Find the general solution to non homogeneous linear equation

Question

The question: State the general solutuon of: $$5x_1-2x_2+4x_3=5$$

My attempt: I tried following closely an example provided for homogeneous equations without any luck:

$$ x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bmatrix}= x_2\begin{bmatrix}2\\1\\0\end{bmatrix} +x_3\begin{bmatrix}-4\\0\\1\end{bmatrix} +\begin{bmatrix}5\\1\\1\end{bmatrix} $$

And after this step I get stuck since I don't know whether is correct and should proceed with the solution space (and also how to find the solution space for non homogenous equations)

As for who might be concerned, the question is homework related.

It would be nice to know why I got downvoted so I can improve my question

Based on the feedback received from @Isham the system should have been written as follows:

$$ x=\begin{bmatrix}5x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bmatrix}= \begin{bmatrix}2/5x_2-4/5x_3+1\\x_2\\x_3\end{bmatrix}= x_2\begin{bmatrix}2/5\\1\\0\end{bmatrix} +x_3\begin{bmatrix}-4/5\\0\\1\end{bmatrix} +\begin{bmatrix}1\\0\\0\end{bmatrix} $$

Answer 1

According to your equation Lucian, it should be rather this:

$x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix}2/5x_2-4/5x_3+1\\x_2\\x_3\end{bmatrix}= x_2\begin{bmatrix}2/5\\1\\0\end{bmatrix} +x_3\begin{bmatrix}-4/5\\0\\1\end{bmatrix} +\begin{bmatrix}1\\0\\0\end{bmatrix}$

Answer 2

Consider the pattern:

For $$5x_1-2x_2+4x_3=5$$ then $x_{3} = \frac{1}{4} \, (5 - 5 \, x_{1} + 2 \, x_{2})$ and \begin{align} x &=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix} x_{1} \\ x_2 \\ \frac{1}{4} \, (5 - 5 \, x_{1} + 2 \, x_{2}) \end{bmatrix} = \frac{1}{4} \, \begin{bmatrix} x_{1} \\ x_2 \\ 5 - 5 \, x_{1} + 2 \, x_{2} \end{bmatrix} \\ &= \frac{x_1}{4} \, \begin{bmatrix} 1 \\ 0 \\ -5 \end{bmatrix} + \frac{x_2}{4} \, \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} + \frac{5}{4} \, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}. \end{align}