anonymous
  • anonymous
using the gauss-jordan method to solve system of equations for: x=1-y 2x=z 2z=-2-y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
this is what i have so far.. |dw:1332720397518:dw|
phi
  • phi
|dw:1332720488237:dw|
anonymous
  • anonymous
manipulating matrix to obtain |dw:1332720478178:dw|

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anonymous
  • anonymous
its a -2?
phi
  • phi
Your top entry is +1 so the 2 should be -2
phi
  • phi
sometimes people draw a line to remind themselves |dw:1332720640332:dw|
phi
  • phi
to clear the 0 in position row 2, col 1 (2,1) multiply the 1st row by -2 and add to the 2nd row. (Of course, leave the 1st row unchanged)
anonymous
  • anonymous
im confused already
phi
  • phi
|dw:1332720850068:dw|
phi
  • phi
you want to get rid of all the numbers below 1 in the first column (namely, the 2) you copy the 1st row to a new matrix. to get the new 2nd row, multiply the 1st row by -2 and add to the 2nd row. Copy the answer to the new matrix. Copy the 3rd row over to the new matrix. Now we are ready to continue with the new matrix.
anonymous
  • anonymous
so y is the 2rd 2 listed as a neg? shouldnt it be positive
phi
  • phi
We start with the 1st row 1 1 0 1 multiply by -2 to get -2 -2 0 -2 add this to the 2nd row 2 0 -1 0 ---------- 0 -2 -1 -2 the new 2nd row We are not finished. But the new matrix is 1 1 0 1 0 -2 -1 -2 0 1 2 -2
phi
  • phi
now divide all the numbers in the 2nd row by -2. This changes the -2 to a 1 which is what we will want eventually: 1 1 0 1 0 -2 -1 -2 <--- divide this row only by -2 0 1 2 -2 we get 1 1 0 1 0 1 1/2 1 0 1 2 -2
phi
  • phi
now you want to get rid of the 1 in that last row. 1st. copy the 1st row to a new matrix 2nd copy the 2nd row to a new matrix 3rd. calculate a new 3rd row by multiplying the 2nd row by -1 and adding to the 3rd row. Copy the answer to the new matrix.
anonymous
  • anonymous
1 1 0 1 0 1 1/2 1 0 0 1 2/3
phi
  • phi
to get the last row, start with the 2nd row 0 1 0.5 1 times -1 to get 0 -1 -0.5 -1 add this row to the last row 0 1 2 -2 0 0 1.5 -3 <-- new last row You got 0 0 1 2/3 but I am not sure how? It shoud be 0 0 3/2 -3 (1.5 is the same as 3/2)
phi
  • phi
so we now have 1 1 0 1 0 1 1/2 1 0 0 3/2 -3 <-- multiply this row by 2/3 (changes 3/2 to 1) we get 1 1 0 1 0 1 1/2 1 0 0 1 -2 we are getting there!
anonymous
  • anonymous
theres more?
phi
  • phi
It has to look like your picture with just 1's on the diagonal. so now work up copy the last row to a new matrix calculate a new 2nd row by multiplying the last row by -1/2 and adding to the 2nd row copy the answer to the new matrix copy the 1st row to the new matrix. Post what you get.
anonymous
  • anonymous
0 0 1 -2 0 0 -.5 1
anonymous
  • anonymous
im lost again
phi
  • phi
This is good 0 0 -.5 1 add it the to 2nd row to get a new 2nd row
anonymous
  • anonymous
how do i get a new second row
anonymous
  • anonymous
thats the part that confuses me is where you get the numbers for the next row
phi
  • phi
1 1 0 1 <-- copy to a new matrix 0 1 1/2 1 <-- calculate a new 2nd row HOW. See below. 0 0 1 -2 <-- copy to a new matrix multiply the last row by -1/2 you get 0 0 -0.5 +1 0 1 +0.5 +1 <-- add old 2nd row 0 1 0 2 <-- new 2nd row. COPY to new matrix
anonymous
  • anonymous
so the new matrix starts out with 0 1 0 2 then?
phi
  • phi
We start with 1 1 0 1 0 1 1/2 1 0 0 1 -2 we keep the last row with out changing it. (It matches what we want right?) copy it to the 3rd row in a new matrix. We want to get rid of the 1/2 in the 2nd row so it has a 0. So we change it by multiplying by the last row by -1/2 and adding to the 2nd row. Copy the answer to the 2nd row in the new matrix. then copy the 1st row to the 1st row in the new matrix
phi
  • phi
You would get 1 1 0 1 0 1 0 2 0 0 1 -2 now only the 1st row has an extra one that we want to get rid off, by making it a 0. so copy the 3rd row to the 3rd row of a new matrix copy the 2nd row to the 2nd row of a new matrix change the 1st row by multiplying the 2nd row by -1 0 -1 0 -2 <-- 2nd row multiplied by -1. Add to old 1st row 1 1 0 1 <-- old 1st row. Add together 1 0 0 -1 <-- new first row . copy to 1st row of new matrix. So here is the final answer 1 1 0 1 ---change--> 1 0 0 -1 0 1 0 2 --- copy ---> 0 1 0 2 0 0 1 -2 --- copy ---> 0 0 1 -2 this says that x= -1, y =2 and z = -2 we can check them in the original equations to make sure we did not make a mistake.
anonymous
  • anonymous
how do you find this info out using graphic calc?

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