Algebra Know-It-ALL

We need to turn the 9,152 in the first row into 0. We’ll have to work with the third row

to make it happen. Suppose 9,152 divides cleanly by 143? Let’s give it a try. A calculator

tells us that 9,152 / 143 = 64. Let’s multiply the third row by −64 to get the matrix

5,720 0 9,152 −1,976

0 1,716 0 −3,120

00 −9,152 48,256

Adding the first row to the third and then replacing the first row with the sum, we have

5,720 0 0 46,280

0 1,716 (^0) −3,120
(^00) −9,152 48,256
This matrix is unwieldy, but it’s in diagonal form.

6. We want to get the absolute values of the numbers in solution to Prob. 5 as small as
   possible and still have all integers. That means we must find largest common divisors for
   each row. This process is something like reducing fractions to lowest form. Let’s divide
   the first row by 520, the second row by 156, and the third row by −832. That gives us

11 0 0 89

0110 − 20

(^0011) − 58
That’s as far as we can reduce the matrix, but it’s quite an improvement! This is not a coin-
cidence, even though it may appear that way at first. In the matrix morphing process, we
did a lot of multiplying. Those multiples “went along for the ride,” inflating the numbers.
They have served their purpose. It’s good to be rid of them.

7. To reduce the matrix in solution to Prob. 6 to unit diagonal form, we divide each row
   by 11. That gives us

1 0 0 89/11

010 −20/11

001 −58/11

The solution to the original linear system, stated at the end of solution to Prob. 3, is

apparently

x= 89/11

y=−20/11

z=−58/11

Chapter 19 659