This will likely have infinitely many solutions,
since there are more letters, variables, than equations.
2w - 3x + 4y + z = 7
w - x + 3y - 5z = 10
3w + x - 2y - 2z = 6
Swap rows 1 and 2, to get a 1 in the
upper left corner:
abbreviated R1<->R2
To get a 0 where the 2 is, multiply row 1
by -2 and add it to row 2:
abbreviated -2R1+R2->R2
To get a 0 where the bottom left 3 is,
multiply row 1 by -3 and add it to row 3:
abbreviated -2R1+R2->R2
To get a 1 where the -1 is on Row 2,
multiply row 2 by -1
abbreviated -R2->R2
To get a 0 where the 4 is,
multiply row 2 by -4 and add it to row 3:
abbreviated -4R2+R3->R3
To get a 1 where the -19 is on Row 3,
multiply row 3 by -1/19
Then we change the matrix back to equations:
w - x + 3y - 5z = 10
x + 2y - 11z = 13
y - 3z = 4
Solve each equation for the first letter
w = 10 + x - 3y + 5z
x = 13 - 2y + 11z
y = 4 + 3z
Now we do back substitution:
Sunstitute the expression for y in the middle equation:
x = 13 - 2(4 + 3z) + 11z
x = 13 - 8 - 6z + 11z
x = 5 + 5z
Substitute the expressions for x and y in the top equation:
w = 10 + x - 3y + 5z
w = 10 + (5 + 5z) - 3(4 + 3z) + 5z
w = 10 + 5 + 5z - 12 - 9z + 5z
w = 3 + z
So the solution is
(w,x,y,z) = (3+z, 5+5z, 4+3z, z)
Some teachers will tell you to use a different letter than z for z,
such as "a", or "k". If they use "a" the solution would be:
(w,x,y,z) = (3+a, 5+5a, 4+3a, a)
Edwin
