1. ## Matrix Algebra

I attempted this problem but the answers I got seem sort of unlikely so if somebody could check what I did and tell me where I went wrong (if I did) then that would be amazing.

Solve for x, y, and z.

xy-2sqrt(y) + 3zy = 8
2xy - 3sqrt(y) + 2zy = 7
-xy + sqrt(y) + 2zy = 4

And my solution:
[1 -2 3 | 8] [1 -2 3 | 8] [1 -2 3 | 8]
[2 -3 2 | 7] [0 1 -4 | -9] [0 1 -4 | -9]
[-1 1 2 | 4] [0 -1 5 | 12] [0 0 1 | 12]

This gives that zy = 12

Then I continued row-echelon on the matrix:
[1 -2 3 | 8]
[0 1 0 | 39]
[0 0 1 | 12]

This gives that sqrt(y) = 39, thus squaring both sides, y = 1521

Then since zy = 12, and y = 1521, yz = 12/1521 = 0.00789

Then further reducing the matrix:
[1 -2 0 | -28] [1 0 0 | 56]
[0 1 0 | 39] [0 1 0 | 39]
[0 0 1 | 12] [0 0 1 | 12]

This means that xy = 56 and since y = 1521, x = 56/1521 = 0.0368

I don't think this can be right given the decimal numbers for x and z, so if somebody could see where I went wrong that would be great, and sorry for the not-so-neat matrices.

Thanks

2. I don't know how this was posted twice, I went to edit one of them and it ended up making a new thread somehow, sorry.

I attempted this problem but the answers I got seem sort of unlikely so if somebody could check what I did and tell me where I went wrong (if I did) then that would be amazing.

Solve for x, y, and z.

xy-2sqrt(y) + 3zy = 8
2xy - 3sqrt(y) + 2zy = 7
-xy + sqrt(y) + 2zy = 4

And my solution:
[1 -2 3 | 8] [1 -2 3 | 8] [1 -2 3 | 8]
[2 -3 2 | 7] [0 1 -4 | -9] [0 1 -4 | -9]
[-1 1 2 | 4] [0 -1 5 | 12] [0 0 1 | 12]

This gives that zy = 12

Then I continued row-echelon on the matrix:
[1 -2 3 | 8]
[0 1 0 | 39]
[0 0 1 | 12]

This gives that sqrt(y) = 39, thus squaring both sides, y = 1521

Then since zy = 12, and y = 1521, yz = 12/1521 = 0.00789

Then further reducing the matrix:
[1 -2 0 | -28] [1 0 0 | 56]
[0 1 0 | 39] [0 1 0 | 39]
[0 0 1 | 12] [0 0 1 | 12]

This means that xy = 56 and since y = 1521, x = 56/1521 = 0.0368

I don't think this can be right given the decimal numbers for x and z, so if somebody could see where I went wrong that would be great, and sorry for the not-so-neat matrices.

Thanks

RonL

4. I'm not sure what font is needed, or if you're talking about the form that my matrices are in, if the ladder is the case, then I'm not sure how to fix that.

Is it showing up as a bunch of symbols? Because it does that when I go to edit the post, but when I view it normally, it's in regular letters again.

I'm not sure what font is needed, or if you're talking about the form that my matrices are in, if the ladder is the case, then I'm not sure how to fix that.

Is it showing up as a bunch of symbols? Because it does that when I go to edit the post, but when I view it normally, it's in regular letters again.
The majority of your text showns up as Greek. Presumably you have pasted it
from somewere else and the font has become confused.

RonL

6. ## Converted font

OK I will convert the font for you, here is your question:

I attempted this problem but the answers I got seem sort of unlikely so if somebody could check what I did and tell me where I went wrong (if I did) then that would be amazing.

Solve for x, y, and z.

xy-2sqrt(y) + 3zy = 8
2xy - 3sqrt(y) + 2zy = 7
-xy + sqrt(y) + 2zy = 4

And my solution:

[1 -2 3 | 8] [1 -2 3 | 8] [1 -2 3 | 8]
[2 -3 2 | 7] [0 1 -4 | -9] [0 1 -4 | -9]
[-1 1 2 | 4] [0 -1 5 | 12] [0 0 1 | 12]

This gives that zy = 12

Then I continued row-echelon on the matrix:

[1 -2 3 | 8]
[0 1 0 | 39]
[0 0 1 | 12]

This gives that sqrt(y) = 39, thus squaring both sides, y = 1521

Then since zy = 12, and y = 1521, yz = 12/1521 = 0.00789

Then further reducing the matrix:

[1 -2 0 | -28] [1 0 0 | 56]
[0 1 0 | 39] [0 1 0 | 39]
[0 0 1 | 12] [0 0 1 | 12]

This means that xy = 56 and since y = 1521, x = 56/1521 = 0.0368

I don't think this can be right given the decimal numbers for x and z, so if somebody could see where I went wrong that would be great, and sorry for the not-so-neat matrices.

Thanks

7. Hey, I found the problem with my work, thanks for the help.