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Math 2270 Quiz 10 1. Consider a consistent linear system A~x = ~b. As part of your homework, you showed that this system has a solution ~x0 ∈ ker(A)⊥ . Show that there is only one solution in ker(A)⊥ . Suppose that ~x0 , ~x1 ∈ ker(A)⊥ satisfy A~x0 = A~x1 = ~b. Then A(~x0 − ~x1 ) = A~x0 − A~x1 = ~b − ~b = ~0, so ~x0 − ~x1 ∈ ker(A). But ~x0 − ~x1 ∈ ker(A)⊥ as well, since it is a subspace. We also know that ker(A) ∩ ker(A)⊥ = {~0}, so ~x0 − ~x1 = 0. 2. Find the determinant of the matrix 1 1 A= 2 2 3 3 1 2 . 3 Since the rows of this matrix are all constant multiples of one another, we can see that det(A) = 0 from inspection. However, we can calculate it explicitly to see that 2 2 2 2 2 2 det(A) = det − det + det = (6 − 6) − (6 − 6) + (6 − 6) = 0. 3 3 3 3 3 3