18.06 Quiz 1
Professor Strang
February 28, 2011
Your PRINTED name is
1.
Your Recitation Instructor (and time) is
2.
Instructors: (Pires)(Hezari)(Sheridan)(Yoo)
3.
1. (a) By elimination nd the rank of A and the pivot columns of A (in its column space):
1 2
18.06 Problem Set 1 Solutions
1. Find a solution for x,y,z to the system of equations
2
3e + + 2
1 2 3
x
4 5 6 y = 6e + 4 + 5 2
7 8 10
z
10 e + 7 + 8 2
Solution x = , y =
2 and z = e is a solution.
2. Do problem 11 from section 2.1.
Solution by rows:
18.06 Problem Set 2 Solutions
1. Do problem 5 from section 2.6.
Solution Doing the elimination process, with matrices:
2
A= 0
6
1
4
3
0
1
2 E31 A = 0
5
3
0 0
2
1 0 A = 0
0 1
0
Thus in this case
E = E31
1
= 0
3
0 0
1 0
0 1
With
L = E 1
1
= 0
3
1
A = LU =
Solutions to 18.06 Problem Set 4
1. The matrix is already given in LU form:
1 0 0
0 1 2 3 4
1 1 0 0 0 0 1 2
0 1 1
0 0 0 0 0
The row space and null space for A and for U are the same, so we do it for U. From U,
we can read o the row space as the span of (0
1
18.06 Solutions to PSet 4
3.5:
16: These bases are not unique!
(a) (1, 1, 1, 1) for the space of all constant vectors
(c, c, c, c)
(b) (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1) for the space of vectors with sum
of components = 0
(c) (1, 1, 1, 0), (1, 1,
1
18.06 Solutions to PSet 3
3.1.15 (a) Two planes through (0, 0, 0) probably intersect in a line through (0, 0, 0)
(b) The plane and line probably intersect in the point (0, 0, 0)
(c) If x and y are in both S and T , x + y and cx are in both subspaces.
3.
18.06 F10
Problem Set 5 Solutions
Due Thursday, 14 Oct.
1. Problem 14, section 8.2, p.430.
Solution: Suppose AT CAx = 0. Set y:= CAx. Then |y|2 = xT ( CA)T
( CA)x = xT AT CAx = 0. So y = 0. We assume no diagonal entry of C is
0. Hence Ax = 0. Conversely,
1
18.06 Solutions to PSet 5
4.2:
11: (a) p =
A(AT A)1 AT b = (2, 3, 0), e = (0, 0, 4), AT e = 0 (b) p = (4, 4, 6), e = 0.
1 0 0
12: P1 = 0 1 0 = projection matrix onto the column space of A (the xy plane)
0 0 0
0.5 0.5 0
Projection matrix onto the second
18.06 Problem Set 3 Solutions
1. Find the LU and the reduced row echelon matrix form for
2 4 6 8
A = 4 11 15 24
2 10 12 28
4
14 .
24
Compute the column space C(A) and the null space N (A) for A. Give all solutions for the system
8
Ax = 25 .
34
Solution Th
1
18.06 Solutions to PSet 6
5.1:
3: (a) False: det(I + I) is not 1 + 1
(b) True: The product rule extends to ABC (use
0 0
0 1
it twice) (c) False: det(4A) is 4 det A (d) False: A =
,B =
,
0 1
1 0
0 1
AB BA =
is invertible.
1
0
15: The rst determinant is 0
Fall 2010 18.06 Problem Set 7 Solutions
(1) (a) False, take A = I (2 2 matrices). Then det(I + A) = det(0) = 0 and 1 + det(A) = 2.
(b) True, det(ABC) = det(AB) det(C) = det(A) det(B) det(C).
(c) False, in general det(4A) = 4n det(A) if A is n n. For an ex
18.06 F10
Problem Set 6 Solutions
Due Thursday, 21 Oct.
1. Problem 5, section 4.3, p.226.
Solution: In matrix form, the unsolvable equations
become A = b with A = [1; 1; 1; 1] and b =
x
T
20
e4
T
[0; 8; 8; 20]. So A A = A b is 4C = 36 . Thus
x
the best he
1
18.06 Solutions to PSet 7
8.3:
4
1 1
3: = 1 and .8, x = (1, 0); 1 and .8, x = ( 5 , 9 ); 1, 1 , and 1 , x = ( 3 , 3 , 1 ).
