# Linear Algebra infinite number of solutions

1. Aug 19, 2010

### _Bd_

1. The problem statement, all variables and given/known data
just started this semester, any help will be appreciated.

4x + 12y - 7z - 20w = 22
3x + 9y - 5z - 28w = 30

solve the system

2. Relevant equations

well since I have more variables than equations I know theres an infinite number of solutions therefore Im gonna have to use parameters later on. (BTW can someone give me tips on how to do parameters?)

3. The attempt at a solution

I have my matrix as follows
4 12 -7 -20 22
3 9 -5 -28 30

multiplying row_2 by -1 and adding it to row_1 I get
1 3 2 8 -8
3 9 -5 -28 30
multiplying row_1 by -3 and adding it to row_2 yields
1 3 2 8 -8
0 0 -11 -52 -54

then what? if I divide by 11 I start getting fractions, and the answer in the back doesnt have any fractions.

AFAIK you have to make the first one in each row 1 and have a zero below it. (condition on row_1 is met)

2. Aug 19, 2010

### Staff: Mentor

This is wasted effort. The idea is to make the leading entry of a row 1, and the corresponding entry of a row below 0. What you should do instead is to add (-3) times row 1 to 4 times row 2. That will make the 2nd row
0 0 1 -52 54

After that, use the leading 1 in row 2 to eliminate the entry above it.

3. Aug 20, 2010

### _Bd_

what do you mean? If i have 0 0 1 -52 54 I'd have the top row as the original 4 12 -7 -20 22. . .I cant do anything to the 4 (unless I make fractions) to get it into a 1

cause thats what I did actually, I started by making the leading coefficient on row_1 1, then I planned on using that to cancel out the zeros below it. . .and so on, I dont really get what you're doing, Im 'ma try it out on paper tho.

4. Aug 20, 2010

### Staff: Mentor

You missed what I said, which was to use the leading 1 in the 2nd row to eliminate the -7 in the row above it.

Then replace row 1 by 1/4 of itself. I don't end up with any fractions.

5. Aug 20, 2010

### _Bd_

oh ok ok, I got it! so I did that and came up with:
1 3 0 96 100
0 0 1 -52 54

so then. . .now my problem is to do the parameters. . .

I can say that

x_4 = 1/52 {54-x_3}
and
x_4 = 1/96 {100-x_1-3x_2}

. . .altho I think im tripping out and going in a tangent. . .how should I do to make parameters (like. . .any tips overall?)

and again thank you.

. . .I was thinking if I solve for x_1 on the first equation and replug back and plug back again I can get x_4 and x_1 in both in terms of x_2 and x_3. . .I guess that would be the answer?

===== on another problem =====

3x + 3y + 12z = 6
x + y + 4z = 2
2x + 5y + 20z = 10
-x + 2y + 8z = 4

I started the way I usually do (from top to bottom, dividing the top row by 3 and using that leading 1 to cancel the 1 on the second row and so forth)

but when I noticed I was getting the exact same lanes. . .

1 1 4 2
0 1 4 2
0 1 4 2

I erased everything so I cant copy all but I remember one of the lines just got ocmpletely elimiated into zeros. . .

Last edited: Aug 20, 2010
6. Aug 20, 2010

### Staff: Mentor

I think that's what I got, but I don't have my work with me, so can't say for sure.
Edit: You're off in one sign in the top row. For your augmented matrix, you should have
1 3 0 -96 100
0 0 1 -52 54

I have edited what I wrote earlier to account for this difference.
Anyway, read off the two lines you have, keeping in mind that this is an augmented matrix.

x1 + 3x2 - 96x4 = 100
x3 -52x4 = 54

Now solve for the first variable in each equation.
x1 = -3x2 + 96x4 + 100
x3 = 52x4 + 54

Since there are no equations for x2 and x4, they are arbitrary.

Last edited: Aug 20, 2010
7. Aug 20, 2010

### Staff: Mentor

That's not a problem. It just means you will have two free variables, hence an infinite no. of solutions.

8. Aug 24, 2010

### _Bd_

I have now to do modeling, I have to find the equation that passes trough given points:

we know its a cubic equation with horizontal tangents at (1,-2) and (-1,2)
since the points seem "simetric" I assumed the graph passes trough the origin (0,0)

so I get the following

a_0 + a_1 (x_1) + a_2 (x_1)^2 + a_3 (x_1)^3 = y_1
a_0 + a_1 (x_2) + a_2 (x_2)^2 + a_3 (x_2)^3 = y_1
a_0 + a_1 (x_3) + a_2 (x_3)^2 + a_3 (x_3)^3 = y_1

when I do the matrix I cant instantly notice that a_0 is zero (therefore there is no constant)

once I do the matrix I have
1 1 1 -2
-1 1 -1 2
0 0 0 0

so my third equation cancels. . .I have 3 variables and only 2 equations... AFAIK you cant solve it (unless you parametize)
I continued and tried to reduce and ended up with
a_1 a_2 a_3
1 0 1 -2
0 1 0 0
which tells me that a_0 =0 and a_1 =0

the answer at the back shows p(x) = -3x + x^3