Systems and Matrices - Beginning Pre-Calculus for Game Developers - page 264

is 5. Accordingly, your work proceeds in this way:

4
3

1
3

:

1 First row

20
3

5
3

5 First row multiplied by 5 :

You then add the first row as transformed to the second row:

5 2 19 Second row :

20
3

5
3

5 Add the new first row :

26
3

52
3

0 Result of the addition :

Your matrix now assumes this form:

4
3

1
3

1

52
3
At this point, you evaluate the second row to discover a way to reduce the value in
the second column to 1. To accomplish this, you can multiply the elements in the
row by 26 , the multiplicative inverse of 2 3 . Here is the form this activity assumes:

26
3

0

26
3

52
3

0 The original row :

01 156
78

3
26 :

Multiply by

01 2 ð 78 Þ
ð 78 Þ

Factor :

0 1 2 Eliminate the fractions :

You now have a new matrix, which appears this way:

4
3

1
3

1

0 1 2