Pretty much complete, with one
or two exceptions where the computations were
too messy. Make sure you
have the most recent version, as I am correcting
various mistakes/typos over
time.
Same deal as before: please email
me with any typos/errors you find!
I am again writing all vectors horizontally, since it's easier to type.
1)
a) rank = 3
b) nullity = 5-3 = 2
c) Basis = { (1 -1 0 1), (0 1 2 1), (3 -3 1 4) }
d) Another basis = {(1 0 0 0), (0 1 0 -1), (0 0 1 1)} (non-zero
rows of transpose)
e) Basis = {1 0 -2 1 3), (-1 1 5 -1 -3), (0 2 6 0 1)}
f) Another basis = {(1 0 -2 1 0), (0 1 3 0 0), (0 0 0 0 1) }
g) Basis = { (2 -3 1 0 0), (-1 0 0 1 0)}
h) Same as g), the basis of the nullspace Why?? Convince yourseves!
2)
a) rank = 2
b) nullity = 2
c) Basis = { (1 2 -1 0 3), (1 1 0 1 2)}
d) Another basis = { (1 0 1 2 1), (0 1 -1 -1 1)}
e) Basis = {(1 2 1 -1),(2 4 1 -4)}
f) Another basis = {(1 2 0 -3), (0 0 1 2)}
g) Basis = {(-2 1 0 0), (3 0 -2 1)}
h) Same as g)
3)
b - c -a = 0
4)
a) no solution if a = sqrt(2) or -sqrt(2)
b) unique solution for all other values of a.
c) it can never have infinitely many solutions.
5) L(v) = Av, where:
A =
-3 2 -1
1 -2
2
2 1
4
6) The quick way is this: notice that (1 1 -1) = -(-1 -1 1), so
L((1 1 -1)) = (1, -2, -4)
7) L(v) = Av, where:
A =
2 0
-1
1 2
1
3 -1
0
8)
3x+4y-6z -3 = 0 (same normal, since the planes are parallel)
9) Solution #1:
The plane passes through (3,-1,2), and the normal to the plane should
be perpendicular
to both lines. So if n=(a,b,c), then:
2a -3b +4c = 0
2a +b -c = 0
normal = (-1 10 8) (or any
multiple of this)
Eqn of the plane: -x + 10y +8z -3 = 0.
Solution #2:
Simply pick three points on the two lines, and write the equation of
the plane
passing through 3 points (use the determinant form).
10) Form the matrix containing the 4 vectors:
A =
1 -1
2 6
1 1
3 6
0 2
1 0
1 0
-1 -1
Then:
>> rref(A)
ans =
1 0
0 1
0 1
0 -1
0 0
1 2
0 0
0 0
So v = v1-v2+2v3
11)
Linearly independent for any value except 2.
12),13) ---> check yourself with Matlab!
14)
Any diagonal matrix similar to A will have the eigenvalues on the diagonal,
so:
D1 =
2 0
0 6
or:
D2 =
6 0
0 2
15) A has eigenvalues -4 (repeated twice), and 8.
D =
-4 0 0
0 -4 0
0 0 8
16) The directions are given by:
v1 = (2 -1 -2)
v2 = (2 -1 2)
cos(alpha) = 1/sqrt(6)
17) Note first that only 3 out of the 4 vectors are linearly, independent,
the first 3, since:
A =
1 2
3 1
1 -1 -3
-2
0 0
0 0
0 1
-2 -3
>> rref(A) =
1 0
0 0
0 1
0 -1
0 0
1 1
0 0
0 0
Then apply Gram-Schmidt for just the first three vectors -- the results
are not particularly
clean and I'll leave it at that. You can simplify your life somewhat
by noting that
the first and third vectors are perpendicular, so you can take:
v1 = (1 1 0 0)
v2 = (3 -3 0 -2)
v3 <--- compute with Gram-Schmidt, then divide all vectors by their
norms.
18)
a) Basis = {(1 0 1), (0 1 1)}
b) On basis = {(1/sqrt(2) 0 1/(sqrt(2)), (-1/sqrt(6),
2/sqrt(6), 1/sqrt(6))}
19) First check if the 3 given vectors are orthogonal - they are.
