Answer:
The solution can be defined as follows:
Step-by-step explanation:
Augmented matrix for the given system is
[tex]\left[\begin{array}{ccccc}2&1&-2&|&4\\4&0&2&|&10\\ -4&5&-17&|&-15 \end{array}\right][/tex]
Apply elementary row transformations on the matrix:
[tex]R_1\to R_2 (2) \sim \left[\begin{array}{ccccc}4&2&-4&|&8\\4&0&2&|&10\\ -4&5&-17&|&-15 \end{array}\right] \\\\\\R_2\to R_2-R_1\\R_3\to R_3 \frac{1}{7} \sim \left[\begin{array}{ccccc}4&2&-4&|&8\\0&-1&3&|&1\\ 0&1&-3&|&-1 \end{array}\right] \\\\R_3\to R_3+R_2 \sim \left[\begin{array}{ccccc}4&2&-4&|&8\\0&-1&3&|&1\\ 0&0&0&|&0 \end{array}\right][/tex]
Matrix Ranking = 2 = Matrix Ranking So the method in question is consistent
And there are endless options. To be the equation,
We're having
[tex]\to 4x_1+2x_2-4x_3=8 ..........(1)\\\\ \to -x_2+3x_3=1............(2)\\\\take\ x_3=s[/tex]
from equation 2 we get [tex]-x_2+ 3s = 1[/tex]
Substituting [tex]x_3= 3s- 1[/tex]in equation 1 we get
[tex]\to 4x_1 +6s-2-4s=8 \\\\\to 4x_1=-2s+10\\\\\to x_1=-\frac{1}{2}s+\frac{5}{2}[/tex]
[tex]\left[\begin{array}{c}X_1\\x_2\\x_3\end{array}\right]= \left[\begin{array}{c}-\frac{1}{2}s +\frac{5}{2} \\3s-1\\s\end{array}\right][/tex]
