Gaussian elimination

Question

solve using gaussian elimination with matrices 

x+y-3z=4

4x+5y+z=1

2x+3y+7z=-7

Answer 1

X   Y    Z          C

1   1   -3          4

4   5    1           1

2   3    7          -7

 

First, do -4*Row1 + Row2 and  -2*row1 + row3

 

X    Y     Z           C
1    1    -3           4
0    1    13        -15
0    1    13        -15

 

At this point, notice the last two rows are the same. So if you do -Row2 + Row 3 

they cancel.

 

X Y  Z   C
1 1 -3    4
0 1 13 -15
0 0   0   0

 

So the system has infinitely many solutions.

The two remaining equations are:

  X + Y - 3Z =  4

      Y + 13Z = -15

 

Letting Z be the free variable, Y = -13Z - 15

plugging this into the first equation:

  X + -13Z - 15 - 3Z = 4

   X =   13Z + 15 + 3Z + 4

   X = 16Z + 19

 

With one degree of freedom,  the solutions are X = 16Z + 19, Y = -13Z - 15  

 

Check:  let z = 0. Then X=19 and Y = -15. plugging these values into the all three of the equations, they work.

             let z=1. Then X = 35 ad Y = -28.  They also work.

 

 

 

 

 

 

 

 

X    Y     Z           C
1    1    -3           4
0    1    13        -15
0    1    13        -15