1703 Lecture: Introduction to Matlab

Matrix Representation

$x-2y=1$

Can someone give me some values of x and y that satisfy this equation?

x	y
3	1
1	0
2	${\frac {1}{2}}$
101	50

How many combinations of x and y will satisfy this equation? $\infty$

How else can we represent this equation?

As a line:

(insert figure here)

Is the slope positive or negative?

What's the slope?

$3x+2y=11$

Some values of x and y that satisfy this equation?

x	y
3	1
1	4
75	-107

How many combinations of x and y will satisfy this equation? $\infty$

We can also represent this equation as a line

(insert figure here - both lines on same plot)

What's the slope?

We can see that the point x=3, y=1 is where these two lines meet - which means it is the combination of x and y that satisfies both of these equations.

How else can we represent these equations?

In column form:

$\left[{\begin{array}{cc}1\\3\end{array}}\right]x+\left[{\begin{array}{cc}-2\\2\end{array}}\right]y=\left[{\begin{array}{cc}1\\11\end{array}}\right]$

After pushing these two columns together, we get:

$\left[{\begin{array}{cc}1&-2\\3&2\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}1\\11\end{array}}\right]$

What about the equations:

${\begin{aligned}3x+5y+z&=&16\\2x-y-z&=&3\\x+4y-2z&=&3\end{aligned}}$

How to represent this graphically?

Instead of lines, use planes

Let's look at the first 2 equations only:

(Insert figure here)

The two planes intersect to form a line

Now the third equation: Also a plane

The line that represents the intersection of these first two equations will intersect the third equation's plane at one point

We can also represent these equations in matrix form:

$\left[{\begin{array}{cc}3\\2\\1\end{array}}\right]x+\left[{\begin{array}{cc}5\\-1\\4\end{array}}\right]y+\left[{\begin{array}{cc}1\\-1\\-2\end{array}}\right]z=\left[{\begin{array}{cc}16\\3\\3\end{array}}\right]$

$\left[{\begin{array}{ccc}3&5&1\\2&-1&-1\\1&4&-2\end{array}}\right]\left[{\begin{array}{c}x\\y\\z\end{array}}\right]=\left[{\begin{array}{c}16\\3\\3\end{array}}\right]$

How do we represent 4 equations graphically?

5 equations?

6 equations?

We run into a limit using graphical methods

What about the matrix representation of 4 equations? 5 equations? 6 equations?

The matrix representation is easy and flexible

Solving Systems of Equations

Now let's talk about how you actually solve these systems...

Going back to original example:

${\begin{aligned}x-2y&=&1\\3x+2y&=&11\end{aligned}}$

or,

$\left[{\begin{array}{cc}1&-2\\3&2\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}1\\11\end{array}}\right]$

Option A:

Use elimination to eliminate one variable, solve for the other variable

Then plug that into one of these equations to find the other variable

Example:

Eliminate y by adding the two equations:

${\begin{aligned}x-2y&=&1\\{\underline {+\left(3x+2y\right)}}&=&{\underline {+(11)}}\\4x+0y&=&12\end{aligned}}$

and therefore

$x=3$

Then solve for y by plugging $x=3$ into original equations:

${\begin{aligned}x-2y&=&1\\(3)-2y&=&1\\-2y&=&-2\\y&=&1\end{aligned}}$

Let's see what's happening in matrix form:

$\left[{\begin{array}{cc}1&-2\\3&2\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}1\\11\end{array}}\right]$

We're adding equation (1) to equation (2), and using that as our new equation (2)

So in the matrix, we're replacing row(2) with ( row(1) + row(2) )

The matrix becomes:

$\left[{\begin{array}{cc}1&-2\\4&0\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}1\\12\end{array}}\right]$

Option B:

What's another way we can solve this?

(Cramer's Rule homework assignment)

Cramer's Rule:

If we have a system like:

$\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}e\\f\end{array}}\right]$

then the solution is:

${\begin{aligned}x&=&{\frac {ed-bf}{ad-bc}}={\frac {(1)(2)-(-2)(11)}{(1)(2)-(-2)(3)}}\\&=&{\frac {2+22}{2+6}}\\x&=&3\end{aligned}}$

and for y:

${\begin{aligned}y&=&{\frac {af-ec}{ad-bc}}={\frac {(1)(11)-(1)(3)}{(1)(2)-(-2)(3)}}&=&{\frac {11-3}{2+6}}y&=&1\end{aligned}}$

One more 2x2 example:

${\begin{aligned}2x+4y&=&1\\-3x-2y&=&3\end{aligned}}$

or,

$\left[{\begin{array}{cc}2&4\\-3&-2\end{array}}\right]\left[{\begin{array}{c}x\\y\end{array}}\right]=\left[{\begin{array}{c}1\\3\end{array}}\right]$

Let's try eliminating x or y and then solving for the remaining variable

So try (1) + 2*(2):

${\begin{aligned}2x+4y&=&1\\{\underline {+2\left(-3x-2y\right)}}&=&{\underline {+2(3)}}\\-4x+0y&=&7\end{aligned}}$

and solving for x,

$x=-{\frac {7}{4}}$

Then we can plug this into our remaining equation and solve for y:

${\begin{aligned}2x+4y&=&1\\2\left(-{\frac {7}{4}}\right)+4y&=&1\\-{\frac {14}{4}}+4y&=&1\\4y&=&{\frac {4}{4}}+{\frac {14}{4}}={\frac {18}{4}}\\\end{aligned}}$

$y={\frac {9}{8}}$

Look at what we did with our matrix:

We changed row (2) to be row (1) + 2*row(2)

(Can you tell me what I should change in my matrix?)

$\left[{\begin{array}{cc}2&4\\-3&-2\end{array}}\right]\rightarrow \left[{\begin{array}{cc}2&4\\-4&0\end{array}}\right]$

Now an example for a 3x3 system:

${\begin{aligned}x+3y-z&=&1\\-2x-6y+z&=&-3\\3x+5y-2z&=&4\end{aligned}}$

or,

$\left[{\begin{array}{ccc}1&3&-1\\-2&-6&1\\3&5&-2\end{array}}\right]\left[{\begin{array}{c}x\\y\\z\end{array}}\right]=\left[{\begin{array}{c}1\\-3\\4\end{array}}\right]$