Question

Consider this system of linear equations, where k is a real parameter
x-2ky=-3
3x+(1-6k)y-(k+2)z=k-9
-2x+(5k-1)y+(1-k)z=2k^2+6
In each case, select all of the following which are true.
a) k=-1
b)k=0
c)k=1
d) k<-1
e) k>1
f) k in in the interval (-1,0)
g) k is in the interval (0, -1)

Any help would be greatly appreciated because I'm completely lost.

Answer 1

what has to happen with the system.
does it have to have one or infinite solutions?
does it have to be inconsistent?

Answer 2

After those choices are listed, it asks for values of k such that the system has an infinite solutions, one solution and values for when the system is inconsistent.

Answer 3

do u know determinants?

Answer 4

No. Its only my second day in the class and we've only covered the basics of linear systems and this question is on the homework to be graded

Answer 5

ok. then you KNOW gaussian reduction of a matrix. right??

Answer 6

Yes, I know that

Answer 7

great, then tha matrix of your system is
\[ \large \left[\begin{array}{ccc|c}
1 & -2k & 0 & -3\\ 3 & 1-6k & -(k+2) & k-9\\
-2 & 5k-1 & 1-k & 2k^2+6
\end{array}\right] \]
apply gaussian reduction to it and post the final matrix.
don't do gauss-jordan reduction, you r not looking for the solution of the system, just trying to find out if it is or not consistent.

Answer 8

it should take u no more than 5-10 minutes.

Answer 9

I'm not sure that I did this right but this is what i got:
\[\left[\begin{matrix}1 & -2k & 0 &3 \\ 0 & 1 & -k+2 & k\end{matrix}\right]\]
\[\left[\begin{matrix}0 & 9k-1& 1-k &2k ^{2} \end{matrix}\right]\]
(Thes last row is part of the matrix. I couldn't figure out how to type it into one.

Answer 10

no way.

Answer 11

i got this
\[ \large \left[\begin{array}{ccc|c}
1 & -2k & 0 & -3\\ 0 & 1 & -k-2 & k\\
0 & 0 & -(1-k)^2 & k^2-k
\end{array}\right] \]
check your work (honestly) and tell me if you get the same.

Answer 12

if it helps you, the operations i used are:
(-3)*first row + second row
2*first row + third row

(1-k)*second row + third row

Answer 13

I'm going over the work and I realized I forgot to write the negative sign for the negative 3 in the first row. I made a similar mistake in the second row where I forgot to keep the parentheses around k+2 so my signs were off.
The last row is giving me trouble still. Can you explain why you used (1-k)*second row + third row? I feel like I'm missing something obvious.

Answer 14

Never mind, Now I understand why I have to use (1-k).
Now I just find values for k that which make the system have an infinite number of solutions as well as inconsistent and a unique solution right?

Answer 15

yes

Answer 16

the worst that can happen is to lose one of the pivots.
for which value(s) of k do u lose the thirs pivot?

Answer 17

sorry i made a mistake the last rows should be
\[ \large `[0\quad0\quad-(1-k^2)\quad|\quad k^2-k] \]
check this.

Answer 18

have to go for a while.
if u still need help, i'll be back in 15 minutes.

Answer 19

I get that for the last row as well.
I saw you mentioned pivots, my professor has never even mentioned that word before...
thank you so much for your help so far!

Answer 20

i'm back

Answer 21

a "pivot" is the entry in the matrix u use to do the reduction.
for instance u used the 1 in the first row to introduce zeros below it, this 1 is your first pivot.

Answer 22

Ok, I get what that means. But how do you know which values lose their pivots?

Answer 23

ahh.
the last pivot which is \(-(1-k^2)\) could turn zero if \(k=\pm1\)
agree??

Answer 24

Yes, I agree there. So does that mean there is infinitely many solutions for that value?

Answer 25

wait.
now look at the independent term of the third row: \(k^2-k\)
if you put
\[ \large k^2-k=0\qquad\Rightarrow\qquad k=0\text{ or }k=1 \]
agree??

Answer 26

Yes, I follow you

Answer 27

look at the coincidences:
for the pivot u got \(k=\pm1\)
and for the independent term u got \(k=0\vee k=1\)

If k=1 then u lose the pivot and the independent term. so u lose the whole third row, and now u have more variables than equations.

how many solutions do u have in this case??

Answer 28

I think you have no solutions because when you have more variables than equations you cannot have a unique solution or an infinite number of them?

Answer 29

infinite solutions.

Answer 30

now if k=-1 you lose the third pivot but u do NOT lose the independent term. so your third row would be like this
\[ \large (0\quad 0\quad0\quad|\quad2) \]
how many solutions do u have in this case???

Answer 31

You have no solutions because \[0\neq2\]

Answer 32

yeah!! great.

Answer 33

now finally.
if \(k\neq\pm1\) then u don't lose the pivot. this means u have three non-zero pivots in a \(3\times3\) system.
how many solutions do u have in this case??

Answer 34

That means there is one unique solution

Answer 35

yes. fair lady.
and we r done!

Answer 36

ok! so just to make sure I got everything straight,
when k is -1 or 0, there is no solution
when k is 1, there is an infinite number of solution
when k is anything other than 1 or -1, there is one solution.

Right? And on the interval between (-1,0), there is infinite solutions?

Answer 37

when k=1, infinite solutions.
when k=-1, no solutions.
when k is anything other than 1 or -1, one solution.

regarding the interval, u have to decide according to the three choices above.

Answer 38

All right. I think I got it now. Thank you so much for all your help. You are a way better teacher than my professor. Thank you!!

Answer 39

u r welcome.