MATH 2574H Lecture : Discussion 3273014.pdf
Document Summary
W = set of (x1, x2) w/ x1. Then x1 (cx1) 2 + (cx2) 2 = c2(x1 + x2) = c2 (1, 0) in w (2, 0) not in w. 2 = 1 adj(a) = [aij]t (cofactor matrix transpose) To span rn, don"t have to be linearly independent, but to be a basis, they must be linearly independent ex: also spans r2; Another ex: 3 linearly indpendent vectors spanning r 3 is a basis for r3. If determinant doesn"t equal 0, then the vectors = linearly independent b/c if determinant doesn"t equal 0, then a = invertible then. Thus, the only solution here would be c1= c2 = c3 = 0 --> linearly independent. Ax = 0 has only the trivial solution. Ax = b has only 1 solution: a-1b. A matrix b exists such that ab = i (we know that if ab = i then ba = i)
