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 =

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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)   (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)   (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







First let . 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 , , and 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 . 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: 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 (since we're replacing the 'x' column so to speak).






Now compute the determinant of to get . 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 by the determinant of to get:



So the first solution is




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


We'll follow the same basic idea to find the other two solutions. Let's reset by letting 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 (since we're replacing the 'y' column in a way).






Now compute the determinant of to get .



To find the second solution, divide the determinant of by the determinant of to get:



So the second solution is




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





Let's reset again by letting 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






Now compute the determinant of to get .



To find the third solution, divide the determinant of by the determinant of to get:



So the third solution is




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

Final Answer:




So the three solutions are , , and 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.
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