latex.matrix_delimiters(left='[', right=']')
latex.matrix_column_alignment('c')

$\newcommand{\la}{\langle}$ $\newcommand{\ra}{\rangle}$ $\newcommand{\vu}{\mathbf{u}}$ $\newcommand{\vv}{\mathbf{v}}$ $\newcommand{\vw}{\mathbf{w}}$ $\newcommand{\vzz}{\mathbf{z}}$ $\newcommand{\nc}{\newcommand}$ $\nc{\Cc}{\mathbb{C}}$ $\nc{\Rr}{\mathbb{R}}$ $\nc{\Qq}{\mathbb{Q}}$ $\nc{\Nn}{\mathbb{N}}$ $\nc{\cB}{\mathcal{B}}$ $\nc{\cE}{\mathcal{E}}$ $\nc{\cC}{\mathcal{C}}$ $\nc{\cD}{\mathcal{D}}$ $\nc{\mi}{\mathbf{i}}$ $\nc{\span}[1]{\langle #1 \rangle}$ $\nc{\ol}[1]{\overline{#1} }$ $\nc{\norm}[1]{\left\| #1 \right\|}$ $\nc{\abs}[1]{\left| #1 \right|}$ $\nc{\vz}{\mathbf{0}}$ $\nc{\vo}{\mathbf{1}}$ $\nc{\DMO}{\DeclareMathOperator}$ $\DMO{\tr}{tr}$ $\DMO{\nullsp}{nullsp}$ $\nc{\va}{\mathbf{a}}$ $\nc{\vb}{\mathbf{b}}$ $\nc{\vx}{\mathbf{x}}$ $\nc{\ve}{\mathbf{e}}$ $\nc{\vd}{\mathbf{d}}$ $\nc{\vh}{\mathbf{h}}$ $\nc{\ds}{\displaystyle}$ $\nc{\bm}[1]{\begin{bmatrix} #1 \end{bmatrix}}$ $\nc{\gm}[2]{\bm{\mid & \cdots & \mid \\ #1 & \cdots & #2 \\ \mid & \cdots & \mid}}$ $\nc{\MN}{M_{m \times n}(K)}$ $\nc{\NM}{M_{n \times m}(K)}$ $\nc{\NP}{M_{n \times p}(K)}$ $\nc{\MP}{M_{m \times p}(K)}$ $\nc{\PN}{M_{p \times n}(K)}$ $\nc{\NN}{M_n(K)}$ $\nc{\im}{\mathrm{Im\ }}$ $\nc{\ev}{\mathrm{ev}}$ $\nc{\Hom}{\mathrm{Hom}}$ $\nc{\com}[1]{[\phantom{a}]^{#1}}$ $\nc{\rBD}[1]{ [#1]_{\cB}^{\cD}}$ $\DMO{\id}{id}$ $\DMO{\rk}{rk}$ $\DMO{\nullity}{nullity}$ $\DMO{\End}{End}$ $\DMO{\proj}{proj}$ $\nc{\GL}{\mathrm{GL}}$

Examples of GJE

Example 2. Let's use GJE to solve a HW problem (Problem 6)

Find the values of $h$ and $k$ so that the following system has more than one solutions. (That means the system is consistent and has at least one free-variable. So in the case when the scalar field is infinite, then having more than one solution is equivalent to having infinitely many solutions).

%display latex
var('h,k') # declare h and k as variables
A = matrix([(-6,-7,-6),(5,-7,6),(-28,-7,h)]); b = column_matrix([3,7,k]); 
M = block_matrix([A,b], ncols = 2); M

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} -6 & -7 & -6 & 3 \\ 5 & -7 & 6 & 7 \\ -28 & -7 & h & k \end{array}\right]$$

M.rescale_row(0,-1/6); M

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 1 & \frac{7}{6} & 1 & -\frac{1}{2} \\ 5 & -7 & 6 & 7 \\ -28 & -7 & h & k \end{array}\right]$$

M.add_multiple_of_row(1,0,-5); M.add_multiple_of_row(2,0,28); M

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 1 & \frac{7}{6} & 1 & -\frac{1}{2} \\ 0 & -\frac{77}{6} & 1 & \frac{19}{2} \\ 0 & \frac{77}{3} & h + 28 & k - 14 \end{array}\right]$$

M.rescale_row(1,-6/77); M

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 1 & \frac{7}{6} & 1 & -\frac{1}{2} \\ 0 & 1 & -\frac{6}{77} & -\frac{57}{77} \\ 0 & \frac{77}{3} & h + 28 & k - 14 \end{array}\right]$$

M.add_multiple_of_row(2,1,-77/3); M

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{ccc|c} 1 & \frac{7}{6} & 1 & -\frac{1}{2} \\ 0 & 1 & -\frac{6}{77} & -\frac{57}{77} \\ 0 & 0 & h + 30 & k + 5 \end{array}\right]$$

At this point, we see that the system has a free variable if and only if $h+30 =0$, i.e. $h=-30$. If that happens, then the system is consistent if and only if $k+5$ is also 0, i.e. $k =-5$.

Thus, the system has more than one solution if and only if $h=-30$ and $k = -5$.