SOLUTION: I can't get Cramer's Rule to work for Y. It's OK for X and Z. the matrix is: 1X + 3Y -1Z = 1 -2X -6Y + 1Z = -3 3X + 5Y -2Z = 4 The Determinate is 2. X = 2, Y =

Algebra ->  Matrices-and-determiminant -> SOLUTION: I can't get Cramer's Rule to work for Y. It's OK for X and Z. the matrix is: 1X + 3Y -1Z = 1 -2X -6Y + 1Z = -3 3X + 5Y -2Z = 4 The Determinate is 2. X = 2, Y =      Log On


   

Question 239300: I can't get Cramer's Rule to work for Y. It's OK for X and Z.
the matrix is:
1X + 3Y -1Z = 1
-2X -6Y + 1Z = -3
3X + 5Y -2Z = 4
The Determinate is 2.
X = 2, Y = 1, Z = 1 using Cramer's Rule, but this doesn't check
Using RREF, the solution is 2,0,1 which works.
When I take the determinate of Y in Cramer's rule, it is not 0 which it has to be to make Y=0.
what am I doing wrong?
could it be that Cramer's Rule doesn't work some times?
Thanks
Photonjohn
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I can't get Cramer's Rule to work for Y. It's OK for X and Z.
the matrix is:
1X + 3Y -1Z = 1
-2X -6Y + 1Z = -3
3X + 5Y -2Z = 4
The Determinate is 2.
X = 2, Y = 1, Z = 1 using Cramer's Rule, but this doesn't check
Using RREF, the solution is 2,0,1 which works.
When I take the determinate of Y in Cramer's rule, it is not 0 which it has to be to make Y=0.
what am I doing wrong?
could it be that Cramer's Rule doesn't work some times?
Thanks
----------------------
Cramer's Rule always works.
If the coefficient determinant is zero it means there is no
x,y,z point that satisfies the linear systerm.
----------------------
You have to be careful when using Cramer as a single arithmetic
error will result in a wrong answer.
===========================================
By the way, the answer is x = 2 ; y = 0, z = 1
AND the Y-determinant is zero. Check your
arithmetic.
=======================================
Cheers,
Stan H.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
1x+3y-1z=1
-2x-6y+1z=-3
3x+5y-2z=4

D=-4
Dx=-8
Dy=0
Dz=-4
x=-8/-4= 2
y=0/-4= 0
z=-4/-4= 1
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables



system%281%2Ax%2B3%2Ay%2B-1%2Az=1%2C-2%2Ax%2B-6%2Ay%2B1%2Az=-3%2C3%2Ax%2B5%2Ay%2B-2%2Az=4%29



First let A=%28matrix%283%2C3%2C1%2C3%2C-1%2C-2%2C-6%2C1%2C3%2C5%2C-2%29%29. This is the matrix formed by the coefficients of the given system of equations.


Take note that the right hand values of the system are 1, -3, and 4 and they are highlighted here:




These values are important as they will be used to replace the columns of the matrix A.




Now let's calculate the the determinant of the matrix A to get abs%28A%29=-4. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.



Notation note: abs%28A%29 denotes the determinant of the matrix A.



---------------------------------------------------------



Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).






Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=-8. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.



To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-8%29%2F%28-4%29=2



So the first solution is x=2




---------------------------------------------------------


We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C1%2C3%2C-1%2C-2%2C-6%2C1%2C3%2C5%2C-2%29%29 again (this is the coefficient matrix).




Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).






Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=0.



To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%280%29%2F%28-4%29=0



So the second solution is y=0




---------------------------------------------------------





Let's reset again by letting A=%28matrix%283%2C3%2C1%2C3%2C-1%2C-2%2C-6%2C1%2C3%2C5%2C-2%29%29 which is the coefficient matrix.



Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D






Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=-4.



To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-4%29%2F%28-4%29=1



So the third solution is z=1




====================================================================================

Final Answer:




So the three solutions are x=2, y=0, and z=1 giving the ordered triple (2, 0, 1)




Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.