Question

Solve the system of equations using a matrix equation.
All help is appreciated! :)
(picture below)

Answer 1

First, do you know how to put them into a matrix?

Answer 2

For example:

\(ax + by = c\)
\(dx - ey = f\)

becomes:

\(\left[\begin{array}{cc|c}
a & b & c\\
d & -e & f\\
\end{array}\right]\)

Answer 3

Not really to be honest. D: @e.mccormick

Answer 4

Ahh okay

Answer 5

How would it look for this equation?

Answer 6

Well, try using spaces and [ ]

So:
[ 1 1 1 | 4]

for the first line.

Remember to add a 0 for any variable that is not in an equation.

Answer 7

So line two would be
[4, 5 | 3]
?
@e.mccormick

Answer 8

Don't forget the 0 zs in line 2.

Answer 9

Alright..so line 3 would be like..
[0, 0, 3 | -10]
@e.mccormick

Answer 10

There is a y there, and the 3z is -.

\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & 4\\
4 & 5 & 0 & 3\\
0 & 1 & -3 & -10
\end{array}\right]\)

Answer 11

FYI, how I type that is:
```
\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & 4\\
4 & 5 & 0 & 3\\
0 & 1 & -3 & -10
\end{array}\right]\)
```

Answer 12

Ahh alright. So then what do we do now that we have that part down? @e.mccormick

Answer 13

You do elementary row operations until it is solved. If you are doing Gauss-Jordan method, then you get it to:

\(\left[\begin{array}{ccc|c}
1 & 0 & 0 & x\\
0 & 1 & 0 & y\\
0 & 0 & 1 & z
\end{array}\right]\)

Where what is in the x spot is the answer for x, the y spot for y, and x for z.

Answer 14

Overview of row operations:
http://www.purplemath.com/modules/mtrxrows.htm

Answer 15

Altern8ively, arrange the simultaneous equation \[\left[\begin{matrix}1 & 1 & 1 \\ 4 & 5 & 0\\0 & 1 & -3\end{matrix}\right] *\left(\begin{matrix}x \\ y\\z\end{matrix}\right) = \left(\begin{matrix}4\\3 \\ -10\end{matrix}\right)\]
\[Ax=b\] \[x = A ^{-1}*b\]

Answer 16

Yes, but if a person does not even know how to put it in matrix form, I tend not to assume they know how to invert a matrix.

Answer 17

\[\therefore\left(\begin{matrix}x \\ y\\z\end{matrix}\right) = \left[\begin{matrix}1 & 1 & 1 \\ 4 & 5 & 0\\0 & 1 & -3\end{matrix}\right] ^{-1} *\left(\begin{matrix}4\\3 \\ -10\end{matrix}\right)\\\]

Answer 18

@e.mccormick yes, I see your point. Yet inverse matrix tends to be basic before Guassian elimination or row echelon form. Yh?

Answer 19

do you know how to find the adjoint of a square matrix ?

Answer 20

Not a lot. I haven't done these kinds of equations much before. @EarthCitizen

Answer 21

Actually, elimination is taught with them as linear equations, before they teach matrices, so people tend to already know it. That would be Gaussian with back substitution, but they rarely tell the students that name. Once they get into matrices, this is usually expanded to Gauss-Jordan, which they also call Reduced Row Echelon Form or rref..

Answer 22

seen, fair enough. Give it a go then @e.mccormick

Answer 23

The joy of math. There are like 5 different ways to solve a matrix. Hehe.

Answer 24

Quite true! So many different ways to solve.

Answer 25

tru

Answer 26

Could you finish showing how you were solving this equation? It's starting to make sense.

Answer 27

Did you look at the elementary row operations thing I linked?

Answer 28

yh, idk that depends on can if you transpose a square matrix. To find the inverse of a square matrix we divide the adjoint; the adjoint is the transpose of the co-factors of the square matrix. Then we divide by the determinant of the same square matrix to find the inverse. It's awesome and very useful

Answer 29

@EarthCitizen Yep. Matrices are very useful things. You can find the areas of odd shapes, like an oval, or volume of things like a parallelepiped, all without calculus.

Answer 30

hehe @e.mccormick

Answer 31

Yes I did. @e.mccormick

And I'm not entirely sure how to do that. @EarthCitizen

Answer 32

alryt, so do you want to ?

Answer 33

OK, so, what do you think would be a good operation to do yo get it closer to solved? With two rows already having 0s, it can be a bit odd choosing the right place to start.

Answer 34

Yes please @EarthCitizen

Annnd I'm not sure where to start. Row two? @e.mccormick

Answer 35

Personally I would do:
(3)R1+R3 to new R3

See, that would make rows 2 and 3 into dealing with just the first two columns, which makes it easier to get the x and y spots.

Answer 36

So [1, 0, 0] + [0, 0, 1] ? @e.mccormick

Answer 37

R1 means Row 1, R3 means Row 3. The (3) in front means to multiply 3 through the version of row 1 you use:

\(3\left[\begin{array}{ccc|c}
1 & 1 & 1 & 4
\end{array}\right] + \left[\begin{array}{ccc|c}
0 & 1 & -3 & -10
\end{array}\right]\implies\)

\(\left[\begin{array}{ccc|c}
3 & 3 & 3 & 12
\end{array}\right] + \left[\begin{array}{ccc|c}
0 & 1 & -3 & -10
\end{array}\right]\implies\)

\(\left[\begin{array}{ccc|c}
3+0 & 3+1 & 3+(-3) & 12+(-10)
\end{array}\right]\implies\)

\(\left[\begin{array}{ccc|c}
3 & 4 & 0 & 2
\end{array}\right]\)

That replaces the original row 3, making the matrix as a whole into:

\(\left[\begin{array}{ccc|c}
1 & 1 & 1 & 4\\
4 & 5 & 0 & 3\\
3 & 4 & 0 & 2
\end{array}\right]\)

Now it is much easier to work with rows 2 and 3.

Answer 38

https://www.youtube.com/watch?v=hu6B1d3vvqU hope this helps

Answer 39

patrickJMT is very good in his examples. He also covers Gauss-Jordan.
https://www.youtube.com/watch?v=CsTOUbeMPUo

Answer 40

What is neat is that as you learn more about matrices you can see how the inversion and Gauss-Jordan are related.