00:01
Hello, so here we look at our given matrix a, the matrix here, one, two, one, two, two, three.
00:10
Okay, looking at this by the inspection, we see the second column is going to be two times the first, and therefore the column then, two, two, two is going to be redundant, and then a basis for the image of a is going to be one, one, one, one, and one, two, three.
00:38
Okay, and then to find a basis for the kernel of a, we first need to put a in reduced row echelon form, and set ax equal to zero.
00:48
So the row reduced echelon form of our matrix is going to be 1 ,2 ,0, and then 0 ,0, and then the last row of all 0 ,000, so now we solve the system here, 1 -2001 -0 -0 -0 times the column vector x1, x2, x3 is then going to imply that x1 plus 2 x2 equal to 0, which then gives us that x -1 is going to be equal to negative to x2, and we get that x -3 is just equal to 0...