Math 1600 Lecture 7, Section 2, 19 Sep 2014

$ \newcommand{\bdmat}[1]{\left|\begin{array}{#1}} \newcommand{\edmat}{\end{array}\right|} \newcommand{\bmat}[1]{\left[\begin{array}{#1}} \newcommand{\emat}{\end{array}\right]} \newcommand{\coll}[2]{\bmat{r} #1 \\ #2 \emat} \newcommand{\ccoll}[2]{\bmat{c} #1 \\ #2 \emat} \newcommand{\colll}[3]{\bmat{r} #1 \\ #2 \\ #3 \emat} \newcommand{\ccolll}[3]{\bmat{c} #1 \\ #2 \\ #3 \emat} \newcommand{\collll}[4]{\bmat{r} #1 \\ #2 \\ #3 \\ #4 \emat} \newcommand{\ccollll}[4]{\bmat{c} #1 \\ #2 \\ #3 \\ #4 \emat} \newcommand{\colllll}[5]{\bmat{r} #1 \\ #2 \\ #3 \\ #4 \\ #5 \emat} \newcommand{\ccolllll}[5]{\bmat{c} #1 \\ #2 \\ #3 \\ #4 \\ #5 \emat} \newcommand{\red}[1]{{\color{red}#1}} \newcommand{\lra}[1]{\mbox{$\xrightarrow{#1}$}} \newcommand{\rank}{\textrm{rank}} \newcommand{\row}{\textrm{row}} \newcommand{\col}{\textrm{col}} \newcommand{\null}{\textrm{null}} \newcommand{\nullity}{\textrm{nullity}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \renewcommand{\Arg}{\operatorname{Arg}} \renewcommand{\arg}{\operatorname{arg}} \newcommand{\adj}{\textrm{adj}} \newcommand{\mystack}[2]{\genfrac{}{}{0}{0}{#1}{#2}} \newcommand{\mystackthree}[3]{\mystack{\mystack{#1}{#2}}{#3}} \newcommand{\qimplies}{\quad\implies\quad} \newcommand{\qtext}[1]{\quad\text{#1}\quad} \newcommand{\qqtext}[1]{\qquad\text{#1}\qquad} \newcommand{\svec}[1]{\,\vec{#1}} \newcommand{\query}[1]{\toggle{\text{?}\vphantom{#1}}{#1}\endtoggle} \newcommand{\smallquery}[1]{\toggle{\text{?}}{#1}\endtoggle} \newcommand{\bv}{\mathbf{v}} %\require{AMScd} $

Announcements:

Read Section 2.2 for next class. Remember that the text gives more examples and explanation than I can give in a lecture. Work through recommended homework questions.

Office hour: Monday, 3:00-3:30, and Wednesday, 11:30-noon, MC103B.

Help Centers: Monday-Friday 2:30-6:30 in MC 106.

Review: Section 1.3: Distance from a point to a line

Recall the formula for the projection of $\vv$ onto $\vu$: $$ \proj_{\vu}(\vv) = \left( \frac{\vu \cdot \vv}{\vu \cdot \vu} \right) \vu . $$

Example: Find the distance from the point $B = (1,3,6)$ to the line through $P = (1, 1, 0)$ in the direction $\vd = [0,-1,1]$. We derived the formula $d(B, \ell) = \| \vv - \proj_\vd(\vv) \|$, where $\vv = \vec{PB}$.

If the line was in $\R^2$ and had been described in normal form, one could instead compute $\| \proj_\vn(\vv) \|$, which saves one step.

New material

Last bit of Section 1.3: Distance from a point to a plane

Example: Find the distance from the point $B = (1,3,6)$ to the plane $\mathcal{P}$ whose equation is $2x + y - z = 2$. Solution on board, leading to $$ d(B, \mathcal{P}) = \| \proj_\vn(\vv) \| = \frac{|\vn \cdot \vv|}{\|\vn\|} = |-3|/\sqrt{6} = 3/\sqrt{6} .$$

Section 2.1: Systems of Linear Equations

Definition: A linear equation in the variables $x_1, x_2, \ldots, x_n$ is an equation that can be written in the form $$ a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b , $$ where the coefficients $a_1, \ldots, a_n$ and the constant term $b$ are constants.

Linear equations: $$ \kern-6ex \begin{aligned} 2x - 5y\ &= 10, & r + \frac{1}{2} s\ &= 0.5 t - 2, \\ x\ &= 0, & \quad x_1 - \sqrt{2} \, x_2 - (\sin \frac{\pi}{5}) \, x_3\ &= 0 . \end{aligned} $$ Non-linear equations: $$ \kern-6ex x y + z = 1, \quad x^2 + y^2 = 2, \quad \sin(x) = 0, \quad 2^y + z = 16. $$

A solution to $ a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b $ is a vector $[s_1, \ldots, s_n]$ such that the equation is true when we substitute $x_1 = s_1, \ldots, x_n = s_n$. For example, $[ 10, 2 ]$ is a solution to $2x-5y=10$.

When a linear equation has two unknowns, its solutions form a line in $\R^2$. (The linear equation is the general form of this line.) To describe the solutions in parametric form, we can solve for one of the variables in terms of the other.

For example, for $2x-5y=10$, we can write $y = \frac{2}{5} x - 2$. If we set $x$ to a parameter $t$, we get parametric solutions $$ \begin{aligned} x\ &= t \\ y\ &= \frac{2}{5} t - 2 \end{aligned} $$ or, more concisely, $[t, \frac{2}{5} t - 2]$.

