Math211.1.4

129 days ago by Jesse_Berwald

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