4030.sagews

# Math 211. Section 1.2

# 3

A = matrix(QQ,[[1,2,4,8],[2,4,6,8],[3,6,9,12]]);A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 2 & 4 & 8 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \end{array}$$\right)

A.add_multiple_of_row(1,0,-2)
A.add_multiple_of_row(2,0,-3)
A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 2 & 4 & 8 \\ 0 & 0 & -2 & -8 \\ 0 & 0 & -3 & -12 \end{array}$$\right)

A.rescale_row(1,-1/2);A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 2 & 4 & 8 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & -3 & -12 \end{array}$$\right)

A.add_multiple_of_row(2,1,3);A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 2 & 4 & 8 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 \end{array}$$\right)

A.add_multiple_of_row(0,1,-4);A

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 2 & 0 & -8 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 \end{array}$$\right)

# Pivot positions are (0,0) and (1,2), in 0-based indexing. Pivot columns can be read off of these as 0 and 2. 
# If you just want to check your answers after computing with pencil and paper, use A.rref() ("reduced row echelon form").

# 7

B = matrix(QQ,[[1,3,4,7],[3,9,7,6]])
B

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrrr} 1 & 3 & 4 & 7 \\ 3 & 9 & 7 & 6 \end{array}$$\right)

B.rref()

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

# The solution is 
# x_1 = -5 -3x_2
# x_2 free
# x_3 = 3
# There is no equation such as "0=1", so the system is consistent. The solution exists, but is _not_ unique, owing to the free variable x_2.

# 17

var('h', domain=RR)

\newcommand{\Bold}[1]{\mathbf{#1}}h

M = matrix(SR,[[1,-1,4],[-2,3,h]])
M

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

M.rref()

\newcommand{\Bold}[1]{\mathbf{#1}}\left($$\begin{array}{rrr} 1 & 0 & h + 12 \\ 0 & 1 & h + 8 \end{array}$$\right)

# x_1 = h+12
# x_2 = h+8
# M has a solution for all values of h