9
4
4
3
T
4: A always has the eigenvector (1, 1, . . . , 1) for = 1, because each row of AT adds to
1. (Note again that many autho
1
18.06 Solutions to PSet 8
6.4:
2
1
2
5: Q =
3
1
1
7: (a) (a)
2
1
2
The columns of Q are unit eigenvectors of A
2 1 .
Each unit eigenvector could be multiplied by 1
2
2
2
has = 1 and 3 (b) The pivots have the same signs as the s
1
1 2
(c) trace = 1 + 2 =
1
18.06 Solutions to PSet 9
6.7:
3: If A has rank 1 then so does AT A. The only nonzero eigenvalue of AT A is its trace,
which is the sum of all a2 . (Each diagonal entry of AT A is the sum of a2 down one
ij
ij
column, so the trace is the sum down all col
18.06 Problem Set 10 Solutions
1. Do problem 5 from section 6.5.
Solution f = (x + 3y)(x + y) = (x + 2y + y)(x + 2y y) = (x + 2y)2 y 2 . There are many
points where this is negative, say (1, 2), where the above is 02 22 = 4.
This goes to show that not eve
1
18.06 Solutions to PSet 1
1.1.12 A four-dimensional cube has 24 = 16 corners and 2 4 = 8 three-dimensional faces
and 24 two-dimensional faces and 32 edges in Worked Example 2.4 A.
1.1.26 Two equations come from the two components: c + 3d = 14 and 2c + d
18.06 Pset 9 Solutions
Problem #25: With positive pivots in D, the factorization A = LDLT
6.5,
becomes L D DLT . (Square roots of the pivots give D = D D.) Then C =
DLT yields the Cholesky factorization A = C T C which is symmetrized LU .
From
C=
3 1
0
18.06 Quiz 2
Professor Strang
April 6, 2011
Your PRINTED name is
1.
Your Recitation Instructor (and time) is
2.
Instructors: (Pires)(Hezari)(Sheridan)(Yoo)
3.
Please show enough work so we can see your method and give due credit.
1. (8 pts. each) Suppose
1 (30 pts.)
(a) (25 pts.) Compute the determinant (as a function of
x
x
A=
x
x
(Note that all the entries
x
x)
44
of the
x x x
matrix
x 0 0
0 x x
0 x 1
in the matrix represent the same number.)
There are several ways to do this. One way is to use the cofa
18.06 Problem Set 2 Solution
Due Thursday, 18 February 2010 at 4 pm in 2-106.
Total: 100 points
Section 2.5. Problem 24: Use Gauss-Jordan elimination on [U I] to
U 1 :
1 a b
1 0
0 1 c x1 x2 x3 = 0 1
U U 1 = I
0 0 1
0 0
Solution (4 points): Row
1 a b
1 0
18.06 Problem Set 6 Solutions
Due Thursday, 18 March 2010 at 4 pm in 2-108.
Total: 100 points
Section 4.3. Problem 4: Write down E = Ax b 2 as a sum of four squares
the last one is (C + 4D 20)2 . Find the derivative equations E/C = 0 and
E/D = 0. Divide b
18.06 PSET 8 SOLUTIONS
APRIL 15, 2010
Problem 1. (6.3, #14) The matrix in this question is skew-symmetric (AT = A):
u = cu2 bu3
1
0
c b
du
a u
= c 0
or u = au3 cu1
2
dt
b a 0
u = bu1 au2
3
(a) The derivative of u(t) 2 = u2 + u2 + u3 is 2u1 u + 2u2 u + 2u
18.06 Problem Set 10 Solution
Due Thursday, 29 April 2009 at 4 pm in 2-106.
Total: 100 points
Section 6.6. Problem 12. These Jordan matrices have eigenvalues 0, 0, 0, 0. They
have two eigenvectors (one from each block). But the block sizes dont match and
18.06 Fall 2010 Exam 1 solutions
1 (30 pts.)
Start with the 3 by 4 matrix:
0 0 0 0
A = 0 1 2 3 .
0 2 4 6
(a) (10 pts.) What are all the special solutions to Ax = 0, and describe the nullspace A.
of
0 1 2 3
0 0 0 0. The free
Solution: We can do row operati
18.06 Fall 2010 Exam 3 Solutions
1 0
(1) (a)
0 1
(b) No example exists. There are many ways to see this.
First way: Negative denite means that all upper left submatrices have negative determinant.
In particular, the (1, 1) entry needs to be negative, but