Then you
can apply the formula for the projection, and get:
proj = (1 -2 1 3)
20)
a) false
b) false
c) false
d) true
e) true
21)
a) false
b) true
c) true
d) true
e) false
22) a) yes b) yes (hint: x^3 = y^3 if and only
if x = y)
c) yes d) yes e) no
23) a) Independent b) Independent c) Dependent d) Dependent e) Independent
24) a) det = 3 b) det = 21 c) det = 0 d) det = 3 e) det = 1/(125*3)
25) a) true b) true c) true d) true e) false
26)
a) evalues 2, -2
b) evalue 2, evector (5 1) ; evalue -2, evector
(1 1)
c) D =
2 0
0 -2
d) Let P =
5 1
1 1
Then A^(47) = P (D^(47)) P^(-1),
and
D^(47) =
2^47 0
0 -2^47
Compute P(-1), and multiply everything through to get the answer.
27)
a) eigenvalues 1 1 4
b) basis of eigenvectors for eigenvalue 1 : {(-1 0 1),
(-1 1 0)}
eigenvector for eigenvalue 4: (1 1
1)
c) D =
1 0 0
0 1 0
0 0 4
P =
-1/sqrt(2) -1/sqrt(6)
1/sqrt(3) <--- find a basis
of orthonormal eigenvectors.
0
sqrt(2/3) 1/sqrt(3)
The ones corresponding to different eigenvalues
1/sqrt(2)
1/sqrt(6) 1/sqrt(3)
are guaranteed to be orthogonal. The ones
corresponding to the same eigenvalue you
have to make orthonormal using Gram-Schmid.
28)
A^ =
4 1
1 2
b^ = (4 2)
least square soln: (6/7 4/7)
29)
y = (1/2)x + 1
30) a)Be warned, the computations are a bit wild here...
A =
1 1
1
9 3
1
16 4
1
36 6
1
b= 1
0
3
4
b) A ^ =
1634
308 62
308
62 14
62
14 4
b^ =
193
37
8
c) The last column would give you the coefficients, so:
>> rref([A'*A A'*b])
ans =
1.0000
0 0 0.1667
0
1.0000 0 -0.4744
0
0 1.0000 1.0769
SO, the answer in all its gory: y = .1667 x^2 - .4744 x + 1.0767
31)
a) eigenvalue 4 with eigenvector v1=(-1 2); eigenvalue
6 with eigenvector v2=(-1 4)
b) general solution: x(t) = C1e^(4t)*v1 + C2e^{6t)*v2
c) C1 = -1, C2 = -2 in the above, so solution is:
x1 = e^(4t) + 2e^(6t)
x2 = -2e^(4t) -8e^(4t)
32)
a) eigenvalues : i sqrt(3), -i sqrt(3)
eigenvector for isqrt(3) is
v =
1+ i sqrt(3)
2
b) Take the real and imaginary part of x(t) = e^(i sqrt(3)t ) v. and
then take a linear combination of them
to get the general solution, Real soln = C1 * Re(x(t)) + C2 * Im (x(t))
Where, denoting b= sqrt 3:
Re(x(t))= [ cos(bt) - b sin(bt), 2 cos(bt) ]
Im(x(t))= [ sin (bt) + b cos(bt), 2 sin(bt)]
c) C1 = 1/2, C2 = (3 sqrt3)/2
33)
a) eigenvalue 1 with eigenvector v1 = (-1 1 2)
eigenvalue 2 with eigenvector v2 = (-2 1 4)
eigenvalue 3 with eigenvector v3 = (-1 1 4)
b) x(t) = C1 e^(t)v1 + C2 e^(2t) v2 + C3 e^(3t) v3 <--- you would have to replace v1,v2,v3 from above of course.
c) To find C1,C2,C3:
A =
-1 -2 -1
-2
1 1
1 1
2 4
4 2
>> rref(A)
ans =
1 0
0 1
0 1
0 1
0 0
1 -1
The last column tells you C1=1, C2=1, C3=-1