I can see no difficult in reducing it- it seems easier than most of this kind.
Start by subtracting the first row from the second row and subtracting three times the first row from the third row.
Can you finish that?
Hello,
I need some help in this problem....
A = [1 -2 -1, 1 -3 0, 3 -8 -1] and let linear transformation T: R^3--> R^3 be defined by T(x) = Ax, where x = [ x1 x2 x3].
a) Determine rank(T) and nullity (T)
b) Determine bases for the range R(T) and the null space N(T).
(If N(T) = {[0 0 0]} state that N(T) has no basis)...
I know we have to use the Gauss Jordan to figure that matrix out but I am unable to reduce it properly, perhaps there is another way of doing it??
please help!
thanks!
thanks for the response Sir. When I started simplifying using Gauss Jordan I ended up getting---
x = 2(y) + z
z = y...If I consider 1st column as x, 2nd as y and 3rd column as z and then set three of these equations equal to 0...
So the N(A), null space of A = Span {z [3 1 1] }
but then this is what I get when I simplify A, then how do I implement it on T(x) = Ax...and then find the rank and null space of T?...
Halls, your first row is incorrect. a13 should be -1.
vedicmath, how many non-zero rows did you end up with in rref(A)? that is the rank of A, so any two columns of A would be a basis (it is customary to choose the rows of A corresponding to the pivot columns of rref(A)).
what is the nullity of A, by the rank-nullity theorem? you have a spanning set, does it have the same number of elements as the dimension of the null space (that is one criterion for a basis)?