$$\left\{\begin{align}x_{1} + x_{2} +4x_{3} &= 1 \\ 3x_{1} +2x_{2} -5x_{3} &= -8 \end{align}\right.$$

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \quad \\ \quad \\ \quad \\ \end{bmatrix} + s\begin{bmatrix} \quad \\ \quad \\ \quad \\ \end{bmatrix}$$

So when I try to solve this, I turn it into an augmented/coefficient matrix. Then I row reduce. Which gives me the reduced matrix:

$$\left[\begin{array}{ccc|c} 1 & 0 & -13 & -10 \\ 0 & 1 & 17 & 11 \end{array}\right]$$

Then I turned this back into an equation
$$\begin{align*} x_{1} + 0x_{2} -13 x_{3} &= -10 \\ 0x_{1} + x_{2} +17 x_{3} &= 11 \end{align*}$$

Then basically I get that
$$\begin{align*} x_{1} &= -10 + 13s \\ x_{2} &= 11 - 17s \end{align*}$$

The answer has $3$ rows per vector. So I'm just confused as to what goes into the $x_3$ spot for the row or if I even did this correctly.

  • 2
    $\begingroup$ Please format your question properly. $\endgroup$ – lightxbulb Feb 7 at 1:57
  • $\begingroup$ I am not sure how to do so. I tried my best $\endgroup$ – K Jay Feb 7 at 2:02
  • $\begingroup$ But the x1 is x sub one, but I don't know how to make it look like that $\endgroup$ – K Jay Feb 7 at 2:03
  • 2
    $\begingroup$ Here is a page that shows how to format mathematical symbols and operations for the site... $\endgroup$ – Eleven-Eleven Feb 7 at 2:06
  • $\begingroup$ @KJay I've edited your question. If my interpretation is not correct, please feel free to edit it again, or simply revert back to the previous version. In either case, I would advise pressing "edit" to see how I've formatted your question. Have a play around with it, so you can take the opportunity to learn to do this for yourself. $\endgroup$ – Theo Bendit Feb 7 at 2:44

By the looks of it, you've let $x_3$ be a free variable $s$. Hence, in addition to $x_1 = -10 + 13s$ and $x_2 = 11 - 17s$, you also have $x_3 = 0 + 1s$. That is, $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -10 \\ 11 \\ 0 \end{bmatrix} + s\begin{bmatrix} 13 \\ -17 \\ 1 \end{bmatrix}.$$

  • $\begingroup$ I'm still a little confused why does $x_{3}$ = 0 +1s $\endgroup$ – K Jay Feb 7 at 3:02
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    $\begingroup$ @KJay The method you're following, even if you don't completely understand it, is to take certain variables in your system of equation (specifically, ones corresponding to a column without pivots, in this case, just $x_3$) and make them free variables. By choosing a value of $x_3$, you obtain unique values for $x_1, x_2$ that produce a solution. Conversely, any solution must assign $x_3$ a value! So, we can parameterise all solutions by letting $x_3$ roam free over the reals. $\endgroup$ – Theo Bendit Feb 7 at 3:10
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    $\begingroup$ @KJay Conventionally, we give our free parameters their own name, typically $t, s, r$, etc. You've called $x_3$ the free parameter $s$, and found $x_1$ and $x_2$ in terms of $s$. The only relationship you missed was $x_3 = s$, which was basically the definition of $s$. $\endgroup$ – Theo Bendit Feb 7 at 3:11

When you "turned it back into an equation " you made an error. You should get: $x_1-13x_3=-10\\x_2+17x_3=11$. You can choose $x_3$ freely, then solve for $x_1$ and $x_2$. The solution space is $1$-dimensional.

You get $\{\begin{pmatrix}-10+13s\\11-17s\\s\end{pmatrix}: s\in\Bbb R\}$.


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