Question

Does a/o know linear algebra?

Answer 1

what is a/o?

Answer 2

anyone

Answer 3

okies

Answer 4

lol

Answer 5

umm what class of algebra are you in

or

what are you learning right now?

Answer 6

Did u see the file i attached?

Answer 7

ya

Answer 8

I just learnt solving linear systems and a bit of matrices involving elementary operations

Answer 9

ahh ok

Answer 10

Do u know linear algebra? What grade r u in?

Answer 11

lol i forgot most of this stuff

Answer 12

did you learn cramers rule yet or determinants?

Answer 13

ohhhh well basically they gave the answer but like i can't figure out how they did it.
No i didnt learn that yet. Still in the way begginning. Only in chapter 1. Skewl started like 2 days ago

Answer 14

they subsituted the x values and then there was a system of linear equations and somehow they got the answer what e/t equaled

Answer 15

whtvr i am blabbing too much

Answer 16

lemme see
ill try ok ?

Answer 17

hehe

Answer 18

k thanks :D

Answer 19

you want to graph this thing right?

Answer 20

um well that wld be afterwards i really want to understand how they got the equation. Once i have the equation it is simple to graph

Answer 21

so what happened?

Answer 22

in the end you want to graph it though

so

the matrix you have is a 3x3

1 | a b c |
2 | a b c |
3 | a b c |

Answer 23

im not done

Answer 24

LOL i am still here don't worry

Answer 25

you ssaid beginning class right?

Answer 26

ya lol just started like on monday

Answer 27

can you use combination?
i didnt try it yet?

Answer 28

wait

Answer 29

i see that you have a solution to the problem

Answer 30

what is combination

Answer 31

so what do you want again?

Answer 32

lol ya i just dont know how they got the answer

Answer 33

o ok ill tell you

Answer 34

Sorry abt that i thought i had told that i just wanted the steps

Answer 35

they just substituted

Answer 36

sorrrrry

Answer 37

lol np i just didnt understand the question my fault also

anyways

Answer 38

do you know how to put a quadratic equation in standard form?

Answer 39

like vertex form?

Answer 40

?

Answer 41

no just the equation in standard form

Answer 42

show me an example

Answer 43

do you know this format

ax^2 + by +c = 0

standard for for the quadratic equation ?

Answer 44

ya ya

Answer 45

but they used diff format

Answer 46

do you notice any thing in any of the problems similar to this ?

Answer 47

Whoa whoa hang on. Are you trying to figure out how to solve the system of equations, or how they came up with the system of equations? Or both?

Answer 48

lol maybe both

Answer 49

?

Answer 50

nah

Answer 51

basically how to solve the system

Answer 52

how they came up with the solution

Answer 53

like i know the equation of a polynomial

Answer 54

try to see if there is a likeness to the standard form of

ax^2 + by+c = 0

Answer 55

Alright. There are a large number of ways by which you could get the solution to some arbitrary system of equations. Given that you say this is your first experience with linear algebra and class started two days ago, I assume that you should be doing this either by adding or subtracting the equations from each other or by row-reducing a matrix. Which would you like me to go over?

Answer 56

i dont think they solved it like that

Answer 57

i prefer row reducing matrix

Answer 58

you dont need that yet

Answer 59

at least not for this problem

Answer 60

but for practice in the future
its good to start early

Answer 61

:)

Answer 62

what do u mean how shld u solve it?

Answer 63

many ways

you just dont need to put so much effort in to it if you just use my way :P

all i know

Answer 64

Okay. I assume then that you can represent the problem as the following matrix equation:

\[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9\end{matrix}\right] \left(\begin{matrix} a_0 \\ a_1 \\ a_2 \end{matrix}\right)= \left(\begin{matrix}4 \\ 0 \\ 12\end{matrix}\right)\]

Reasonable assumption?

Answer 65

ya but like i only know elementary operations so nothing too fancy over there ok

Answer 66

its good to take notes for future references though :)

Answer 67

true ;D

Answer 68

Sure. So what you want to do is tack the column on the right hand side of the equals sign on to the matrix, like this:
\[\left[\begin{matrix}1 & 1 & 1 & | & 4 \\ 1 & 2 & 4 & | & 0 \\ 1 & 3 & 9 & | & 12\end{matrix}\right]\]

Answer 69

ill just type my my effort less way

and call it quits

Answer 70

ok

Answer 71

Thanks mth3v4

Answer 72

Now, you want to add and subtract rows from each other so you get the identity matrix on the left. For example, if you subtract the first row from the second row and the third row, you get the following matrix:
\[\left[\begin{matrix}1 & 1 & 1 & |& 4 \\ 0 & 1 & 3 & | & -4 \\ 0 & 2 & 8 & | & 8\end{matrix}\right]\]

Answer 73

ok ya i get that

Answer 74

Next, I'd subtract 2 x the second row from the third row, and get this:
\[\left[\begin{matrix}1 & 1 & 1 & | & 4 \\ 0 & 1 & 3 & | & -4 \\ 0 & 0 & 2 & | & 16\end{matrix}\right]\]

Answer 75

nm that im getting yelled at :(

Answer 76

gn peoples

Answer 77

who is yelling at you :((((

Answer 78

parents :P

Answer 79

gn gn

Answer 80

good night thanks for ur help :D

Answer 81

Dividing the third row by two yields\[\left[\begin{matrix}1 & 1 & 1 & | & 4\\ 0 & 1 & 3 & | & -4 \\ 0 & 0 & 1 & | & 8\end{matrix}\right]\]

Answer 82

ya still following HEHE

Answer 83

Almost done. Lets subtract the second row from the first row:
\[\left[\begin{matrix}1 & 0 & -2 & | & 8\\ 0 & 1 & 3 & | & -4 \\ 0 & 0 & 1 & | & 8\end{matrix}\right]\]

Answer 84

oh we have two more steps to go

Answer 85

And then add 2x the third to the first:
\[\left[\begin{matrix}1 & 0 & 0 & | & 24\\ 0 & 1 & 3 & | & -4 \\ 0 & 0 & 1 & | & 8\end{matrix}\right]\]

Answer 86

ok

Answer 87

And finally, subtract 3x the third from the second to yield
\[\left[\begin{matrix}1 & 0 & 0 & | &24\\ 0 & 1 & 0 & | & -28 \\ 0 & 0 & 1 & | & 8\end{matrix}\right] \]

Answer 88

ooh YAY tralalaaa

Answer 89

Behold the solution via row reduction :) There are quite a number of ways to solve such systems but this is a good first start and something you can always fall back on if the others get too complicated.

Answer 90

that is what they got

Answer 91

ya this is theonly way we learnt so far

Answer 92

Thanks u r a savior :DDDDD XD

Answer 93

Seriously u were really clear

Answer 94

No problem, good luck. I love linear algebra, it's one of my favorite subjects, so I hope it treats you well :)

Answer 95

so far i am actually really liking it

Answer 96

i am embarrassed to say that i am even liking it better than calc 2