When there are three variables, the solutions form a plane, and it can be described in parametric form by solving for one variable in terms of the other two.

The same works when there are $n$ variables: we can solve for one in terms of all of the others, and get a solution with $n-1$ parameters.

 

Systems of linear equations

Definition: A system of linear equations is a finite set of linear equations, each with the same variables. A solution to the system is a vector that satisfies all of the equations.

Example: $$ \begin{aligned} x + y\ &= 2\\ -x + y\ &= 4 \end{aligned} $$ Is $[1, 1]$ a solution? How about $[-1, 3]$? How can we find all solutions? What's happening geometrically?

Example: $$ \begin{aligned} x + \phantom{2} y\ &= 2\\ 2 x + 2 y\ &= 4 \end{aligned} $$ Is $[1, 1]$ a solution? How about $[-1, 3]$? How can we find all solutions? What's happening geometrically?

Example: $$ \begin{aligned} x + y\ &= 2\\ x + y\ &= 3 \end{aligned} $$ Is $[1, 1]$ a solution? How about $[-1, 3]$? How can we find all solutions? What's happening geometrically?

A system is consistent if it has one or more solutions, and inconsistent if it has no solutions. We'll see later that a consistent system always has either one solution or infinitely many.

 

Solving a system

We started with the system on the left and produced the system on the right: $$ \begin{aligned} x + y\ &= 2, & \qquad x + \phantom{2} y\ &= 2\\ -x + y\ &= 4, & 2y\ &= 6 \end{aligned} $$ The system on the right was easy to solve. These two systems are said to be equivalent because they have exactly the same solutions. (The geometry is different, though!)

Example 2.5: Similarly, a large system such as $$ \begin{aligned} x - y - \phantom{3} z\ &= 2 \\ y + 3 z\ &= 5 \\ 5 z\ &= 10 \end{aligned} $$ is easy to solve, because of its triangular structure. The method is called back substitution: $$ \begin{aligned} z\ &= 2\\ y\ &= 5 - 3z = 5 - 6 = -1\\ x\ &= 2 + y + z = 2 - 1 + 2 = 3. \end{aligned} $$ So the unique solution is $[3, -1,\, 2]$.

Let's see how a general system can be converted into a system with a triangular form.

Example 2.6: We'll solve the system on the left $$ \begin{aligned} \ph x - \ph y - \ph z\ &= 2 \\ 3 x - 3 y + 2 z\ &= 16 \\ 2 x - \ph y + \ph z\ &= 9 \end{aligned} \qquad\qquad \bmat{rrr|r} 1 & -1 & -1 & 2 \\ 3 & -3 & 2 & 16 \\ 2 & -1 & 1 & 9 \emat $$ but to save time, we can write it as the augmented matrix on the right.
Today, we'll show the equations as well.

To put it into triangular form, the first step is to eliminate the $x$s in equations 2 and 3. We write $R_i$ for the $i$th equation or the $i$th row of the augmented matrix.

Replace $R_2$ with $R_2 - 3 R_1$: $$ \begin{aligned} \ph x - \ph y - \ph z\ &= 2 \\ 5 z\ &= 10 \\ 2 x - \ph y + \ph z\ &= 9 \end{aligned} \qquad\qquad \bmat{rrr|r} 1 & -1 & -1 & 2 \\ 0 & 0 & 5 & 10 \\ 2 & -1 & 1 & 9 \emat $$ Replace $R_3$ with $R_3 - 2 R_1$: $$ \begin{aligned} \ph x - \ph y - \ph z\ &= 2 \\ 5 z\ &= 10 \\ y + 3 z\ &= 5 \end{aligned} \qquad\qquad \bmat{rrr|r} 1 & -1 & -1 & 2 \\ 0 & 0 & 5 & 10 \\ 0 & 1 & 3 & 5 \emat $$ Now we can exchange rows 2 and 3, to end up in triangular form: $$ \begin{aligned} \ph x - \ph y - \ph z\ &= 2 \\ y + 3 z\ &= 5 \\ 5 z\ &= 10 \end{aligned} \qquad\qquad \bmat{rrr|r} 1 & -1 & -1 & 2 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 5 & 10 \\ \emat $$ Hey! This is the system we solved earlier, so now we know that the solution is $[3, -1, 2]$.

This system and the original system have exactly the same solutions. (Explain.) We say they have the same solution set and therefore that they are equivalent systems.

Questions

True/false: The equation $$ \frac{x}{\sin(2)} + y = z $$ is linear.

True/false: The system $$ \begin{aligned} 2 x + 3 y + 4 z\ &= 7 \\ 4 x + 6 y + 8 z\ &= 9 \end{aligned} $$ has no solutions.

True/false: The system $$ \begin{aligned} 2 x + 3 y + 4 z\ &= 7 \\ 4 x + 6 y + 8 z\ &= 14 \end{aligned} $$ has a unique solution.

Question: Solve the system $$ \begin{aligned} 2 x + 3 y\ &= 2 \\ x + 2 y\ &= 2 \end{aligned} $$ geometrically and algebraically.

Question: How many solutions does the system $$ \begin{aligned} 2 x + 3 y\ &= 2 \\ x + 2 y\ &= 2 \\ x + 4 y\ &= 2 \end{aligned} $$ have?

This about this for next class.