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anonymous
 one year ago
help me plz again
anonymous
 one year ago
help me plz again

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1how far did you get? were you able to set up the matrix?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was not i was so confused on this like i dont know where to start

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1well you first pull out the coefficients of each equation. Do you see what they are?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1they are simply the numbers in front of the variables

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1example: `4x` has the coefficient of `4`

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the `y` is really `1y`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so four and five right

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1along the first equation, what are the 3 coefficients?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1read across the row

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1do you see what I mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh ok i got you hold on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 4 and three for the first one right

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1don't forget about the 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the first equation is really `4x  1y + 3z = 12`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the seciond one is 1x+4y+6z=12

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.132 on the right side, not 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0opps my bad i looked at it wrong haha i acciedently looked up at the end sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05x+3y+9z=20 for the last

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1So if we just focus on the coefficients on the left side, we have this 3x3 matrix \[ \left[\begin{array}{ccc} 4 & 1 & 3\\ 1 & 4 & 6\\ 5 & 3 & 9 \end{array} \right] \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1agreed? or no?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the next matrix is simply the matrix of variables written as a column like this \[ \left[\begin{array}{ccc} x\\ y\\ z \end{array} \right] \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1and then finally we have a third matrix of the values on the right side of the equation the next matrix is simply the matrix of variables written as a column like this \[ \left[\begin{array}{ccc} 12\\ 32\\ 20 \end{array} \right] \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1putting this all together, we form this matrix equation \[ \left[\begin{array}{ccc} 4 & 1 & 3\\ 1 & 4 & 6\\ 5 & 3 & 9 \end{array} \right] * \left[\begin{array}{ccc} x\\ y\\ z \end{array} \right] = \left[\begin{array}{ccc} 12\\ 32\\ 20 \end{array} \right] \] making sense so far?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1next we form an augmented matrix by ignoring the variables temporarily. We just combine the first and third matrix like this \[ \left[\begin{array}{cccc} 4 & 1 & 3 & 12\\ 1 & 4 & 6 & 32\\ 5 & 3 & 9 & 20 \end{array} \right] \] the vertical bar separates the two matrices

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1this is where gauss jordan elimination begins

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1do you know how to do gauss jordan elimination? or no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i really dont i have treid and i keep faling at it this is whay im here now hahaha

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1The first thing I would do is swap rows 1 and 2. We denote this with notation R1 <> R2. I'm doing this to make the element in row1,column1 be a `1` \[ \left[\begin{array}{cccc} 4 & 1 & 3 & 12\\ 1 & 4 & 6 & 32\\ 5 & 3 & 9 & 20 \end{array} \right] \] \[ \left[\begin{array}{cccc} 1 & 4 & 6 & 32\\ 4 & 1 & 3 & 12\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \phantom\\ R_1 \leftrightarrow R_2\\ \phantom\\ \end{array} \] I couldn't get the thing to line up properly, but the `R1 <> R2` should either be in row 1 or row 2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1let me know when you're ready for the next step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK GIVE ME A SECOND TO GET THIS DOWN sorry for the caps i for got i had them on

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Now we want to make that `4` in row2,column1 to be a `0`. We do this by computing `R24*R1` and replacing all of R2 with that result \[ \left[\begin{array}{cccc} 1 & 4 & 6 & 32\\ 0 & 17 & 21 & 140\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \\ R_2  4R_1 \rightarrow R_2\\ \\ \end{array} \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Next we want to make that `5` in row3,column1 to be a `0`. We do this by computing `R35*R1` and replacing all of R3 with that result \[ \left[\begin{array}{cccc} 1 & 4 & 6 & 32\\ 0 & 17 & 21 & 140\\ 0 & 17 & 21 & 180\\ \end{array} \right] \begin{array}{c} \\ \\ R_3  5R_1 \rightarrow R_3\\ \end{array} \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1We could divide the second row by 17 to make that first '17' in that row equal to '1', but I'm going to skip that step. Instead I'm going to subtract R2 and R3 and replace R3 R3R2 > R3 \[ \left[\begin{array}{cccc} 1 & 4 & 6 & 32\\ 0 & 17 & 21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \begin{array}{c} \\ \\ R_3  R_2 \rightarrow R_3\\ \end{array} \]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1The last row of this matrix \[ \left[\begin{array}{cccc} 1 & 4 & 6 & 32\\ 0 & 17 & 21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \] tells us that `0x + 0y + 0z = 40` which essentially turns into `0+0+0 = 40` and that turns into `0 = 40`

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1but `0 = 40` is NEVER true regardless of what x,y,z are. Since we have a contradiction, we have an inconsistent system that has NO solutions at all.
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