Clarence
  • Clarence
How would I go abouts solving a 3x4 matrix system?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Clarence
  • Clarence
\[\left[\begin{matrix}5 & 3 & -9 & 2 \\ -4 & -4 & -8 & 5 \\ 21 & 11 & -53 & 15\end{matrix}\right]\]
Clarence
  • Clarence
Specifically one such as above, except with a line down the middle before the last column?
ganeshie8
  • ganeshie8
\[\left[\begin{matrix}5 & 3 & -9 &|& 2 \\ -4 & -4 & -8 &|& 5 \\ 21 & 11 & -53 &|& 15\end{matrix}\right]\] like this ?

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Clarence
  • Clarence
Yes, how did you manage to do that on here by the way?
ganeshie8
  • ganeshie8
you may right click and see latex commands : |dw:1440239723612:dw|
Clarence
  • Clarence
Oh, okay then.
ganeshie8
  • ganeshie8
btw, we call it a 3x3 system the size of matrix \(A\) is \(3\times 3\) in the system of equations : \[Ax=b\]
ganeshie8
  • ganeshie8
you'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
  • ganeshie8
familiar with elimination/row opearations ?
Clarence
  • Clarence
Familiar but not well enough to solve the problem at hand unfortunately.
ganeshie8
  • ganeshie8
first 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
  • ganeshie8
if 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
  • ganeshie8
first convince yourself that below two systems will have same solutions : system1 ``` x + y = 2 x+2y = 3 ``` system2 ``` x + y = 2 2x+3y = 5 ```
Clarence
  • Clarence
Okay..
ganeshie8
  • ganeshie8
your goal in elimination is to get a triangular matrix
Clarence
  • Clarence
So, not 4x + 6y = 10?
ganeshie8
  • ganeshie8
that is a valid row operation too, you're allowed to multiply entire row by a constant
ganeshie8
  • ganeshie8
|dw:1440240774738:dw|
ganeshie8
  • ganeshie8
you want to change the starting matrix to above form
ganeshie8
  • ganeshie8
start 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 ?
Clarence
  • Clarence
Yes, at least I think so anyway
ganeshie8
  • ganeshie8
do it
Clarence
  • Clarence
So now the second and third row match?
ganeshie8
  • ganeshie8
no, what do you get after doing that first step
Clarence
  • Clarence
\[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]\]
ganeshie8
  • ganeshie8
how ?
Clarence
  • Clarence
5 times the 1st row added into the 2nd row?
ganeshie8
  • ganeshie8
yeah you could do that!
Clarence
  • Clarence
Was I not supposed to do that?
ganeshie8
  • ganeshie8
that is not a standard method, but what you did is a better one..
ganeshie8
  • ganeshie8
so lets keep going
ganeshie8
  • ganeshie8
for second step, maybe subtract second row from third row
ganeshie8
  • ganeshie8
\[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]\] R3-R2 gives : \[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 0& 0& 0& |0\end{matrix}\right]\]
Clarence
  • Clarence
Okay then, what's the next step?
ganeshie8
  • ganeshie8
next look at the element at 2,1 position you want to make it 0 so that you get a triangular matrix
Clarence
  • Clarence
So 21 needs to be 0?
ganeshie8
  • ganeshie8
yes how to do that using row operations ?
Clarence
  • Clarence
Multiply the 2nd row by 5 and then add 21x1st row to the second row?
ganeshie8
  • ganeshie8
Multiply the 2nd row by 5 and then "subtract" 21x1st row to the second row?
ganeshie8
  • ganeshie8
you mean that right ?
Clarence
  • Clarence
Yes, sorry I was thinking in terms of -21
ganeshie8
  • ganeshie8
that works perfectly! do it
Clarence
  • Clarence
\[\left[\begin{matrix}5 & 3 & -9 & |2 \\ 0 & -8 & -76 & |33 \\ 0 & 0 & 0 & |0\end{matrix}\right]\]
ganeshie8
  • ganeshie8
Looks good! so above reduced system is equivalent to : ``` 5x + 3y - 9z = 2 - 8y - 76z = 33 ```
ganeshie8
  • ganeshie8
let \(z=0\) and find a particular solution
Clarence
  • Clarence
x-y=7?
ganeshie8
  • ganeshie8
how ? plugin \(z=0\) in the bottom equation first
Clarence
  • Clarence
Oh, I just plugged 0 in for z for both equations, my bad.
ganeshie8
  • ganeshie8
the advantage of triangular matrix is just that solving is trivial from bottom
Clarence
  • Clarence
So y = -33/8 for the bottom solution?
ganeshie8
  • ganeshie8
yes, find x also
ganeshie8
  • ganeshie8
plugin z=0, y=-33/8 in first equation
Clarence
  • Clarence
x = 23/8
ganeshie8
  • ganeshie8
right, so our particular solution is (23/8, -33/8, 0)
ganeshie8
  • ganeshie8
save that, we need to find nullspace
ganeshie8
  • ganeshie8
for nullspace, you want to solve : ``` 5x + 3y - 9z = 0 - 8y - 76z = 0 ```
ganeshie8
  • ganeshie8
plugin \(z=1\) and solve y, x values
Clarence
  • Clarence
x = 15/2 & y = -19/2?
ganeshie8
  • ganeshie8
Yes, so the nullspace is `t(15/2, -19/2, 1)`
ganeshie8
  • ganeshie8
the complete solution is given by `particular` + `nullspace`
Clarence
  • Clarence
So the one with z being 0 + what we just got?
ganeshie8
  • ganeshie8
Complete solution : ``` (23/8, -33/8, 0) + t(15/2, -19/2, 1) ```
Clarence
  • Clarence
Oh I see, that makes sense!

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