1. Solve system of linear equations using Reduced Row Echelon Form, or rref.
\begin{equation} \left\{ \begin{aligned} x_1 - 2x_2 + x_3 &= 0 \\ 2x_2 - 8x_3 &= 8\\ -4x_1 + 5x_2 + 9x_3 &= -9 . \end{aligned} \right. \end{equation}Notice the two ways to display output:
We see from the SAGE output that the solution set of the original system of linear equations is \begin{equation} \left\{ \begin{aligned} x_1 &= 29 \\ x_2 &= 16 \\ x_3 &= 3 . \end{aligned} \right. \end{equation}
2. Find all $a$ such that the system of equations has no solution:
\begin{equation} \left\{ \begin{aligned} 2x_1 + a x_2 - x_3 &= 2 \\ x_1 + 2x_2 + 2x_3 &= 6\\ x_1 + 5x_2 + 3x_3 &= -2 . \end{aligned} \right. \end{equation}It appears that possibly $n$ cannot equal $4$, however, we should simplify a last element in the last column to get a better idea of the restrictions on $a$. (Note: indices for Sage and Python begin at 0, so the rows are 0, 1, 2, and the columns are 0, 1, 2, 3.)
From this output, we can see that $a\ne-11$. Replace $a$ with $-11$ and reduce matA again to see what happens. (Also try $a=4$.)
Use the above methods to investigate the solution set of each indicated system of linear equations.