# Solve the system (x1, x2, x3)

\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.

• Please format your question properly. – lightxbulb Feb 7 at 1:57
• I am not sure how to do so. I tried my best – K Jay Feb 7 at 2:02
• But the x1 is x sub one, but I don't know how to make it look like that – K Jay Feb 7 at 2:03
• Here is a page that shows how to format mathematical symbols and operations for the site... – Eleven-Eleven Feb 7 at 2:06
• @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. – 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, $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -10 \\ 11 \\ 0 \end{bmatrix} + s\begin{bmatrix} 13 \\ -17 \\ 1 \end{bmatrix}.$$
• I'm still a little confused why does $x_{3}$ = 0 +1s – K Jay Feb 7 at 3:02
• @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. – Theo Bendit Feb 7 at 3:10
• @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 $s$. – 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 $$\{\begin{pmatrix}-10+13s\\11-17s\\s\end{pmatrix}: s\in\Bbb R\}$$.