Matrices

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Questions

well, i got a question about multiplying matrices... say you have 2 with the same size like 2x2 

 [1    3]   [3    2]
 [4    2] x [1    6]

where would i start? would i just match em up and multiply

  • the 1 to the 3
  • the 3 to the 2
  • the 4 to the 1
  • and the 2 to the 6?

i have this same question about adding and subtracting matrices.


What does the 0 matrix mean? i saw it on our last quiz and forgot what it was. 

is a zero matrix. In this case it's 2x2 matrix with zeros in each cell like: . It could also be called the additive identity matrix or null matrix.

today

rows go this way ----> like when you row a boat


columns go this way down

like Mr. H's house


When adding or multiplying if the inside dimensions dont match up you can not solve the problem


Matrices are not commutative because- B*C doesnt equal C*B


Matrices are associative because (AB)C = A(BC)


The Identity Matrix is a matrix the is square(2X2,3X3,4X4 etc..) and has ones going diagonal starting in the left corner and going down to the right and the rest of the cells are filled with zeros

  • Example
    • \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}
    • \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Writing a system of equations as a matrix equation and solving (by hand-the long pointless way):

1. Get all the variables to one side. in this example, 2x=6+3y-4z, it needs to be changed to 2x-3y+4z=6.

2. Set up the first matrix. The number of variables and the number of equations equal the dimensions. If there are 3 variables and 2 equations, then the dimensions will be 3x2.

3. The first matrix will be multiplied by another matrix containing the variables. this second matrix , for example, could be 3x1, x, y, and z.

4. These two matrices equal the answers/numbers without a variable.(the number of equations by one- it will most likely be 3x1)

Example: 2x+3y-4z=1, 4y+5z=25, z=1

    • \begin{bmatrix} 2 & 3 & 4 \\ 0 & 4 & 5 \\ 0 & 0 & 1 \end{bmatrix} * \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix}1\\25\\1 \end{bmatrix}
Coefficients Variables Solutions
[2 3 4 [x [1
0 4 5 * y = 25
0 0 1] z] 1]

5. Next you have to multiply both sides by the inverse of the first matrix, which we can call matrix A.

6. now to find the inverse-on paper-the second matrix has to be the same dimensions as the first matrix. This is done by multiplying matrix A by its inverse and setting that quantity equal to the correct identity matrix (in this case, 3x3).

For the inverse, use rows (a, b, c), (d, e, f), and (g, h, i).

7. continue finding the inverse by multiplying row 1 with column 1, then row 2 column 1, row 3 column 1, row 1 column 2, etc. 2a+3d-4g=1 4d+5g=0 <<< this is one system of equations. use them to solve for column one of the inverse. g=0

8. after you have found the whole inverse matrix, go back to the original (step 5) and plug it in for A inverse.

9. get out yo handy calculator. go to the matrix application and plug in all the numbers for A and B, B being the answer from step 5 (3x1 column matrix containing 1,25,1)

10. after you've got those numbers entered in there and are at the blank screen, type [A]then the inverse button,parenthesis,[B],end parenthesis, and hit enter.

and that is how you solve a system of equations through matrices. using a calculator is WAY faster and easier and i'd say use that way. except you gotta know how to do this by hand for the test.


Example: Linda has a total of 225 points on three tests. The sum of the first and second exceeds her third score by 61. Her first scare exceeds her second by 6. Find her three test scores.

    • \begin{bmatrix} 1 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \\ \end{bmatrix}
    • \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}

Equals

    • \begin{bmatrix} 61 \\ 225 \\ 6 \\ \end{bmatrix}

The equations are: x + y = 61 x + y = 225 x - y = 6

Answer**\begin{bmatrix} 74.5 \\ 68.5 \\ 82 \\ \end{bmatrix}

== Multiplication With Matrices: The Easy Way ==

When multiplying matrices, its easier to remember that the far left column is always going to be the top row of what ever you're multiplying it to. Each column as you go along corresponds to this situation. For example, the 1st column will ALWAYS be multiplied to what ever is in the 1st row, the 2nd column will always be multiplied with the 2nd row, etc. After multiplying these numbers you add whatever you have in the 1st row of the 1st matrices. This solution will be apart of your solution by putting it wherever you started the problem. If I knew how to show it I would. If anybody can give a visual of this statement feel free to do so.

Here's some good info on Matrices. This is a search result from a website that does a good job explaining math. The first few links is the kind of stuff we are doing now. Take a look.
also if you are wondering if matrices have any real world impacts like me they are used in air port traffic according to mr Hust

A realy good use for matrices is for solving for a system of equations. such as

3x + 2y -7z = 28 2x * 4z = 16 4y + z = 8

you do this by making two matrices the first one is all the known numbers in the quation which would look like this 


[[3 2 -7]            
  2 0  4         
  0 4  1]]

If one of the variables is not in one of the equations just plug in a 0 for that variable in the matrix

the second matrix is the product for each equation

[[28 

 16
  8]]

from there you are able to solve the problem for all the variables by multiplying the inverse of the first matrix by the second matrix. In problem form it would look like this

[[3 2 -7 [[x [[28 

 2 0  4   *     y    =    16
 0 4  1]]       z]]        8]]



Direct Variation:

y = kx y varies directly with x

[x= 4 y= 6] [x=8 y=?] y=kx 6=k4 k=3/2 

       y=3/2x
       y=3/2(8)
        y=12

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Contributors

AndrewW, CameronC, ChelseyM, DanielleF, DavidW, EmelieF, JoelR, MeaganT, MichaelM, MichaelP, Mr. H, NajiebM, RudyR, TaylorB, TimD

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