Question

How do you use the matrix method?

Answer 1

do you mean the metric system?

Answer 2

did you

Answer 3

yes, sorry had to step away from the computer.

Answer 4

But no Matrics like in Algebra 2

Answer 5

It's similar to the substitution method

Answer 6

oh i was going to say goin slow

Answer 7

Wouldn't happen to know how to do that anyway would you? lol

Answer 8

no im in the 8th grade sowwy

Answer 9

but mostlikely if you google it it will know

Answer 10

Lol s'okay. Good luck with your work then. :)

Answer 11

thanks you too goos luck finding it

Answer 12

thnx

Answer 13

np

Answer 14

Matrix method for solving systems of equations?

Answer 15

Yea

Answer 16

How many equations are you talking about?

Answer 17

3

Answer 18

But I need to be taught how to do it with two as well (unless by some miricle I understand when you explain how to do three. lol)

Answer 19

If you can do it with three equations, then you can definitely do it with two!

Answer 20

Alright, do you have an example you want to work on, or should I make something up?

Answer 21

Those aren't equations, and the question is just asking for the element in position 1,2

Answer 22

We can use that as a coefficient matrix if you like and make up a constant matrix to go with it.

Answer 23

Sure.

Answer 24

Ok, I'll make this easy, the constant matrix will be
\[\large \left(\begin{matrix}-2 \\ 5 \\ 37\end{matrix}\right)\]

Answer 25

So we'll use
\[\left[\begin{matrix}8 & -2 & -2 \\ 2 & 6 & -3 \\ 9 & 5 & 6 \end{matrix}\right]\left[\begin{matrix}x & y & z \end{matrix}\right]=\left[\begin{matrix}-2 \\ 5 \\ 37 \end{matrix}\right]\]
\[\large \matrix \]

Answer 26

This represents the equations
8x -2y -2z = -2
2x +6y -3z = 5
9x +5y +6z = 37

Cool so far?

Answer 27

Yeah I get that.

Answer 28

Great, so I'm going to condense the matrix equation into a single augmented matrix, then we are going to do a series of row operations to get it what is called reduced-row echelon form.

Answer 29

\[\large \left[ \begin{matrix} 8 & -2 & -2 & -2 \\ 2 & 6 & -3 & 5 \\ 9 & 5 & 6 & 37\end{matrix} \right]\]

Answer 30

The first step is to make the number in the upper left corner into 1.

Answer 31

The easiest way to do that is to divide everything in row one by 8.

Answer 32

Alright =-0.25

Answer 33

Right, but I'll show another trick that might make things easier.

Answer 34

Wait is row 1 going down?

Answer 35

What do you mean by "going down?"

Answer 36

oh okay

Answer 37

I'm going to swap rows one and two to get a smaller number in the first position to start
\[\large \left[ \begin{matrix} 2 & 6 & -3 & 5 \\ 8 & -2 & -2 & -2 \\ 9 & 5 & 6 & 37\end{matrix} \right]\]

Answer 38

Now, to get a 1 in the first pivot, I'll divide by 2.
\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 8 & -2 & -2 & -2 \\ 9 & 5 & 6 & 37\end{matrix} \right]\]

Answer 39

The next step is to turn everything else in that first column into a zero. This will be done by subtracting multiples of row one from rows two and three.

Answer 40

Can you see a way to do that?
Remember, this is very similar to elimination method with systems of equations. (actually it is exactly the same thing, just in a different notation.)

Answer 41

(Also, a warning: if your arithmetic skills are a little shaky, you might not like where this is going, there is going to be a lot of operations with fractions)

Answer 42

Damn. xD Well I'll try to keep up with it. Would it still be simple if I converted the fractions to decimal tho?

Answer 43

That depends on your preference and if you're using a calculator to help you. I use fractions and do the arithmetic in my head.

To move this along, I'll just show the next step.
To get a zero under that first pivot, I'm going to subtract 8 times row one from frow two:

\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & -26 & 10 & -22 \\ 9 & 5 & 6 & 37\end{matrix} \right]\]

Make sure you understand how I got the above matrix.

Answer 44

So you used the same 8 from earlier to do that?

Answer 45

Sortof, it works out like that coincidentally here. The important thing is to see that I had an 8 in the first position of row two and I wanted to eliminate it. There is a 1 right above it so if I multiply that 1 by 8 and subtract that from the 8 in the second row, I get the zero that I want, but I have to do the same thing to everything else in the row.

Answer 46

I think it's also a good time to show what the ultimate goal looks like.
We want to continue to do this series of row operations until the matrix looks like this.
\[\large \left[ \begin{matrix} 1 & 0 & 0 & c_1 \\ 0 & 1 & 0 & c_2 \\ 0 & 0 & 1 & c_3\end{matrix} \right]\]

Answer 47

(c1, c2, c3) will be the solutions for (x, y, z)

Answer 48

Alright, so we still have
\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & -26 & 10 & -22 \\ 9 & 5 & 6 & 37\end{matrix} \right]\]

What do you suppose we should do to get a zero in the first position of row three?

