SOLUTION: Hi all I asked a similiar question recently and got a reaaly good response, but this imiliar problem is still giving me trouble. I need to solve the following Matrices and comment

Algebra ->  Matrices-and-determiminant -> SOLUTION: Hi all I asked a similiar question recently and got a reaaly good response, but this imiliar problem is still giving me trouble. I need to solve the following Matrices and comment      Log On


   

Question 200660: Hi all I asked a similiar question recently and got a reaaly good response, but this imiliar problem is still giving me trouble.
I need to solve the following Matrices and comment on the outcomes.For the following:
A =
(2 3 0)
1 4 -3
B =
(1 -1 4)
2 3 5
C =
(2 -1)
3 5
Find and state why or why not the following is possible.
i) A+B
ii) C^-1
iii) AB
1v) B^C
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!


Table of Contents:
Part i)
Part ii)
Part iii)
Part iv)


Given Matrices:


A=%28matrix%282%2C3%2C2%2C3%2C0%2C1%2C4%2C-3%29%29


B=%28matrix%282%2C3%2C1%2C-1%2C4%2C2%2C3%2C5%29%29


C=%28matrix%282%2C2%2C2%2C-1%2C3%2C5%29%29




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Part i)


Start with the addition of the two given matrices


A%2BB=%28matrix%282%2C3%2C2%2B1%2C3%2B%28-1%29%2C0%2B4%2C1%2B2%2C4%2B3%2C-3%2B5%29%29 Add up the matrices by adding the corresponding entries


A%2BB=%28matrix%282%2C3%2C3%2C2%2C4%2C3%2C7%2C2%29%29 Add

===========================================

Answer:

So A%2BB=%28matrix%282%2C3%2C3%2C2%2C4%2C3%2C7%2C2%29%29




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Part ii)

To find the inverse of the matrix C=%28matrix%282%2C2%2C2%2C-1%2C3%2C5%29%29, we can follow these steps:


Step 1) Find the determinant


The determinant of %28matrix%282%2C2%2C2%2C-1%2C3%2C5%29%29 is abs%28matrix%282%2C2%2C2%2C-1%2C3%2C5%29%29=13. So this means that d=13

note: since the determinant is NOT equal to zero, the matrix inverse exists.

--------------------------------
Step 2) Swap the values


Now switch the highlighted values %28matrix%282%2C2%2Chighlight%282%29%2C-1%2C3%2Chighlight%285%29%29%29 to get %28matrix%282%2C2%2Chighlight%285%29%2C-1%2C3%2Chighlight%282%29%29%29


--------------------------------


Step 3) Change the sign


Now change the sign of the highlighted values %28matrix%282%2C2%2C5%2Chighlight%28-1%29%2Chighlight%283%29%2C2%29%29 to get %28matrix%282%2C2%2C5%2Chighlight%281%29%2Chighlight%28-3%29%2C2%29%29


--------------------------------


Step 4) Multiply by the inverse of the determinant


Multiply by 1%2Fd to get %281%2Fd%29%28matrix%282%2C2%2C5%2C1%2C-3%2C2%29%29


Plug in d=13 to get %281%2F13%29%28matrix%282%2C2%2C5%2C1%2C-3%2C2%29%29


--------------------------------

Step 5) Perform scalar multiplication and simplify.


Multiply 1%2F13 by EVERY element to get


Multiply to get %28matrix%282%2C2%2C5%2F13%2C1%2F13%2C-3%2F13%2C2%2F13%29%29


=================================================================

Answer:
So the inverse of %28matrix%282%2C2%2C2%2C-1%2C3%2C5%29%29 is %28matrix%282%2C2%2C5%2F13%2C1%2F13%2C-3%2F13%2C2%2F13%29%29


This means that if C=%28matrix%282%2C2%2C2%2C-1%2C3%2C5%29%29 then C%5E%28-1%29=%28matrix%282%2C2%2C5%2F13%2C1%2F13%2C-3%2F13%2C2%2F13%29%29


note: to verify your work, multiply the matrices C and C%5E%28-1%29 and you should get the matrix %28matrix%282%2C2%2C1%2C0%2C0%2C1%29%29 (the identity matrix).





