Answer:
![x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/148j5w97tf6j140euxnkpgwcw816u9muhk.png)
Explanation:
According to the given situation, The computation of all x in a set of a real number is shown below:
First we have to determine the
so that

![\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/vzke34f6mg8p701vjgq2nge8nmnsh7b261.png)
Now the augmented matrix is
![\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/w18uiynp11vdbddridwxz7cgfy087nputw.png)
After this, we decrease this to reduce the formation of the row echelon
![R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/gogq7j8zsm5fd1gcczivd29hkx379hyiff.png)
![R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/wbpdgzuewhtg3j5y0kmrbgm3y95v9z8jbz.png)
![R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/125gy0iuq527utf2bbl0le7dkpm8fy3h8n.png)
![R_2 = (R_2)/(4), R_3 = (R_3)/(-4) \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/664qasyx69n54htrkhnug4hj9o9beyin7z.png)
![R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/6ivk1ked6z7lvegv3awvksepaqh4kse38b.png)
![R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/cc0luxk3jgfz2e1t15amh8rzcz88qq7y6a.png)

![x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/h90n178aurkhbgc2v37p5lxvj4wgi2yvk0.png)
By applying the above matrix, we can easily reach an answer