Linear Algebra infinite number of solutions

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Homework Statement


just started this semester, any help will be appreciated.

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

solve the system

Homework 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?)



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)
 

Answers and Replies

  • #2
35,238
7,058

Homework Statement


just started this semester, any help will be appreciated.

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

solve the system

Homework 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?)



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
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.
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)
 
  • #3
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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
35,238
7,058
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
109
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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:
  • #6
35,238
7,058
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}
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.
. . .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:
  • #7
35,238
7,058
_bd_ said:
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. . .
That's not a problem. It just means you will have two free variables, hence an infinite no. of solutions.
 
  • #8
109
0
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
 

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