MATH511 Linear Algebra Chapter 1: Difference between revisions
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== Section 1.1 == | == Section 1.1 == | ||
Insert lecture notes here! | |||
<math> | ===Assignment 1.1=== | ||
\begin{ | |||
1 | We'll be doing 1(b), 5(c), 6(e), 8*, 10 this round from the source textbook. | ||
For '''1(b)''', consider the following: | |||
\end{ | |||
We'll need to use back substitution to solve this system of linear equation. | |||
<math display=block> | |||
\begin{align} | |||
x_{1}+x_{2}+x_{3}=8 \\ | |||
2x_{2}+x_{3}=5 \\ | |||
3x_{3}=9 | |||
\end{align} | |||
</math> | |||
We'll first need to use substitution to simply and find the values x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>. | |||
A good rule of thumb is to find the easiest value to simplify that can then be used to find the other values. In this example, x<sub>3</sub> can be used to help find x<sub>1</sub>, x<sub>2</sub>: | |||
<math> | |||
\begin{align} | |||
3x_{3}=9 \\ | |||
x_{3}=\frac{9}{3} | |||
\end{align} | |||
</math> | |||
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: | |||
<math> | |||
\begin{align} | |||
2x_{2}+x_{3}=5 \\ | |||
x_{3}=\frac{9}{3} \\ | |||
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\ | |||
2x_{2}+\frac{9}{3}=5 \\ | |||
2x_{2}=2 \\ | |||
x_{2}=1 | |||
\end{align} | |||
</math> | </math> | ||
Latest revision as of 15:36, 24 January 2022
Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve.
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically.
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works.
Prologue
Hey how are you. I will insert something here later. :)
Section 1.1
Insert lecture notes here!
Assignment 1.1
We'll be doing 1(b), 5(c), 6(e), 8*, 10 this round from the source textbook.
For 1(b), consider the following:
We'll need to use back substitution to solve this system of linear equation.
We'll first need to use substitution to simply and find the values x1, x2, x3. A good rule of thumb is to find the easiest value to simplify that can then be used to find the other values. In this example, x3 can be used to help find x1, x2:
We'll take this and plug it into to find x2: