$$\left\{\begin{align}x_{1} + x_{2} +4x_{3} &= 1 \\ 3x_{1} +2x_{2} -5x_{3} &= -8 \end{align}\right.$$
$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \quad \\ \quad \\ \quad \\ \end{bmatrix} + s\begin{bmatrix} \quad \\ \quad \\ \quad \\ \end{bmatrix}$$
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.
$\endgroup$ 92 Answers
$\begingroup$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,$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -10 \\ 11 \\ 0 \end{bmatrix} + s\begin{bmatrix} 13 \\ -17 \\ 1 \end{bmatrix}.$$
$\endgroup$ 3 $\begingroup$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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