anonymous
  • anonymous
help me plz again
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
@Nnesha
anonymous
  • anonymous
@paki

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anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
how far did you get? were you able to set up the matrix?
anonymous
  • anonymous
i was not i was so confused on this like i dont know where to start
jim_thompson5910
  • jim_thompson5910
well you first pull out the coefficients of each equation. Do you see what they are?
anonymous
  • anonymous
i do not sorry
jim_thompson5910
  • jim_thompson5910
they are simply the numbers in front of the variables
jim_thompson5910
  • jim_thompson5910
example: `4x` has the coefficient of `4`
jim_thompson5910
  • jim_thompson5910
the `-y` is really `-1y`
anonymous
  • anonymous
so four and five right
jim_thompson5910
  • jim_thompson5910
along the first equation, what are the 3 coefficients?
anonymous
  • anonymous
4 5 and 1x right
jim_thompson5910
  • jim_thompson5910
read across the row
jim_thompson5910
  • jim_thompson5910
not column
jim_thompson5910
  • jim_thompson5910
do you see what I mean?
anonymous
  • anonymous
ohh ok i got you hold on
anonymous
  • anonymous
so 4 and three for the first one right
jim_thompson5910
  • jim_thompson5910
don't forget about the -1
jim_thompson5910
  • jim_thompson5910
the first equation is really `4x - 1y + 3z = 12`
anonymous
  • anonymous
ohh yeah sorry lol
anonymous
  • anonymous
so the seciond one is 1x+4y+6z=12
jim_thompson5910
  • jim_thompson5910
-32 on the right side, not 12
anonymous
  • anonymous
opps my bad i looked at it wrong haha i acciedently looked up at the end sorry
anonymous
  • anonymous
5x+3y+9z=20 for the last
jim_thompson5910
  • jim_thompson5910
So 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
  • jim_thompson5910
agreed? or no?
anonymous
  • anonymous
yes i agree
jim_thompson5910
  • jim_thompson5910
the 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
  • jim_thompson5910
and 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
  • jim_thompson5910
putting 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?
anonymous
  • anonymous
yeah
jim_thompson5910
  • jim_thompson5910
next we form an augmented matrix by ignoring the variables temporarily. We just combine the first and third matrix like this \[ \left[\begin{array}{ccc|c} 4 & -1 & 3 & 12\\ 1 & 4 & 6 & -32\\ 5 & 3 & 9 & 20 \end{array} \right] \] the vertical bar separates the two matrices
anonymous
  • anonymous
makes since now
jim_thompson5910
  • jim_thompson5910
this is where gauss jordan elimination begins
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
do you know how to do gauss jordan elimination? or no?
anonymous
  • anonymous
i really dont i have treid and i keep faling at it this is whay im here now hahaha
jim_thompson5910
  • jim_thompson5910
that's ok
jim_thompson5910
  • jim_thompson5910
The 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` \[ \left[\begin{array}{ccc|c} 4 & -1 & 3 & 12\\ 1 & 4 & 6 & -32\\ 5 & 3 & 9 & 20 \end{array} \right] \] \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 4 & -1 & 3 & 12\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \phantom\\ R_1 \leftrightarrow R_2\\ \phantom\\ \end{array} \] 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
  • jim_thompson5910
let me know when you're ready for the next step
anonymous
  • anonymous
OK GIVE ME A SECOND TO GET THIS DOWN sorry for the caps i for got i had them on
anonymous
  • anonymous
OKim ready
jim_thompson5910
  • jim_thompson5910
Now we want to make that `4` in row2,column1 to be a `0`. We do this by computing `R2-4*R1` and replacing all of R2 with that result \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \\ R_2 - 4R_1 \rightarrow R_2\\ \\ \end{array} \]
anonymous
  • anonymous
OK GOTCHA
jim_thompson5910
  • jim_thompson5910
Next we want to make that `5` in row3,column1 to be a `0`. We do this by computing `R3-5*R1` and replacing all of R3 with that result \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & -17 & -21 & 180\\ \end{array} \right] \begin{array}{c} \\ \\ R_3 - 5R_1 \rightarrow R_3\\ \end{array} \]
anonymous
  • anonymous
ok i got that
jim_thompson5910
  • jim_thompson5910
We 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 R3-R2 --> R3 \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \begin{array}{c} \\ \\ R_3 - R_2 \rightarrow R_3\\ \end{array} \]
jim_thompson5910
  • jim_thompson5910
The last row of this matrix \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \] tells us that `0x + 0y + 0z = 40` which essentially turns into `0+0+0 = 40` and that turns into `0 = 40`
anonymous
  • anonymous
wow that was alot
jim_thompson5910
  • jim_thompson5910
but `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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