00:01
Hello students, here we have been given with a matrix a that is 339 minus 1 -11226 minus 357.
00:19
This will be 4 -1 and 6.
00:23
In the first one now, solving for a, the row echelon form of the matrix is the row, eccolon form of matrix is 3 .00 minus 1 -20 -2 -0 -2 -0 -0 -mines 3 -80 -4 -3 -3 -0.
00:57
The column space is a space spanned by the columns of the initial matrix which corresponds to the pi -watt column of the reduced matrix and thus it can be said that, the column space it is equal to 339 comma minus 1 1 and 1.
01:26
Now for the b part it can be said that we get the we now solve the matrix the reduced row equal on make it to the reduced row equalon form so the reduced rho echelon form, it will be 1 -0 -0 -1 -0, 2 by 3 -0, 1 -3 -4 -0, 5 by 6 minus 3 by 2 and 0.
02:13
So we take x3 is equal to t, x4 is equal to s, x5 as u, then x1 will be minus minus s by 3 minus 2 t by 3 minus 5 u by 6 x2 is minus 4 s plus 3 u divided by 2 so that means we have x bar is equal to minus s by 3 minus 2 t by 3 minus 5 u by 6 minus 4 s plus 3 u by 2 t s this is equal to minus 2 by 3 0 1 0 0 0 0 0 into 2.
02:55
Plus minus 1 by 3 minus 4 010 into s plus minus 5 by 6 3 by 2 001 into you...