00:01
We have been given matrix a equal to 1, 3, 0, 3, minus 1, minus 1, minus 1, 1, 0, minus 4, 2, minus 8, 2, 0, 3, minus 1.
00:17
Now, we have to find how many rows of a contains a pivot position and does the equation ax equal to b have a solution for each b in r4.
00:51
Now, we can write a as 1, 3, 0, 3, minus 1, minus 1, minus 1, 1, 0, minus 4, 2, minus 8, 2, 0, 3, minus 1.
01:05
This is similar to matrix 1, 3, 0, 3, 0, 2, minus 1, 4, 0, minus 4, 2, minus 8, 0, minus 6, 3, minus 7.
01:24
By applying the elementary row operation r2, r2 plus r1 and r4 to r4 minus 2r1 which is similar to 1, 3, 0, 3, 0, 1, minus 1, 2, 0, minus 4, 2, minus 8, 0, minus 6, 3, minus 7.
01:52
When r2 is 1 by 2r2, this is similar to 1, 0, 3 upon 2, minus 3, 0, 1, minus 1 by 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 5.
02:14
When r1 is r1 minus 3r2, r3 is r3 plus 4r2 and r4 is r4 plus 6r2.
02:29
This is similar to 1, 0, 3 upon 2, minus 3, 0, 1, minus 1 upon 2, 2, 0, 0, 0, 5, 0, 0, 0, 0...