Math211.1.2

129 days ago by Jesse_Berwald

# 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)
\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)
\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)
\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)
\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)
\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)
\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)
\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