SOLUTION: solve the system of linear equations x+y+2z+3w=13, x-2y+z+w=8, 3x+y+z-w=1, a)find all solutions if any exist, by using gaussian elemination method. b)find all solutions

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: solve the system of linear equations x+y+2z+3w=13, x-2y+z+w=8, 3x+y+z-w=1, a)find all solutions if any exist, by using gaussian elemination method. b)find all solutions      Log On


   



Question 1158779: solve the system of linear equations x+y+2z+3w=13, x-2y+z+w=8, 3x+y+z-w=1, a)find all solutions if any exist, by using gaussian elemination method. b)find all solutions if any exist, by using gauss_jordan method.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve the system of linear equations x+y+2z+3w=13, x-2y+z+w=8, 3x+y+z-w=1, a)find all solutions if any exist, by using gaussian elemination method. b)find all solutions if any exist, by using gauss_jordan method.
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4 variables but only 3 equations.
No unique solution.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!


Put in all the understood 1 coefficients:



erase the letters and put a vertical line where the
equal signs are:



Multiply -1 times the first row and add it to 1 times the
second row to get a 0 where the leftmost 1 is on the 2nd
row.  It makes it easier if you put what you're multiplying
the rows by (even when it's 1) on the left.  The symbol for 
the row operation is -1R1+1R2->R2

  matrix%283%2C1%2C-1%2A%22%22%2C1%2A%22%22%2C%22%22%29

gives:



Multiply -3 times the first row and add it to 1 times the
third row to get a 0 where the leftmost 3 is on the 3rd
row.  Again, it makes it easier if you put what you're 
multiplying the rows by (even when it's 1) on the left.  
The symbol for the row
operation is -3R1+1R3->R3

matrix%283%2C1%2C-3%2A%22%22%2C%22%22%2C1%2A%22%22%29

gives



Multiply -2 times the second row and add it to 3 times the
third row to get a 0 where the leftmost -2 is on the 3rd
row.  Again, it makes it easier if you put what you're 
multiplying the rows by on the left. The symbol for the row
operation is -2R2+3R3->R3

matrix%283%2C1%2C%22%22%2C-2%2A%22%22%2C3%2A%22%22%29

gives



Now to make things easier, we can multiply the 3rd row by -1/13.
The symbol for the row operation is (-1/13)R3->R3

matrix%283%2C1%2C%22%22%2C%22%22%2C%22-1%2F13%22%2A%22%22%29

gives



Multiply 1 times the third row and add it to 1 times the
third row to get a 0 where the leftmost -1 is on the 2nd
row.  Again, it makes it easier if you put what you're 
multiplying the rows by on the left. The symbol for the row
operation is 1R3+1R2->R2

matrix%283%2C1%2C%22%22%2C1%2A%22%22%2C1%2A%22%22%29

gives



Now to make things easier, we can multiply the 2nd row by -1/3.
The symbol for the row operation is (-1/3)R2->R2

matrix%283%2C1%2C%22%22%2C%22%22%2C%22-1%2F13%22%2A%22%22%29


Make that back into a system of equations:



or



Solve the 3rd equation for z

matrix%281%2C3%2Cz%2C%22%22=%22%22%2C8-2w%29

"Solve" the second equation for y.  It's already solved
for y.

matrix%281%2C3%2Cy%2C%22%22=%22%22%2C-1%29


Solve the 1st equation for x

matrix%281%2C3%2Cx%2C%22%22=%22%22%2C13-y-2z-3w%29

Substitute -1 for y and 8-2w for z




Solution



Edwin