\(\newcommand{\markedPivot}[1]{\boxed{#1}}
\newcommand{\IR}{\mathbb{R}}
\newcommand{\IC}{\mathbb{C}}
\renewcommand{\P}{\mathcal{P}}
\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\RREF}{\operatorname{RREF}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\setList}[1]{\left\{#1\right\}}
\newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}}
\newcommand{\unknown}{\,{\color{gray}?}\,}
\newcommand{\drawtruss}[2][1]{
\begin{tikzpicture}[scale=#1, every node/.style={scale=#1}]
\draw (0,0) node[left,magenta]{C} --
(1,1.71) node[left,magenta]{A} --
(2,0) node[above,magenta]{D} -- cycle;
\draw (2,0) --
(3,1.71) node[right,magenta]{B} --
(1,1.71) -- cycle;
\draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle;
\draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle;
\draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle;
\draw[thick,red,->] (2,0) -- (2,-0.75);
#2
\end{tikzpicture}
}
\newcommand{\trussNormalForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, blue,->] (4,0) -- (3.5,0.5);
}
\newcommand{\trussCompletion}{
\trussNormalForces
\draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026);
\draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684);
\draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71);
\draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855);
\draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026);
\draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855);
\draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026);
}
\newcommand{\trussCForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, magenta,->] (0,0) -- (0.4,0.684);
\draw [thick, magenta,->] (0,0) -- (0.5,0);
}
\newcommand{\trussStrutVariables}{
\node[above] at (2,1.71) {\(x_1\)};
\node[left] at (0.5,0.866) {\(x_2\)};
\node[left] at (1.5,0.866) {\(x_3\)};
\node[right] at (2.5,0.866) {\(x_4\)};
\node[right] at (3.5,0.866) {\(x_5\)};
\node[below] at (1,0) {\(x_6\)};
\node[below] at (3,0) {\(x_7\)};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Linear Algebra for Team-Based Inquiry Learning
2023 Edition
| Steven Clontz | Drew Lewis |
|---|---|
| University of South Alabama | |
August 24, 2023
Section 4.3: Solving Systems with Matrix Inverses (MX3)
Activity 4.3.1 (~15 min)
Consider the following linear system with a unique solution:
\begin{equation*}
\begin{matrix}
3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\
2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\
-x_{1} & & & + & x_{3} & & & = & -1 \\
& - & x_{2} & & & - & 2x_{4} & = & -5 \\
\end{matrix}
\end{equation*}
Part 1.
Define
\begin{equation*}
T\left(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\right)=
\left[\begin{matrix}\hspace{1em}\unknown\hspace{1em}\\\hspace{1em}\unknown\hspace{1em}\\
\hspace{1em}\unknown\hspace{1em}\\\hspace{1em}\unknown\hspace{1em}\end{matrix}\right]
\end{equation*}
so that \(T(\vec x)=\left[\begin{matrix}-7\\-1\\-1\\-5\end{matrix}\right]\) has the same solution set as this system.
Activity 4.3.1 (~15 min)
Consider the following linear system with a unique solution:
\begin{equation*}
\begin{matrix}
3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\
2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\
-x_{1} & & & + & x_{3} & & & = & -1 \\
& - & x_{2} & & & - & 2x_{4} & = & -5 \\
\end{matrix}
\end{equation*}
Part 2.
Define
\begin{equation*}
A=
\left[\begin{matrix}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown
\end{matrix}\right]
\end{equation*}
so that \(A\vec x=\left[\begin{matrix}-7\\-1\\-1\\-5\end{matrix}\right]\) has the same solution set as this system.
Activity 4.3.1 (~15 min)
Consider the following linear system with a unique solution:
\begin{equation*}
\begin{matrix}
3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\
2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\
-x_{1} & & & + & x_{3} & & & = & -1 \\
& - & x_{2} & & & - & 2x_{4} & = & -5 \\
\end{matrix}
\end{equation*}
Part 3.
Find
\begin{equation*}
B=
\left[\begin{matrix}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown
\end{matrix}\right]
\end{equation*}
so that \(BA\vec x=\vec x\text{.}\)
Activity 4.3.1 (~15 min)
Consider the following linear system with a unique solution:
\begin{equation*}
\begin{matrix}
3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\
2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\
-x_{1} & & & + & x_{3} & & & = & -1 \\
& - & x_{2} & & & - & 2x_{4} & = & -5 \\
\end{matrix}
\end{equation*}
Part 4.
Find \(\vec x=BA\vec x=B\left[\begin{matrix}-7\\-1\\-1\\-5\end{matrix}\right]\) to solve the system.
Remark 4.3.1
The linear system described by the augmented matrix \([A \mid \vec w]\) has exactly the same solution set as the matrix equation \(A\vec x=\vec w\text{.}\)
Activity 4.3.2
Let \(A\vec x=\vec w\) describe a linear system. When will this linear system have exactly one solution?
When \(A\) is invertible.
When \(A\) is not invertible.
When \(\RREF A\) has a non-pivot column.
When \(\RREF A\) has a non-pivot row.
Fact 4.3.2
When \(A\vec x=\vec w\) has exactly one solution, this solution is given by \(\vec x=A^{-1}\vec w\text{.}\)
Activity 4.3.3
Consider the vector equation
\begin{equation*}
x_{1} \left[\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right] + x_{2} \left[\begin{array}{c} -2 \\ -3 \\ 3 \end{array}\right] + x_{3} \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] = \left[\begin{array}{c} -3 \\ 5 \\ -1 \end{array}\right]
\end{equation*}
with a unique solution. Part 1.
Explain and demonstrate how this problem can be restated using matrix multiplication.
Activity 4.3.3
Consider the vector equation
\begin{equation*}
x_{1} \left[\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right] + x_{2} \left[\begin{array}{c} -2 \\ -3 \\ 3 \end{array}\right] + x_{3} \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] = \left[\begin{array}{c} -3 \\ 5 \\ -1 \end{array}\right]
\end{equation*}
with a unique solution. Part 2.
Use the properties of matrix multiplication to find the unique solution.