# Solve each linear system using row reduction

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.

Homework Helper
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.

HallsofIvy
Homework Helper
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.

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

Homework Helper
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.

HallsofIvy
Homework Helper
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

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.

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