

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