Answer 49

Sorry window froze and we multiply it by the others right?

Answer 50

Have to multiply something by something and add that to row three.

Answer 51

It'll be very similar to what was done to eliminate that 8 from row two.

Answer 52

1x9 like the 8?

Answer 53

Yep, If you multiply row one by 9 and subtract that from row three, then you'll get the following

\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & -26 & 10 & -22 \\ 0 & -22 & 39/2 & 29/2\end{matrix} \right]\]

Make sure you're also checking my arithmetic at each step. Like I said, I'm adding these fractions in my head, so I might be making mistakes...

Answer 54

okay. but what for the steps that don't start with one? Can we use the one to make the three into a one or...?

Answer 55

You mean the other columns? We'll get to those next.

If you can verify for me that those numbers look correct to you, then I'll show another simplifying step to make life easier.

Answer 56

There is an algorithm to it - a repeated process of similar steps until it's all broken down - kinda like long division or synthetic division: multiply, add; multiply, add; multiply, add; etc. until it's done.

Answer 57

Okay then, btw your math seems right to me.

Answer 58

Ok, so here's the simplifying step (this is optional, but might make it look a little prettier before we go on):

See how all the numbers in row two are even? I can divide row two by 2 to make the numbers smaller.

Likewise, in row three, there are some fractions with denominators of 2, so I can multiply row three by 2 to clear the fractions.

Dig it?

Answer 59

Yea. A lot simplilar like that.

Answer 60

So now the matrix looks like
\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & -13 & 5 & -11 \\ 0 & -44 & 39 & 29\end{matrix} \right]\]

Answer 61

Ok, now let's get on with column two. First step is to make that -13 into a 1 (that's the second pivot).

Answer 62

Unfortunately, the best way to do that is to divide everything in row two by -13. That will make fractions with 13 in the denominator, which is a bit of a drag, but whatever; they're numbers and we'll treat them the same as any other numbers.

Answer 63

\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & 1 & -5/13 & 11/13 \\ 0 & -44 & 39 & 29\end{matrix} \right]\]

Ewww, I know, but it gets the job done.

Answer 64

Here is where using a calculator and decimals will fail you as those fractions have endlessly repeating blocks of decimal digits.

Answer 65

Anyway, onward.
Now, usually, since I like to go all the way down to reduced-row echelon form, I'll go ahead and make that 3 in row one into a zero, but I think we'll take care of row three first, so I can talk about row echelon form.

Answer 66

Alright

Answer 67

So, how do you figure we can change that -44 in row three into a zero?

Answer 68

1x-44 + 44? The reverse of what we did for the others.

Answer 69

Be more specific, though; I want to make sure you got the process.

Answer 70

We had to multiply a number by one and then cancle it out by adding it's inverse like the 8 with -8. Now we already have the -44 so we need to cancle it out with +44

Answer 71

Ok, good, but make sure you know that the actual operation is to take 44 times row two and add that to row three.

Answer 72

That operation leads to
\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & 1 & -5/13 & 11/13 \\ 0 & 0 & 287/13 & 861/13\end{matrix} \right]\]
Which looks ugly, but remember: now you can multiply row three by 13 to cancel the fractions.

Answer 73

\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & 1 & -5/13 & 11/13 \\ 0 & 0 & 287 & 861\end{matrix} \right]\]
And for a really nice trick, try dividing everything in row three by 287 to get a 1 in that third pivot.

Answer 74

How did doing that remove 13 asthe denominator from 287?

Answer 75

If you multipy 287/13 by 13, the 13 cancels out.

Answer 76

Ahhh makes sense.

Answer 77

Now, after this step
\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & 1 & -5/13 & 11/13 \\ 0 & 0 & 1 & 3\end{matrix} \right]\]

You can actually quit using matrices, and put this back into equations and do back substitution.

x + 3y -3z/2 = 5/2
y - 5z/13 = 11/13
z = 3.

Answer 78

That process is called Gaussian Elimination and gets the matrix in 'row echelon form.'
You can also continue the same algorithm of row operations to get the matrix into 'reduced-row echelon form' then it'll be completely solved without ever having to look at the equations again.

Answer 79

I hope you have the idea. I need to go now.
If you want to continue to reduced-row echelon form from

\[\large \left[ \begin{matrix} 1 & 3 & -3/2 & 5/2 \\ 0 & 1 & -5/13 & 11/13 \\ 0 & 0 & 1 & 3\end{matrix} \right]\]

First change that 3 in a12 to a zero, then get zeros in a13 and a23. Once you do that, the solutions will be in vector form in the fourth column.

Good luck.

Answer 80

See here
http://tutorial.math.lamar.edu/Classes/LinAlg/SolvingSystemsOfEqns.aspx

and here
http://www.youtube.com/watch?v=ZK3O402wf1c

for more info.

Answer 81

Thanks for all the help.