Question

Find all values of c for which the system
cx+z=0
2y-4z=0
2x-y=0
has a solution other than (0,0,0) and find all solutions for this value c.

Answer 1

find the null space

Answer 2

row reduce echelon the matrix

c 0 1
0 2 -4
2 -1 0

Answer 3

without a value at c how to I find the nul A?

Answer 4

the value of c tends to migrate along the process

Answer 5

So c stays c?

Answer 6

yep, just treat it as some unknown value, other than zero

Answer 7

or, you can swap rows if need be

Answer 8

interchange rows to bottom and r1-->r3,

Answer 9

sure, that sounds like a good step :)

Answer 10

create vector equation
x2=2x3
x1=(1/2)x2

Answer 11

c=-x3

Answer 12

2 -1 0
0 2 -4
c 0 1


1 -1/2 0
0 1 -2
c 0 1


1 -1/2 0
0 1 -2
0 -1/2-c -c


1 0 -1
0 1 -2
0 0 -1-2c

let -1-2c = 0, c = -1/2 should be one solution

Answer 13

im sure somethings wrong in that ....

Answer 14

I had the first portion, but what did you do to get -1/2-c on the reduction

Answer 15

there it is :) i did addition instead of multiplication

Answer 16

nevermind I am an idiot

Answer 17

also, finding a det(A) = 0 seems like a good route

Answer 18

when it is in reduced form right?

Answer 19

nah, just take the original form and run a determinant on it, then solve for c

-(4+4c+0)
c 0 1 c 0
0 2 -4 0 2
2 -1 0 2 -1
+(0+0+0)

-4-4c = 0 when c = -1

Answer 20

\[\begin{pmatrix}x_1&x_2&x_3\end{pmatrix}=x_3\begin{pmatrix}1&2&1\end{pmatrix}\]

Answer 21

.... latex fail

Answer 22

the preview isnt previwing :/
\[\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=x_3\begin{pmatrix}1\\2\\1\end{pmatrix}\]

Answer 23

det A= (c*2*0)+(0*-4*2)+(1*0*1)-((0)+(c*-4*-1)+(1*2*2))=0-4c-4

Answer 24

yeah, you can go that long route if you like

Answer 25

c=-1 and then use that in original matrix to find values?

Answer 26

yep

Answer 27

Ok thanks a ton. This is the easy stuff and I cannot even do it. The worst part is he hard stuff I understand better.

Answer 28

another route we can tak

c 0 1
0 2 -4
2 -1 0


1 0 1/c
0 1 -2
0 -1 -2/c


1 0 1/c
0 1 -2
0 0 (-2c-2)/c

c =-1

Answer 29

good luck :)

Answer 30

Thanks

Answer 31

Cx+z=0
2y-4z=0
2x-y=0
OR AX=0
Where A=[■(c&0&1@0&2&-4@2&-1&0)],X=[■(x@y@z)]
Given system has solutions if |A|=0
|A|=|■(C&0&1@0&2&-4@2&-1&0)|=C|■(2&-4@-1&0)|-0|■(0&-4@2&0)|+1|■(0&2@2&-1)|
|A|=C[(2)(0)-(-4)(-1)]-0+1[(0)(-1)-(2)(2)]
|A|=C(-4)-4=0
-4C=4
C=4/(-4)=-1
Whenc=-1,given system of equation becomes
-x+z=0
2y-4z=0
2x-y=0
Letz=k(≠0) be any number then
-x+k=0 orx=k
2y-4k=0 or 2y=4k or y=2k
Solution is x=k,y=2k, z=k