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lecture_2_slides
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Table of Contents
Linear equations in 3 variables
Definition
If $a,b,c,d$ are any fixed numbers, then equation \[ ax+by+cz=d\] is a linear equation in 3 variables.
When you draw the set of all solutions of a linear equation in 3 variables, you always get a plane in 3-dimensional space, $\mathbb{R}^3$.
Examples
- $x+y+z=1$ <html><iframe scrolling=“no” src=“https://tube.geogebra.org/material/iframe/id/528999/width/800/height/503/border/888888/rc/true/ai/false/sdz/true/smb/false/stb/true/stbh/true/ld/false/sri/true/at/auto” width=“800px” height=“503px” style=“border:0px;”> </iframe></html>
- $x+y=1$ This may be viewed as a linear equation in 3 variables, since it is equivalent to $x+y+0z=1$. <html><iframe scrolling=“no” src=“https://tube.geogebra.org/material/iframe/id/529043/width/800/height/503/border/888888/rc/true/ai/false/sdz/true/smb/false/stb/true/stbh/true/ld/false/sri/true/at/auto” width=“800px” height=“503px” style=“border:0px;”> </iframe></html>
- $z=1$, viewed as the equation $0x+0y+z=1$ <html><iframe scrolling=“no” src=“https://tube.geogebra.org/material/iframe/id/529069/width/800/height/503/border/888888/rc/true/ai/false/sdz/true/smb/false/stb/true/stbh/true/ld/false/sri/true/at/auto” width=“800px” height=“503px” style=“border:0px;”> </iframe><br /></html>This plane is horizontal (parallel to the $x$-$y$ plane).
Linear equations (in general)
A linear equation in $m$ variables (where $m$ is some natural number) is an equation of the form \[ a_1x_1+a_2x_2+\dots+a_mx_m=b\] where $a_1,a_2,\dots,a_m$ and $b$ are fixed numbers (called coefficients) and $x_1,x_2,\dots,x_m$ are variables.
Example
\[ 3x_1+5x_2-7x_3+11x_4=12\] is a linear equation in 4 variables.
- A typical solution will be a point $(x_1,x_2,x_3,x_4)\in \mathbb{R}^4$ so that $3x_1+5x_2-7x_3+11x_4$ really does equal $12$.
- For example, $(-2,0,-1,1)$ is a solution.
- The set of all solutions is a 3-dimensional object in $\mathbb{R}^4$, called a hyperplane.
- Since we can't draw pictures in 4-dimensional space $\mathbb{R^4}$ we can't draw this set of solutions!
Systems of linear equations
A system of linear equations is just a list of several linear equations. By a solution of the system, we mean a common solution of each equation in the system.
Example
Find the line of intersection of the two planes $ x+3y+z=5$ and $ 2x+7y+4z=17$.
- <html><iframe scrolling=“no” src=“https://tube.geogebra.org/material/iframe/id/529147/width/800/height/503/border/888888/rc/true/ai/false/sdz/true/smb/false/stb/true/stbh/true/ld/false/sri/true/at/auto” width=“800px” height=“503px” style=“border:0px;”> </iframe></html>
Intersection of $ x+3y+z=5$ and $ 2x+7y+4z=17$
- To find the equation of the line of intersection, we must find the points which are solutions of both equations at the same time.
- Eliminating variables, we get $x=-16+5z$, $y=7-2z$
- The line of intersection consists of the points $(-16+5z,7-2z,z)$, where $z\in\mathbb{R}$
A detailed look at the last example
- $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
- Find solutions of this system by applying operations
- Aim to end up with a very simple sort of system where we can see the solutions easily.
- $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
- Replace equation (2) with $(2)-2\times (1)$:
- $\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
- Now replace equation (1) with $(1)-3\times (2)$
- $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
- $\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
- can easily rearrange (1) to find $x$ in terms of $z$
- can easily rearrange (2) to find $y$ in terms of $z$
- Since $z$ can take any value, write $z=t$ where $t$ is a “free parameter”
- (which means $t$ can be any real number, or $t\in \mathbb{R}$).
- Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*}
- Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*}
- Can also write this in “vector form”:
- $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.$
- This is the equation of the line where the two planes described by the original equations intersect.
- $\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}$
- For each value of $t$, we get a different solution (a different point on the line of intersection).
- e.g. take $t=0$ to see that $(-16,7,0)$ is a solution
- take $t=1.5$ to see that $(-16+1.5\times 5,7+1.5\times (-2),1.5) = (-8.5,4,1.5)$ is another solution
- etc.
- This works for any value $t\in\mathbb{R}$, and every solution may be written in this way.
lecture_2_slides.txt · Last modified: by rupert