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anonymous
 one year ago
How would I go abouts solving a 3x4 matrix system?
anonymous
 one year ago
How would I go abouts solving a 3x4 matrix system?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}5 & 3 & 9 & 2 \\ 4 & 4 & 8 & 5 \\ 21 & 11 & 53 & 15\end{matrix}\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Specifically one such as above, except with a line down the middle before the last column?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\left[\begin{matrix}5 & 3 & 9 && 2 \\ 4 & 4 & 8 && 5 \\ 21 & 11 & 53 && 15\end{matrix}\right]\] like this ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, how did you manage to do that on here by the way?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you may right click and see latex commands : dw:1440239723612:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1btw, we call it a 3x3 system the size of matrix \(A\) is \(3\times 3\) in the system of equations : \[Ax=b\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you're just attaching that "b" also as the last column to get that 3x4 matrix that 3x4 matrix with "b" attached is called "augmented matrix"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1familiar with elimination/row opearations ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Familiar but not well enough to solve the problem at hand unfortunately.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1first see that the solutions are not affected by row operations because you're just adding/subtracting a multiple of one row from the other... this doesn't change the solutions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if you have x + y = 2 x+2y = 3 then, adding one equation to other won't change the solutions because you're adding the same thing to both sides of a particular equation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1first convince yourself that below two systems will have same solutions : system1 ``` x + y = 2 x+2y = 3 ``` system2 ``` x + y = 2 2x+3y = 5 ```

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1your goal in elimination is to get a triangular matrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, not 4x + 6y = 10?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that is a valid row operation too, you're allowed to multiply entire row by a constant

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440240774738:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you want to change the starting matrix to above form

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1start by working the element at 2,1 \[\left[\begin{matrix}5 & 3 & 9 && 2 \\ 4 & 4 & 8 && 5 \\ 21 & 11 & 53 && 15\end{matrix}\right]\] adding 4/5 times first row to second row does the job, right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, at least I think so anyway

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So now the second and third row match?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1no, what do you get after doing that first step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}5 & 3 & 9 &  2 \\ 21 & 11 & 53 & 15 \\ 21 & 11 & 53 & 15\end{matrix}\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05 times the 1st row added into the 2nd row?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah you could do that!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Was I not supposed to do that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that is not a standard method, but what you did is a better one..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for second step, maybe subtract second row from third row

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\left[\begin{matrix}5 & 3 & 9 &  2 \\ 21 & 11 & 53 & 15 \\ 21 & 11 & 53 & 15\end{matrix}\right]\] R3R2 gives : \[\left[\begin{matrix}5 & 3 & 9 &  2 \\ 21 & 11 & 53 & 15 \\ 0& 0& 0& 0\end{matrix}\right]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay then, what's the next step?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1next look at the element at 2,1 position you want to make it 0 so that you get a triangular matrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 21 needs to be 0?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yes how to do that using row operations ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Multiply the 2nd row by 5 and then add 21x1st row to the second row?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Multiply the 2nd row by 5 and then "subtract" 21x1st row to the second row?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you mean that right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, sorry I was thinking in terms of 21

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that works perfectly! do it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}5 & 3 & 9 & 2 \\ 0 & 8 & 76 & 33 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Looks good! so above reduced system is equivalent to : ``` 5x + 3y  9z = 2  8y  76z = 33 ```

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let \(z=0\) and find a particular solution

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1how ? plugin \(z=0\) in the bottom equation first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I just plugged 0 in for z for both equations, my bad.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1the advantage of triangular matrix is just that solving is trivial from bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So y = 33/8 for the bottom solution?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1plugin z=0, y=33/8 in first equation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1right, so our particular solution is (23/8, 33/8, 0)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1save that, we need to find nullspace

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1for nullspace, you want to solve : ``` 5x + 3y  9z = 0  8y  76z = 0 ```

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1plugin \(z=1\) and solve y, x values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x = 15/2 & y = 19/2?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Yes, so the nullspace is `t(15/2, 19/2, 1)`

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1the complete solution is given by `particular` + `nullspace`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the one with z being 0 + what we just got?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Complete solution : ``` (23/8, 33/8, 0) + t(15/2, 19/2, 1) ```

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see, that makes sense!
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