Stating and reducing the solution
The solution to the system is now clear. We have no fractions to reduce, because we kept
the sizes of the numbers down as we went along. (Otherwise we’d have fractions with large
numerators and denominators, although they would divide out cleanly and leave us with
integers.) Evidently,
x= 2
y=− 1
z= 3
Are you confused?
If you think we’ve finished solving this problem, you’re indeed confused. We must check our work! Only
then can we be totally confident that our solution is correct, because it’s easy to make mistakes in the
matrix morphing game. When we plug the values for x,y, and z into the first original equation, here’s
what happens:
3 x+z= 2 y+ 11
3 × 2 + 3 = 2 × (−1)+ 11
6 + 3 =− 2 + 11
9 = 9
Check one. Now for the second equation:
4 y+ 2 z=x
4 × (−1)+ 2 × 3 = 2
− 4 + 6 = 2
2 = 2
Check two. Finally, the third equation:
− 5 x+y= 3 z− 20
− 5 × 2 + (−1)= 3 × 3 − 20
− 10 − 1 = 9 − 20
− 11 =− 11
Check three. Mission accomplished.
Here’s an extra-credit challenge!
Play the matrix morphing game with the linear system we just got done solving, but take a different route
this time. Don’t reduce the sizes of any of the integers. Let the absolute values grow as large as they “want”!
A Sample Problem 305