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Part iii)

Take note that the dimensions of matrices A and B are 2x3 and 2x3 respectively. Since the inner dimensions do NOT match, this means that the matrix product AB is NOT defined. In other words, AB does NOT exist.

note: if A and B are mxn and nxp matrices, then the product AB is defined. It is only defined if the number of columns of A equals the number of rows of B. Otherwise, AB does NOT exist.

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Answer:

So the matrix product AB is not defined (ie it doesn't exist).







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Part iv)

I'm sure you meant to write B%5ET%2AC correct? If so, then...


Step 1: Transpose matrix B to get B%5ET. You do this by converting each row into a column (and vice versa).




Step 2: Multiply matrices B%5ET and C. Since the inner dimensions are equal, this means that the matrix product B%5ET%2AC is defined.



Since the first matrix is a 3 by 2 matrix and the second matrix is a 2 by 2 matrix, this means that the resulting matrix will be a 3 by 2 matrix.

So the final resulting matrix will look like:


B%5ET%2AC=%28matrix%283%2C2%2Cx%2Cx%2Cx%2Cx%2Cx%2Cx%29%29


note: the "x"s are just placeholders for now



--------------------------------------------------




Multiply the corresponding entries from the 1st row of the first matrix by the 1st column of the second matrix. After multiplying, add the values:


1st row, 1st column:
%281%29%2A%282%29%2B%282%29%2A%283%29=2%2B6=8


So the element in the 1st row, 1st column of the resulting matrix is 8. Now let's update the matrix:

B%5ET%2AC=%28matrix%283%2C2%2C8%2Cx%2Cx%2Cx%2Cx%2Cx%29%29
--------------------------------------------------




Multiply the corresponding entries from the 1st row of the first matrix by the 2nd column of the second matrix. After multiplying, add the values:


1st row, 2nd column:
%281%29%2A%28-1%29%2B%282%29%2A%285%29=-1%2B10=9


So the element in the 1st row, 2nd column of the resulting matrix is 9. Now let's update the matrix:

B%5ET%2AC=%28matrix%283%2C2%2C8%2C9%2Cx%2Cx%2Cx%2Cx%29%29




================================================================================




Multiply the corresponding entries from the 2nd row of the first matrix by the 1st column of the second matrix. After multiplying, add the values:


2nd row, 1st column:
%28-1%29%2A%282%29%2B%283%29%2A%283%29=-2%2B9=7


So the element in the 2nd row, 1st column of the resulting matrix is 7. Now let's update the matrix:

B%5ET%2AC=%28matrix%283%2C2%2C8%2C9%2C7%2Cx%2Cx%2Cx%29%29
--------------------------------------------------




Multiply the corresponding entries from the 2nd row of the first matrix by the 2nd column of the second matrix. After multiplying, add the values:


2nd row, 2nd column:
%28-1%29%2A%28-1%29%2B%283%29%2A%285%29=1%2B15=16


So the element in the 2nd row, 2nd column of the resulting matrix is 16. Now let's update the matrix:

B%5ET%2AC=%28matrix%283%2C2%2C8%2C9%2C7%2C16%2Cx%2Cx%29%29




================================================================================




Multiply the corresponding entries from the 3rd row of the first matrix by the 1st column of the second matrix. After multiplying, add the values:


3rd row, 1st column:

%284%29%2A%282%29%2B%285%29%2A%283%29=8%2B15=23


So the element in the 3rd row, 1st column of the resulting matrix is 23. Now let's update the matrix:

B%5ET%2AC=%28matrix%283%2C2%2C8%2C9%2C7%2C16%2C23%2Cx%29%29
--------------------------------------------------




Multiply the corresponding entries from the 3rd row of the first matrix by the 2nd column of the second matrix. After multiplying, add the values:


3rd row, 2nd column:

%284%29%2A%28-1%29%2B%285%29%2A%285%29=-4%2B25=21


So the element in the 3rd row, 2nd column of the resulting matrix is 21. Now let's update the matrix:

B%5ET%2AC+=+%28matrix%283%2C2%2C8%2C9%2C7%2C16%2C23%2C21%29%29








==============================================================================


Answer:


So



In other words,

B%5ET%2AC+=+%28matrix%283%2C2%2C8%2C9%2C7%2C16%2C23%2C21%29%29


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