00:05
Chapter 3, section 1, question 12.
00:11
The first thing i'm going to do is i'm going to subtract row 2 from row 1, where i'll have 1 ,02, 012, 0100, 0 ,0, where v sub 1 equals negative 2 v .3, v sub 2, v .2 equals negative 2 v .3, and v.
00:44
Sub 3 is a free variable.
00:49
So then we have null a equal to t negative 2, negative 2, 1 is an element of all real numbers.
01:10
And a basis for null space a is negative 2, negative 2, 1.
01:31
So since it's rank equals 2, then the vectors are 1 -0 -2 -012, negative 2, negative 2, 1.
02:08
This forms a basis for 3 space.
02:22
All right, so now we have 1 -0 -2 -0 -1 -2 -2 -1 -3 -3...