Solve each linear system using row reduction

  • Thread starter ND3G
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  • #1
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3x+3y+3z=3
x+y+z=-1

3 3 3 | 3
1 1 1 |-1

1 1 1 | 0
0 0 0 | 1

x = -y -z
y = 1

Substitute y into x, x = -1 -z
Let z = t, tER

Then x = -1 -t, y = 1

The corresponding vector equation of the line intersection is:

(x,y,z) = (-1, 1, 0) + t(-1, 0, 1)

Does that look right? This is my first time dealing with these.
 

Answers and Replies

  • #2
radou
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No, it isn't right. It can easily be seen that this system has no solution.

Btw, when you obtain a "solution", it is useful to plug it back into your equation. For example, for t = 0, you have (x, y, z) = (-1, 1, 0), and it can easily be seen that it doesn't satisfy the equations.
 
  • #3
HallsofIvy
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3x+3y+3z=3
x+y+z=-1
If you divide the first equation by 3, that becomes x+ y+ z= 1! Obviously, you can't have x+ y+ z= 1 and x+ y+ z= -1 for the same x,y,z.

3 3 3 | 3
1 1 1 |-1

1 1 1 | 0
0 0 0 | 1

x = -y -z
y = 1
How did you get that first row? If you divide each number in the first row by 3, you get 1 1 1 | 1, not 1 1 1 | 0. And if you then subtract that new first row from the second row, you get 0 0 0 | -2. Of course, that is equivalent to 0x+ 0y+ 0z= -2 which is not true for any values of x, y, and z.
I don't see how you could get "y= 1". That would be 0 1 0 | 1

Substitute y into x, x = -1 -z
Let z = t, tER

Then x = -1 -t, y = 1

The corresponding vector equation of the line intersection is:

(x,y,z) = (-1, 1, 0) + t(-1, 0, 1)

Does that look right? This is my first time dealing with these.

These two planes are parallel and do not intersect.
 
  • #4
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3 3 3 | 3
1 1 1 |-1

3 3 3 | 3
0 0 0 | 6

1 1 1 | 1
0 0 0 | 1

1 1 1 | 0 This last part is R1 - R2
0 0 0 | 1
 
  • #5
radou
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I suggest you go through your lecture notes on row reduction once again (i.e. what can be done and how can it be done) and read HallsofIvy's post once again carefully.
 
  • #6
HallsofIvy
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3 3 3 | 3
1 1 1 |-1

3 3 3 | 3
0 0 0 | 6
I will have to ask again how you got this. In order to get "0" in the first place of the second row by a row operation, you will have "subtract 1/3 the first row from the second row so as to get 1- (1/3)(3)= 1- 1= 0. That will give 0 for the next two places also but -1- (1/3)(3= -1-1= -2 as I said before. The second row becomes 0 0 0 | -2, not 0 0 0 | 6.

1 1 1 | 1
0 0 0 | 1

1 1 1 | 0 This last part is R1 - R2
0 0 0 | 1
 
  • #7
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Here is what I did.

3 3 3 | 3
1 1 1 |-1

3 3 3 | 3
0 0 0 | 6 (R1-3R2) --> 3 - (-3) = 6

1 1 1 | 1 (R1/3)
0 0 0 | 1 (R2/6)

1 1 1 | 0 (R1-R2)
0 0 0 | 1

This matches exactly with the answer my TI-83 gives me.
 
Last edited:
  • #8
radou
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3 3 3 | 3
0 0 0 | 6 (R1-3R2) --> 3 - (-3) = 6

This isn't "row 1 - 3*row2".
 
  • #9
HallsofIvy
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He has replaced row 2 with "row1- 3*row2". Unfortunately that loses all the information in row1!
 

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