Yes, so AX = 0
where X = ( x1; x2; x3) and A = (1 0 2; 0 3 -3; 4 2 6)
which can be expressed as ( -2; 1; 1)*x3 after row reduction and back substitution.
I need to know whether I can leave it at that, or express it as the line: -x = 2y = 2z
Hi there,
just a pretty straight forward query I need cleared up...
If a question asks for the null space of A in cartesian form how do I set it out?
This is what I've got:
X = ( -2; 1; 1)*x3 for all values of x3
Therefore, Null(A) corresponds geometrically to the line through the...