4034.sagews

# Section 1.4

# 13 (Use the following to check your paper/pencil row reductions)
a1 = vector(QQ,[3,-2,1]); a2 = vector(QQ,[-5,6,1])
a1,a2

A = matrix(QQ,[a1,a2]).transpose();A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rr} 3 & -5 \\ -2 & 6 \\ 1 & 1 \end{array}$$\right)

u = vector(QQ,[0,4,4]); u

\newcommand{\Bold}[1]{\mathbf{#1}}\left(0,\,4,\,4\right)

Ax = A.augment(u)
Ax

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrr} 3 & -5 & 0 \\ -2 & 6 & 4 \\ 1 & 1 & 4 \end{array}$$\right)

Ax.rref()

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrr} 1 & 0 & \frac{5}{2} \\ 0 & 1 & \frac{3}{2} \\ 0 & 0 & 0 \end{array}$$\right)

# Ax = u has a solution. Therefore, u is in the plane spanned by the columns of A.

# 29 (This might be a "cheap" way to answer the question, but it provides an example!)
M = matrix(QQ, [[1,0,0],[0,1,0],[0,0,1]])
M

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}$$\right)

# By performing a row operation, or even just a row swap, the matrix is no longer in echelon form, but the columns still span R^3
M.add_multiple_of_row(2,1,2); M

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}$$\right)