Find all the vectors u that satisfy the equation <1,1,1> X u = <-1,-1,2>
(The answer in the back of the book is <u3-1, u3+1, u3> I just don't know how to get to it!)
Yes that is equivalent (and generally how you do the calculation).
So the determinant with u at the bottom and <1,1,1> in the middle gives
i*(1*u3 - 1*u2) - j*(1*u3 - 1*u1) + k*(1*u2 - 1*u1) = <-1,-1,2>
Now equating components gives you:
u3 - u2 = -1
u1 - u3 = -1
u2 - u1 = 2.
Row reducing this gives:
>> A = [0, -1, 1, -1; 1, 0, -1, -1; -1, 1, 0, 2]
A =
0 -1 1 -1
1 0 -1 -1
-1 1 0 2
>> rref(A)
ans =
1 0 -1 -1
0 1 -1 1
0 0 0 0
So this means that u2 - u3 = 1 and u1 - u3 = -1. So we get
u1 = u3 - 1
u2 = u3 + 1
So plug those in to the following:
<u3 - 1, u3 + 1, u3> = U.