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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?

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The general solution to the second system is the solution you found, plus the general solution to the first system. But you never finished finding the general solution to the first system.

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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

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