Question

I need help once again :D

Answer 1

nice. what's your favorite way to solve systems - do you use substitution, elimination, graphing, tables, matrices, etc, any method will work...

Answer 2

What does nontrvial mean? my textbook doesnt explain

Answer 3

don't worry about it - it basically means don't quit with x=0 y=0 as your only solution

Answer 4

but that is what i am getting LOL

Answer 5

why : what was your first step?

Answer 6

well i subsituted

Answer 7

oh maybe i shld use a matrix lol

Answer 8

i dont see any -lambda y

Answer 9

are you doing\[(\lambda-1)x + 2y = 0\]\[x + \lambda y = 0\]Solve for x on that second one and stick it into the first one

Answer 10

oh man sorry i was looking ta the wrong one sorry abt that

Answer 11

so whaddya do next?

Answer 12

expand everything out? divide everybody by y?

Answer 13

and then u r left with y =0

Answer 14

no, you have stuff * y + other stuff * y = 0, and you divide by y so you can make y go away

Answer 15

oh i see what u did lol

Answer 16

u r left with stuff + other stuff = 0

Answer 17

oh but i am getting left with an ugly number

Answer 18

U R BACK :D

Answer 19

i got -L*(1-L)-2 = 0 or so

Answer 20

it is +2 so when u bring it over to the other side it is -

Answer 21

and then u cant take the sqrt of a negative

Answer 22

or rather (1-L)*(-L)+2 = 0

Answer 23

right

Answer 24

we can find the roots

Answer 25

oh i get what to do now

Answer 26

yah that's no matter, if you were staring at
\[x^2−9x=−8\]
you wouldn't take the square root, since x=1 and x=8 work fine

Answer 27

so L is -2 and 1

Answer 28

ya, if i were your teacher i'd ask you to put those two values into the original system, to make sure that you get a homogeneous system with "non-trivial solutions"

Answer 29

well then i guess i will do that LOL

Answer 30

also i bet (L-2)=0 actually means L = +2 etc.

Answer 31

uh oh

Answer 32

when i did the matrix the last row became all zeros

Answer 33

meaning that it is trivial :(((

Answer 34

How can x and y not equal 0???

Answer 35

oh i think it worked out LOL

Answer 36

homogeneous not same as trivial ... as long as the matrix isn't entirely zeros you should get some nice answers, like x = 20 y = -10 for L = 2 I think.

Answer 37

huh? That didnt happen LOL

Answer 38

L = 2 giving \[x+2y=0\] as the answer

Answer 39

l=-2 btw

Answer 40

\[\lambda=-1,2\]

Answer 41

y am i getting that L=1,-2?

Answer 42

compute

\[\det\left(\left[\begin{matrix}\lambda-1 & 2 \\ 1 & \lambda\end{matrix}\right]\right)\]

set equal to zero and solve

Answer 43

ya i got that thanks. Did s/t wrong with my arithmetic

Answer 44

\[(\lambda-1)\lambda-2=0\]
\[\lambda^2-\lambda-2=0\]
\[(\lambda+1)(\lambda-2)=0\]
\[\lambda=-1,2\]

Answer 45

ok thanks zarkon

Answer 46

Thanks broken fixer