How would I go abouts solving a 3x4 matrix system?

- Clarence

How would I go abouts solving a 3x4 matrix system?

- jamiebookeater

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- Clarence

\[\left[\begin{matrix}5 & 3 & -9 & 2 \\ -4 & -4 & -8 & 5 \\ 21 & 11 & -53 & 15\end{matrix}\right]\]

- Clarence

Specifically one such as above, except with a line down the middle before the last column?

- 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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## More answers

- Clarence

Yes, how did you manage to do that on here by the way?

- ganeshie8

you may right click and see latex commands :
|dw:1440239723612:dw|

- Clarence

Oh, okay then.

- 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

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

familiar with elimination/row opearations ?

- Clarence

Familiar but not well enough to solve the problem at hand unfortunately.

- 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

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

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

Okay..

- ganeshie8

your goal in elimination is to get a triangular matrix

- Clarence

So, not 4x + 6y = 10?

- ganeshie8

that is a valid row operation too, you're allowed to multiply entire row by a constant

- ganeshie8

|dw:1440240774738:dw|

- ganeshie8

you want to change the starting matrix to above form

- 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

Yes, at least I think so anyway

- ganeshie8

do it

- Clarence

So now the second and third row match?

- ganeshie8

no, what do you get after doing that first step

- Clarence

\[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]\]

- ganeshie8

how ?

- Clarence

5 times the 1st row added into the 2nd row?

- ganeshie8

yeah you could do that!

- Clarence

Was I not supposed to do that?

- ganeshie8

that is not a standard method, but what you did is a better one..

- ganeshie8

so lets keep going

- ganeshie8

for second step, maybe subtract second row from third row

- 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

Okay then, what's the next step?

- ganeshie8

next look at the element at 2,1 position
you want to make it 0 so that you get a triangular matrix

- Clarence

So 21 needs to be 0?

- ganeshie8

yes how to do that using row operations ?

- Clarence

Multiply the 2nd row by 5 and then add 21x1st row to the second row?

- ganeshie8

Multiply the 2nd row by 5 and then "subtract" 21x1st row to the second row?

- ganeshie8

you mean that right ?

- Clarence

Yes, sorry I was thinking in terms of -21

- ganeshie8

that works perfectly! do it

- Clarence

\[\left[\begin{matrix}5 & 3 & -9 & |2 \\ 0 & -8 & -76 & |33 \\ 0 & 0 & 0 & |0\end{matrix}\right]\]

- ganeshie8

Looks good! so above reduced system is equivalent to :
```
5x + 3y - 9z = 2
- 8y - 76z = 33
```

- ganeshie8

let \(z=0\) and find a particular solution

- Clarence

x-y=7?

- ganeshie8

how ?
plugin \(z=0\) in the bottom equation first

- Clarence

Oh, I just plugged 0 in for z for both equations, my bad.

- ganeshie8

the advantage of triangular matrix is just that
solving is trivial from bottom

- Clarence

So y = -33/8 for the bottom solution?

- ganeshie8

yes, find x also

- ganeshie8

plugin z=0, y=-33/8 in first equation

- Clarence

x = 23/8

- ganeshie8

right, so our particular solution is (23/8, -33/8, 0)

- ganeshie8

save that, we need to find nullspace

- ganeshie8

for nullspace, you want to solve :
```
5x + 3y - 9z = 0
- 8y - 76z = 0
```

- ganeshie8

plugin \(z=1\) and solve y, x values

- Clarence

x = 15/2 & y = -19/2?

- ganeshie8

Yes, so the nullspace is `t(15/2, -19/2, 1)`

- ganeshie8

the complete solution is given by `particular` + `nullspace`

- Clarence

So the one with z being 0 + what we just got?

- ganeshie8

Complete solution :
```
(23/8, -33/8, 0) + t(15/2, -19/2, 1)
```

- Clarence

Oh I see, that makes sense!

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