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Unformatted text preview: QUIZ 1 ANSWERS 10
1. 4 1 1
2 −1 b1
= b2 .
3 1 0 b1
b2 − 4b1 .
a). (12 points) We reduce: 4 1 b2 → 0 1 b2 − 4b1 → 0 1
2 −1 b3
0 −1 b3 − 2b1
0 0 b3 + b2 − 6b1
So equation becomes:
= b2 − 4b1 .
b3 + b2 − 6b1
b) (6 points) Only when b3 + b2 − 6b1 = 0.
2. a) (8 points) Solutions to A are length 3 column vectors, so A has three columns.
b) (8 points) Any number. In fact consider 1 −c −d . This kills both of the given
vectors. Then, feel free to add any number of rows of zero’s below it.
c) (8 points) We assume that the matrix is not zero as we are given that these are the
only special solutions. To ﬁnd the rank, we note that each row of A must be in the subspace
of R3 which is orthogonal to c 1 0 and d 0 1 , but since these guys span a plane,
this subspace is a line. Thus the rows are all linearly dependant, and so the rank is one.
3. a) (20 points) False. To get A to its reduced form, you need to multiply on the left by
some elimination matrix E ; so Rx = Eb is correct, but there is no reason for Eb − b to be
in the nullspace of A(you can use your favorite non-triangular invertible 2 by 2 matrix to
ﬁnd a counterexample).
b) (10 points) True. It is certainly the case that E 0 = 0.
4. a) (10 points) Permutation.
b) (18 points) Well, Pe = e for any permutation matrix, because each row has eight
zero’s and one one. So by part a), Ae = e + 2e + ... + 9e = 45e. But since the transpose of a
permutation matrix is also a permutation matrix, At e = 45e also. But rank e 45e 45e =
1. 1 ...
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This note was uploaded on 06/21/2011 for the course CIVIL 1011 taught by Professor Juan during the Spring '11 term at HKU.
- Spring '11