Find the general form of the solution to the system of equations below.
\begin{align}
2x_1-x_2+8x_3-x_4&=0\\
-4x_1+3x_2-18x_3+x_4&=0\\
2x_1+x_2+4x_3-3x_4&=0.
\end{align}
My attempt:
$$
\begin{bmatrix}
2 & -1& 8 & -1\\
-4 & 3 & -18 & 1\\
2&1&4&-3\\
\end{bmatrix}\longrightarrow\begin{bmatrix}
1 & -1/2& 3 & -1/2\\
0 & 1 & -2 & -1\\
0&2&-4&-1\\
\end{bmatrix}\longrightarrow\begin{bmatrix}
1 & 0& 3 & -1\\
0 & 1 & -2 & -1\\
0&0&0&0\\
\end{bmatrix}.
$$
Now, the vector $(1 \space 2 \space 0 \space0)^T$ is a solution to the system shown below. Use the answer from the previous system to write down the general solution to this system below.
\begin{align}
2x_1-x_2+8x_3-x_4&=0\\
-4x_1+3x_2-18x_3+x_4&=2\\
2x_1+x_2+4x_3-3x_4&=4.
\end{align}
I'm not sure how to continue. I could make an augmented matrix and reduce it, but I'm just not sure if that would be giving me the correct answer. Also, what would I need to do with the mentioned transformation? Anything?
Best Answer
From your RREF (which I verified as correct):
$$\left[\begin{array}{@{}cccc|c@{}} 1 & 0 & 3 & -1 &0\\ 0 & 1 & -2 & -1 & 0\\ 0 & 0 & 0 & 0 & 0\\ \end{array}\right]$$
We have:
$$\tag 1 x_1+3x_3-x_4=0$$ $$\tag2 x_2-2x_3-x_4=0$$
From $(2)$, we have: $$x_4 = x_2 - 2x_3$$
Substituting this into $(1)$, yields: $$\tag 3 x_1 = x_2 - 6x_3$$
We are now free to choose solutions (we have two free variables) given there are an infinite number of them.
Let $x_3 = a$, $x_2 = b$, yielding, from $(3)$, $x_1 = b - 6a$.
Solving for $x_4$ (from $(2)$), yields: $x_4 = b - 2a$.
Our general solution can now be written as:
$$(x_1 | x_2 | x_3 | x_4)^{T} = (b-6a, b, a, b-2a)^{T}$$
For example, choosing $a = 0, b = 1$, yields:
$$(x_1 | x_2 | x_3 | x_4)^{T} = (1, 1, 0, 1)^{T}$$
Verify that this solves the system.
You did great!
Regards