This is a model for how to work out one of your homework problems using Jupyter notebook. Specifically, we will solve a close analogue of problem 2.4.3 in our textbook. The statement is as follows.
We are given vectors
\begin{equation*} {\bf a}_1 = \begin{bmatrix} \phantom{-}1 \\ -2\\ \phantom{-}2 \end{bmatrix},\quad {\bf a}_2 = \begin{bmatrix} 0\\ 5\\ 5 \end{bmatrix},\quad {\bf a}_3 = \begin{bmatrix} 2\\ 0\\ 8 \end{bmatrix} \end{equation*}
and are asked to decide whether the vector
\begin{equation*} {\bf b}= \begin{bmatrix} -5\\ 11\\ -7 \end{bmatrix} \end{equation*}
is a linear combination of ${\bf a}_1$, ${\bf a}_2$ and ${\bf a}_3$. This means:
Do there exist coefficients $x_1$, $x_2$ and $x_3$ such that \begin{equation*} x_1{\bf a}_1 + x_2{\bf a}_2 + x_3{\bf a}_3 = {\bf b}? \end{equation*}
Or to put it differently, in linear-system language: does the system
\begin{equation*} \begin{aligned} x_1 \phantom{+ 0x_2} + 2x_3 &= -5\\ -2x_1+5x_2 \phantom{+ 0x_3} &= 11\\ 2x_1+5x_2 +8x3 &= -7 \end{aligned} \end{equation*}
have a solution?
We have already seen an example of how to work through this. First, form the extended matrix
\begin{equation*} {\bf E} = \left[ \begin{array}{rrr|r} 1&0&2&-5\\ -2 & 5 & 0 & 11\\ 2 & 5 & 8 & -7 \end{array} \right] \end{equation*}
and then simply outsource the work to SymPy! See below.
from sympy import *
from sympy.solvers.solveset import linsolve
x1, x2, x3 = symbols('x1, x2, x3')
linsolve(Matrix(([1, 0, 2, -5], [-2, 5, 0, 11], [2, 5, 8, -7])), (x1, x2, x3))
That empty-set symbol means the system does not have a solution, i.e. the answer to the original question (of whether ${\bf b}$ is a linear combination of ${\bf a}_i$, $i=1,2,3$) is no.
That's all there is to it.