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So I have found this question to be very vague and I am struggling to understand it and I believe I am close to the solution but I am not sure if it is correct. The question involves a linear system like so:
$x_1+2x_4=1$
$2x_1+x_3+3x_4=2$
$x_1+x_3+x_4=a$
Which I converted into an augmented matrix:
$$\left[\begin{matrix} 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 1 & 3 & 2 \\ 1 & 0 & 1 & 1 & a\end{matrix}\right]$$
So I am asked to reduce this to reduced row echelon form, find which values of a the system has solutions for, and to write those solutions in vector form, e.g. $[x_1,x_2,x_3,x_4]$. The part where this becomes tricky for me is I had reduced the system as far as I could down to this:
$$\left[\begin{matrix} 1 & 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & a-1\end{matrix}\right]$$
And I suppose the system only has one kind of solution, which is the case of infinite solutions where $a=1$.
My question is: am I blind or is this the only value of a for which this system has a solution, and how am I supposed to write the solution to a system for which there is infinite solutions?

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1 Answer 1

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You are right, $a=1$ is the only value of $a$ for which the system has a solution. As for your other question, the matrix you have row-reduced tells you that the system you started with is reduced to the system:

\begin{equation} \begin{split} x_1 + 2x_4 &= 1 \\ x_3 -x_4 &= 0 \end{split} \end{equation}

Now, can you finish the question, and write the solution of this system in vector form? If not, just post a comment, and I'll edit this answer to include the last bit.

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  • $\begingroup$ Thanks so much for your response, was a massive help, I did some research and I think I've got it, not sure how well I can format it in a comment but here goes: $x_4 \left[ -2, 0, 1, 1\right] + x_2 \left[ 0, 1, 0, 0 \right] + \left[ 1, 0 , 0 , 0 \right]$ Thanks again! $\endgroup$ Mar 29, 2020 at 1:11
  • $\begingroup$ Correct! Well done for getting there yourself. Usually, we replace the x-variables with parameters (usually $s, t$) to emphasise that these can take any value. So the answer would be $(1,0,0,0) + s(0,1,0,0) + t(-2,0,1,1).$ And that's how you express the solution of a system with infinite solutions! $\endgroup$ Mar 29, 2020 at 1:34
  • $\begingroup$ Awesome, thanks that was a massive help! $\endgroup$ Mar 30, 2020 at 2:14

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