Now let’s get 0 at the extreme left in the second row. If we multiply the second row by −3/5,
we come up with
3 − 2 111
− 3 12 6 0(^0) − 19 − 13 − 20
Adding the first and second rows and then replacing the second row with the sum, we have
(^3) − 2 111
010711
(^0) − 19 − 13 − 20
Now the number −19 in the third row must somehow be made to vanish, which means we
must turn it into 0. Let’s use the second row to “attack” it. If we multiply the second row
through by 19 and the third row through by 10 (combining two moves), we get
3 − 2 111
0 190 133 209
0 − 190 − 130 − 200
Adding the second and third rows and then replacing the third row with the sum, we obtain
a matrix in echelon form:
3 − 2 111
0 190 133 209
0039
Deriving the diagonal form
To morph the echelon matrix into diagonal form, we simply keep playing the game. In this
case, we want to get 0s in the places that now contain −2, 1, and 133. Let’s start with the third
element in the first row, currently equal to 1. We can use the third row to “attack” it. Let’s
divide the third row by −3 to get
3 − 2 111
0 190 133 209
00 − 1 − 3
Adding the first and third rows and then replacing the first row with the sum, we get
(^3) − 2 08
0 190 133 209
(^00) − 1 − 3
A Sample Problem 303