00:01
Okay, so here we have a given matrix a is the matrix 112 and 33 and 04.
00:16
Looking at this inspection, we see that the column vector 5222 is going to be redundant, and then we get the basis for the image of a is going to be the vectors 1 -0 -0, and then 1 -1 -1, and then 1 -1, and then one, two, four.
00:45
And then to find a basis for the kernel of a, we put a into reduced row echelon form.
00:51
So we're going to have 1 ,0, 0, 0, 0, 0, 0, 0, 0, 0, 0, times the common vector x1, x2...