18.06, Fall 2004, Problem Set 10 Solutions
1. (13 pts.)
Consider the dierential equation
du
dt
= Au where u is 2dimensional and
0 1
1 0
A=
.
(a) The characteristic polynomial of A is
det
1
1
= 2 + 1,
and therefore the eigenvalues are 1 = i and 2 = i. T
18.06, Fall 2004, Problem Set 9 Solutions
1. (5 pts) P is symmetric since P = A(A T A)1 AT (where we have kept only linearly independent
columns of A if A did not have full column rank to start with). In lecture (and in the book),
we saw that symmetric ma
18.06, Fall 2004, Problem Set 8 Solutions
1. (7 pts.) Let
0 1 2
A = 1 0 1 .
4 1 0
(a) Cij is given by (1)i+j det(Mij ) where Mij
column j.
1
C= 2
1
is obtained from A by removing row i and
4 1
8 4 .
2 1
(b) Using the cofactor formula, say along row 1, we
18.06, Fall 2004, Problem Set 1 Solutions
1. (6 pts.)
(a) False. Counterexample: u = (1, 0, 0), v = (0, 1, 0) and w = (0, 0, 1).
(b) True. u (v + 2w) = u v + 2u w = 0 + 0 = 0.
(c) True. u v2 = (u v) (u v) = u u 2u v + v v = 1 + 0 + 1 = 2.
2. (6 pts.)
(A
18.06, Fall 2004, Problem Set 6 Solutions
1. (12 pts.)
(a) F is N (A) where
A = [ 1 2 1 1 ].
To nd a basis of F , we can just take the special solutions for N (A) (the free variables
are x2 , x3 and x4 ):
a1 =
2
1
0
0
1
0
1
0
, a2 =
, a3 =
1
0
0
1
.
18.06, Fall 2004, Problem Set 7 Solutions
1. (11 pts.)
(a) Let F be the subspace of all 2 2 matrices of the form:
a b
c d
,
with a + b + c + d = 0. It is clear that a1 , a2 , a3 and b1 , b2 , b3 are in F .
We claim that any element in F can be expressed a
18.06  Final Exam, Monday May 16th, 2005
solutions
1. (a) We want the coordinates (ai1 , . . . , ain ) of Pi to satisfy the equation c1 x1 + . . . + cn xn = 1.
Thus the system of equations is Ac = ones:
c1 a11 + c2 a12 + . . . + cn a1n = 1
c1 a21 + c2 a2
18.06, Fall 2004, Problem Set 3 Solutions
1. (6 pts.)
(a) No. The set F
0
1 0 =
1
0
is not closed under scalar multiplication. For example, 0 is in F but
1
0
0 is not.
1
(b) No. For a counterexample, consider f (x) = x 2 +x; then f is in our set but 2
Exam 3, Friday May 4th, 2005
Solutions
Question 1. (a) The characteristic polynomial of the matrix A is
3
9
1
1
+ 2 +
= ( 1)
2
16
16
4
2
3
and thus the eigenvalues of A are 1 with multiplicity one and 1 with multiplicity two. Clearly
4
the eigenvectors
Exam 2, Friday April 1st, 2005
Solutions
Question 1. The vector a1 can be any nonzero positive multiple of q1 . The vector a2
can be any multiple of q1 plus any nonzero positive multiple of q2 :
a1 = cq1
a 2 = c 1 q1 + c 2 q2
, with c, c1 > 0.
Question
18.06  Spring 2005  Problem Set 2
Solution to the Challenge Problem
Denote by Cij the n n matrix having a 1 in the (i, j) entry, and 0s everywhere else. In terms of these matrices we may write Eij = I ij Cij and
1
Eij = I + ij Cij . Thus we have
1
M Eij
18.06  Spring 2005  Problem Set 8
Solution to the Challenge Problem
Challenge Problem: Consider the 3 3 matrix
a b c
A = 1 d e
0 1 f
Determine the entries a, b, c, d, e, f so that:
the top left 1 1 block is a matrix with eigenvalue 2;
the top left 2
18.06  Spring 2005  Problem Set 3
Solutions to the Challenge Problems
Problem 1
a) The column space is the space of all vectors whose last m r coordinates
are zero. This is clear since the rank of the matrix R is r and the rst r
columns of R are indepen
Austin Ford
[email protected]
February 11, 2004
18.06 Linear Algebra
Professor Gilbert Strang
Recitation 8
Bianca Santoro
Problem Set 1
Problem
Find the possible failures in the column picture and the row picture, and
match them up. Success would be 3 columns
18.06 Spring 2006  Problem Set 9
SOLUTIONS TO SELECTED PROBLEMS
1. Section 6.6, Problem 4
Answer: A has no repeated eigenvalues so it can be diagonalized: A = SS 1
makes it similar to .
2. Section 6.6, Problem 5
Answer:
1 1 0 0 1 0 0 1
,
,
,
are
18.06 Spring 2006  Problem Set 7
SOLUTIONS TO SELECTED PROBLEMS
1. Section 6.1, Problem 2
Answer: A has 1 = 1 and 2 = 5 with eigenvectors x1 = (2, 1) and
x2 = (1, 1). The matrix A + I has the same eigenvectors with eigenvalues
increased by 1: 1 = 0 and 2
18.06 Spring 2006  Problem Set 8
SOLUTIONS TO SELECTED PROBLEMS
1. Section 6.3, Problem 3
Answer:
y 0 1 y
1
. Then = 2 (5 41).
=
y
4 5
y
2. Section 6.3, Problem 23
Answer: A2 = A so A3 = A.
2
2
t
t
1 0 t 3t t 3t
e 3(e 1)
eAt =
+
+
+ . =
0 1
18.06 Spring 2006  Problem Set 6
SOLUTIONS TO SELECTED PROBLEMS
1. Section 5.1, Problem 8
Answer: There are 5! = 120 permutation matrices. 5!/2 = 60 have det= +1.
A permutation matrix that needs four exchanges to reach the identity matrix:
0 0 0 0 1
1 0
18.06 Spring 2006  Problem Set 5
SOLUTIONS TO SELECTED PROBLEMS
1. Section 4.3, Problem 12
Answer:
a) aT a = m, aT b = b1 + . . . + bm . Therefore aT a = m = b1 + . . . + bm and
x
x
x is the mean of the bs.
m
b) e = b xa, e
2
(bi x)2 .
=
i=1
1 1 1
c) p
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