Answer
The solution of the system is $-a+5,a-2,a$
Work Step by Step
Given: $A=\begin{bmatrix}
1&2 & -1\\
1&0&1\\
4& 4&0
\end{bmatrix} $
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
1&2 & -1|1\\
1&0&1|5\\
4& 4&0|12
\end{bmatrix} \approx^1\begin{bmatrix}
1&2 & -1|1\\
0&-2&2|4\\\
0& -4&4|8
\end{bmatrix} \approx^2 \begin{bmatrix}
1&2 & -1|1\\
0&1&-1|-2\\\
0& -4&4|8
\end{bmatrix} \approx^3 \begin{bmatrix}
1&2 & -1|1\\
0&1&-1|-2\\\
0&0&0|0
\end{bmatrix} $
$1.A_{21}(-1),A_{12}(-4)$
$2. M_2(\frac{-1}{2})$
$3.A_{23}(4)$
The equivalent system for this matrix is now:
$z=a$
$y=a-2$
$x+2(a-2)-a=1 \rightarrow x=-a+5$
The solution of the system is $-a+5,a-2,a$
