Question

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

Answer 1

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

Answer 2

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

Answer 3

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

like this ?

Answer 4

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

Answer 5

you may right click and see latex commands :

Answer 6

Oh, okay then.

Answer 7

btw, we call it a 3x3 system
the size of matrix \(A\) is \(3\times 3\) in the system of equations : \[Ax=b\]

Answer 8

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"

Answer 9

familiar with elimination/row opearations ?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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
```

Answer 14

Okay..

Answer 15

your goal in elimination is to get a triangular matrix

Answer 16

So, not 4x + 6y = 10?

Answer 17

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

Answer 18

you want to change the starting matrix to above form

Answer 19

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 ?

Answer 20

Yes, at least I think so anyway

Answer 21

do it

Answer 22

So now the second and third row match?

Answer 23

no, what do you get after doing that first step

Answer 24

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

Answer 25

how ?

Answer 26

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

Answer 27

yeah you could do that!

Answer 28

Was I not supposed to do that?

Answer 29

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

Answer 30

so lets keep going

Answer 31

for second step, maybe subtract second row from third row

Answer 32

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

R3-R2 gives :
\[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 0& 0& 0& |0\end{matrix}\right]\]

Answer 33

Okay then, what's the next step?

Answer 34

next look at the element at 2,1 position

you want to make it 0 so that you get a triangular matrix

Answer 35

So 21 needs to be 0?

Answer 36

yes how to do that using row operations ?

Answer 37

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

Answer 38

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

Answer 39

you mean that right ?

Answer 40

Yes, sorry I was thinking in terms of -21

Answer 41

that works perfectly! do it

Answer 42

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

Answer 43

Looks good! so above reduced system is equivalent to :

```
5x + 3y - 9z = 2
- 8y - 76z = 33
```

Answer 44

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

Answer 45

x-y=7?

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

yes, find x also

Answer 51

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

Answer 52

x = 23/8

Answer 53

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

Answer 54

save that, we need to find nullspace

Answer 55

for nullspace, you want to solve :

```
5x + 3y - 9z = 0
- 8y - 76z = 0
```

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

Oh I see, that makes sense!