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.Com
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)   (Show Source): You can put this solution on YOUR website!


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


Given Matrices:













Jump to Top of Page


Part i)


Start with the addition of the two given matrices


Add up the matrices by adding the corresponding entries


Add

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

Answer:

So





Jump to Top of Page

Part ii)

To find the inverse of the matrix , we can follow these steps:


Step 1) Find the determinant


The determinant of is . So this means that

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

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


Now switch the highlighted values to get


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


Step 3) Change the sign


Now change the sign of the highlighted values to get


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


Step 4) Multiply by the inverse of the determinant


Multiply by to get


Plug in to get


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

Step 5) Perform scalar multiplication and simplify.


Multiply by EVERY element to get


Multiply to get


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

Answer:
So the inverse of is


This means that if then


note: to verify your work, multiply the matrices and and you should get the matrix (the identity matrix).






Jump to Top of Page

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.

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

Answer:

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








Jump to Top of Page

Part iv)

I'm sure you meant to write correct? If so, then...


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

AMP Rendering Error of 'B^T = (matrix(2,3,1,-1,4,2,3,5))^T = (matrix(3,2,1,2-1,3,4,5)) '.



Step 2: Multiply matrices and C. Since the inner dimensions are equal, this means that the matrix product 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:





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:



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


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




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:



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






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




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:



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


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




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:



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






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




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:




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


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




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:




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










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


Answer:


So AMP Rendering Error of 'B^T*C = (matrix(3,2,1,2,-1,3,4,5))*(matrix(3,2,2,-1,3,5)) = (matrix(3,2,8,9,7,16,23,21))'.




In other words,




Jump to Top of Page

RELATED QUESTIONS

Hi everyone I asked a similiar question recently but got no responce, I may have input... (answered by jim_thompson5910)
Hi everyone I asked a similiar question recently but got no responce, I may have input... (answered by jim_thompson5910)
Hi all, I asked this question recently, but did not get a response. Hopefully someone can (answered by Alan3354)
Hi, I just posted this problem but judging by Ms.MathLover's response, I have confused... (answered by ikleyn,MathLover1)
Hi. I'm very confused In how to get the answer to this problem: t-3+2(t+2) I tried to... (answered by ankor@dixie-net.com)
hi, i an usually good with systems but this one is giving me trouble and i have a test... (answered by Earlsdon)
Hi all, I am still having trouble with some more graphing problems. I need to find the... (answered by solver91311)
Good morning. Please help me solve this question. Approximately .25 of all glass bottles (answered by Cromlix)
Pythagorean Thereom Question. (X+1)^2 + 6^2 = 9^2 It is a professor Leonard problem but... (answered by Alan3354)