SOLUTION: how do i determine whether (1,-5,10) is a solution of the following system. x+y+z=6 3x-y-z=-2 2x-y+4z=37

Algebra ->  Matrices-and-determiminant -> SOLUTION: how do i determine whether (1,-5,10) is a solution of the following system. x+y+z=6 3x-y-z=-2 2x-y+4z=37      Log On


   

Question 651018: how do i determine whether (1,-5,10) is a solution of the following system.
x+y+z=6
3x-y-z=-2
2x-y+4z=37
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
how do i determine whether (1,-5,10) is a solution of the following system.
x+y+z=6
3x-y-z=-2
2x-y+4z=37
--------------
You sub 1 for x, -5 for y & 10 for z and see if it fits.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

to determine whether , plug in values for x=1, y=-5 and z=10. I you get that left side of equation is equal to right side, than (1,-5,10) is a solution of the following system
x%2By%2Bz=6..->..1-5%2B10=6->..11-5=6->..6=6
3x-y-z=-2..->..3%2A1-%28-5%29-10=-2->..3%2B5-10=-2->..
8-10=-2->..-2=-2
2x-y%2B4z=37..->..2%2A1-%28-5%29%2B4%2A10=37.->..2%2B5%2B40=37->..47=37...not true
so, the ordered triple (1, -5, 10) is NOT a solution of this system

here is a solution:

Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables



system%281%2Ax%2B1%2Ay%2B1%2Az=6%2C3%2Ax%2B-1%2Ay%2B-1%2Az=-2%2C2%2Ax%2B-1%2Ay%2B4%2Az=37%29



First let A=%28matrix%283%2C3%2C1%2C1%2C1%2C3%2C-1%2C-1%2C2%2C-1%2C4%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 6, -2, and 37 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=-20. 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=-20. 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=%28-20%29%2F%28-20%29=1



So the first solution is x=1




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


We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C1%2C1%2C1%2C3%2C-1%2C-1%2C2%2C-1%2C4%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=60.



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=%2860%29%2F%28-20%29=-3



So the second solution is y=-3




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





Let's reset again by letting A=%28matrix%283%2C3%2C1%2C1%2C1%2C3%2C-1%2C-1%2C2%2C-1%2C4%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=-160.



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=%28-160%29%2F%28-20%29=8



So the third solution is z=8




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

Final Answer:




So the three solutions are x=1, y=-3, and z=8 giving the ordered triple (1, -3, 8)




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.





as you can see, the three solutions are x=1, y=-3, and z=8 giving the ordered triple (1, -3, 8)