# Math Help - Cramer's Rule Help

1. ## Cramer's Rule Help

Okay so i've spent a good amount of time between yesterday evening and today trying to figure out how to solve this problem using cramer's rule:

x + y +z = 0
7x- y +z=-10
-x+ 4y-z=-20

I heard that Cramer's rule isn't the best way to do this with this many equations and the simple add, substitution and elimination way would be somewhat better but even that way i've gotten the wrong solution. I've tried to eliminate X first by multiplying the first equation by -7 to get -8y-6z=-10 and then i changed the thrid equation to get 27y-6z=-150. I can't remember my exact answer because i erased the rest of it but i solved for z first once i combined them and got a small number and after i finished back substitution everything else i got 2 big numbers that didn't come out as the solution to each of the three equation. I was wondering if anyone could make it clear to me how i should go about solving this problem. Thanks.

2. Originally Posted by MajorJohnson
I was wondering if anyone could make it clear to me how i should go about solving this problem. Thanks.
do an rref on a matrix of the form
[1, 1, 1, 0]
[7, -1, 1, -10]
[-1, 4, -1, -20]
it will output a matrix like so
[1, 0, 0, X]
[0, 1, 0, Y]
[0, 0, 1, Z]
where X, Y, and Z are the answers

3. "I heard that Cramer's rule isn't the best way to do this with this many equations"

The better way is to use Gaussian elimination to reduce the matrix where the diagonal is all one's and the members below the diagonal are all zeros.

4. Originally Posted by 1005
do an rref on a matrix of the form
[1, 1, 1, 0]
[7, -1, 1, -10]
[-1, 4, -1, -20]
it will output a matrix like so
[1, 0, 0, X]
[0, 1, 0, Y]
[0, 0, 1, Z]

where X, Y, and Z are the answers
i don't get where the extra 0's in the output matrix are coming from

5. First, find the determinant of the coefficient matrix:
$\begin{vmatrix}
1 & 1 & 1 \\
7 & -1 & 1 \\
-1 & 4 & -1
\end{vmatrix} = 30$

Set up Cramer's Rule:
$x = \frac {\begin{vmatrix}
0 & 1 & 1 \\
-10 & -1 & 1 \\
-20 & 4 & -1
\end{vmatrix}}{30} = \frac{-90}{30} = -3$

$y = \frac{\begin{vmatrix}
1 & 0 & 1 \\
7 & -10 & 1 \\
-1 & -20 & -1
\end{vmatrix}}{30} = \frac{-120}{30} = -4$

$z = \frac{\begin{vmatrix}
1 & 1 & 0 \\
7 & -1 & -10 \\
-1 & 4 & -20
\end{vmatrix}}{30} = \frac{210}{30} = 7$

Hopefully you can see what I did. The numerators are all determinants of the coefficient matrix with one column replaced by the numbers on the RHS. You replace the 1st column for x, the 2nd column for y, and the 3rd column for z. Yes, Cramer's Rules isn't the best here, especially if you're required to find the determinant of 3x3 matrices by hand. TT_TT

6. Originally Posted by MajorJohnson
i don't get where the extra 0's in the output matrix are coming from
Are you in a matrix algebra class? Unless you're supposed to know where it comes from, just shrug it off and use its useful results. All hail the calculator. If you really want to know the process (it's related to Gaussian elimination, I believe), google reduced row echelon form(rref)

7. If you REALLY want to solve by elimination,
x + y +z = 0
7x- y +z=-10
-x+ 4y-z=-20

Multiply Eq. 1 by -7 and add to Eq. 2:
-7x - 7y - 7z = 0
7x - y + z = -10
----------------
-8y - 6z = -10 [Eq. 4]

Add Eq. 1 and Eq. 3:
Hey! x AND z get eliminated!
x + y +z = 0
-x+ 4y-z=-20
--------------
5y = -20
y = -4

Substitute into Eq. 4:
-8(-4) - 6z = -10
32 - 6z = -10
-6z = -42
z = 7

Substitute into Eq. 1:
x - 4 + 7 = 0
x + 3 = 0
x = -3

8. ok i understand what you did, i messed up and got 28 for the determinate instead of 30. I must have arranged my ascending numbers on the wrong side from my decending ones. Thanks.

Originally Posted by 1005
Are you in a matrix algebra class? Unless you're supposed to know where it comes from, just shrug it off and use its useful results. All hail the calculator. If you really want to know the process (it's related to Gaussian elimination, I believe), google reduced row echelon form(rref)
I'm in a college algebra class, i just learned Cramer's rule yesterday, so i guess she'll teach me the other way at some point idk cause i'm not familar with that way but i might search it later on. Eumyang's method worked for me though i just made some mistakes on getting the determinate that's what though off my answer. Thanks for the input though.