Answer
The solution of the system is $S=\{-\frac{2}{7}+\frac{16}{7}a,-\frac{2}{7}-\frac{33}{7}a,-\frac{41}{7}-\frac{15}{7}a,-\frac{41}{7}-\frac{15}{7}a\}$
Work Step by Step
Given: $A=\begin{bmatrix}
3&0 & -1 & 2&-1\\
1 & 3 & 1 & -3 & 2\\
4& -2&-3&6&-1\\
0 & 0 & 0 & 1 & 4
\end{bmatrix}$
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
3&0 & -1 & 2&-1|1\\
1 & 3 & 1 & -3 & 2 | -1\\
4& -2&-3&6&-1|5\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix}\approx^1\begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
3&0 & -1 & 2&-1|1\\
4& -2&-3&6&-1|5\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
0&-9 & -4 & 11&-7|4\\
0& -14&-7&18&-9|9\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix} \approx^3\begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
0&-27 & -12 & 33&-21|12\\
0& 28&14&-36&18|-18\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix} \approx^4 \begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
0&-27 & -12 & 33&-21|12\\
0& 1&2&-3&-3|-6\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix} \approx^5 \begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
0& 1&2&-3&-3|-6\\
0&-27 & -12 & 33&-21|12\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix} \approx^6 \begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
0& 1&2&-3&-3|-6\\
0&0 & 42 & -48&-102|-150\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix} \approx^7 \begin{bmatrix}
1 & 3 & 1 & -3 & 2 | -1\\
0& 1&2&-3&-3|-6\\
0&0 & 1 & -\frac{8}{7}&-\frac{17}{4}|-\frac{25}{7}\\
0 & 0 & 0 & 1 & 4|-2
\end{bmatrix}$
$1.P_{12}$
$2.A_{12}(-3),A_{13}(-4)$
$3. M_2(3),M_3(-2)$
$4.A_{23}(1)$
$5.P_{23}$
$6.A_{23}(27)$
$7.M_3(\frac{1}{42})$
The equivalent system for this matrix is now:
$x_5=a$
$x_4=-4a-2$
$x_3=-\frac{41}{7}-\frac{15}{7}a$
$x_2=-\frac{2}{7}-\frac{33}{7}a$
$x_1=-\frac{2}{7}+\frac{16}{7}a$
The solution of the system is $S=\{-\frac{2}{7}+\frac{16}{7}a,-\frac{2}{7}-\frac{33}{7}a,-\frac{41}{7}-\frac{15}{7}a,-\frac{41}{7}-\frac{15}{7}a\}$
