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