SOLUTION: Please help me with this, I tried to solve it, but cannot get the solution Solve the system for x,y,z 2x+3y-z=4 3x-y+2z=5 x-4y+3z=1 Thank you.

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Please help me with this, I tried to solve it, but cannot get the solution Solve the system for x,y,z 2x+3y-z=4 3x-y+2z=5 x-4y+3z=1 Thank you.       Log On


   

Question 1090883: Please help me with this, I tried to solve it, but cannot get the solution
Solve the system for x,y,z
2x+3y-z=4
3x-y+2z=5
x-4y+3z=1
Thank you.
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!


This system turns out to be a dependent system which has
infinitely many solutions.  I'll show you below how to
do it on your TI-83 or TI-84 calculator.  If you need
help on how to solve it without a calculator, just say
so in the thank-you note form below and I'll help you
do it by the matrix method without a calculator.




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


Put all those red numbers in a 3x4 augmented matrix
like this:



On your TI-83 or TI-84,

Press CLEAR

Press 2ND
Press x-1 (MATRIX)
Prees the right arrow twice to highlight EDIT
Press ENTER
Press 2
Press ENTER
Press 3
Press ENTER
Type -1, 
Press ENTER
Type 4,
Press ENTER
Type 3,
Press ENTER
Type -1
Press ENTER
Type 2
Press ENTER
...
continue typing in all the rest of the elements 
in the matrix, pressing ENTER after each. Then
...
Press 2ND
Press MODE (QUIT)
Press 2ND
Press x-1 (MATRIX)
Press right arrow to highlight MATH
Press up arrow 5 times to highlight B:rref(
Press ENTER  you should see rref( on the screen
Press 2nd
Press x-1 (MATRIX)
Press ENTER  you should see rref([A]
Type )
Press MATH
Press ENTER
Press Enter
You should see  

rref([A]
    [[1 0 5/11  19/11]
     [0 1 -7/11 2/11 ]
     [0 0       0    ]]

Interpret this as the system:



That's the same as

system%28x%2Bexpr%285%2F11%29z=19%2F11%2C%0D%0Ay-expr%287%2F11%29z=2%2F11%2C%0D%0A0%2Az=0%29

Notice that the third equation 0*z=0 is such that
z can be any number whatever, for any number
substituted for z in 0*z=0 will always give 0=0.
So let a = any number.
Since z = any number, we substitute a for z

system%28x%2Bexpr%285%2F11%29a=19%2F11%2C%0D%0Ay-expr%287%2F11%29a=2%2F11%2C%0D%0A0%2Aa=0%29

Solve the first for x, the second for y,
then write z = a

system%28x=19%2F11-expr%285%2F11%29a%2C%0D%0Ay=2%2F11%2Bexpr%287%2F11%29a%2C%0D%0Az=a%29

So the general solution is



We can get as many solutions as we like by choosing 
different values for the number "a".  For instance,
if we choose a = -5, we have the solution



and if we choose a = 6, we have the solution



Edwin

Answer by AnlytcPhil(1808) About Me  (Show Source):
You can put this solution on YOUR website!
Your solution by calculator is above.  If you need to do
it by the matrix method without a calculator, then tell
me in the thank-you note form below and I'll get back to
you by email.  No charge.

Edwin McCravy