These are just the answers, or just hints, not the full justification. For
full credit, you should give a little more explanation than I gave here.  
I did these fairly quickly; it's always possible that I made a mistake
(professors make mistakes too!)

#1 ((1-y)/2,y,1)

#2 the cofactors are 0,42,-35,0,-21,14,-3,12,3.  The determinant is -21.
The inverse is -1/21 times the transpose of the cofactors.

#3 A is 1 1 1      b is 1 
        1 0 0           0
        1 1 1           1
the solutions are (1 - c) |t| + c|t|^2.
the LU factorization is
    1  0  0        1  1  1
L = 1  1  0   U =  0 -1 -1
    1  0  1        0  0  0
the row-space has basis [1 1 1],[1 0 0]
the null-space has basis [0 1 -1 ]
the col-space has basis [1 1 1] [1 0 1]
the least square solutions are (1/2 - c) |t| + c |t|^2

#4 F T T T T T T T T

#5 1, e^(2\pi i/3) e^(-2\pi i/3), 
S = 1    1              1               D =  1     0             0
    1    e^(2\pi i/3)   e^(-2\pi i/3)        0  e^(2\pi i/3)     0
    1    e^(4\pi i/3)   e^(-4\pi i/3)        0    0           e^(-2\pi/3)
Typo:  should say eigenvalues of A^5.  These are the same
as those for A.
I - A has eigenvalues 1 - 1 = 0, 1 - e^(2\pi i/3), 1 - e^(-2\pi i/3).

#6 1,0 (twice)
In fact A is symmetric, so there is an orthogonal diagonalization
D = 1 0  0   S = 1/\sqrt{3}   1/\sqrt{2}     1/\sqrt{6}
    0 0  0       1/\sqrt{3}  -1/\sqrt{2}     1/\sqrt{6}
    0 0  0       1/\sqrt{3}      0          -2/\sqrt{6}  S^{-1} = S^T
A^5 = A 
I-A has eigenvalues 0,-1

#7 yes,A =  1 -1 3
            1  1 1 
            1  0 0
no, T(v+w) \neq T(v) + T(w)
yes,A= - \sqrt{3}/2      1/2
         -1/2         -\sqrt{3}/2
yes, A = 1 0 0 
         0 1 0 
         0 0 0 

#8 (Answer to b only) In the long run, 3/7 of the customers are with A.

#9 Method:  Apply Gram-Schmidt to e.g. [2 1 0 0], [2 0 1 0], [2 0 0 1]
(b) proj_V v = (v \cdot u1) u1 + .... (v \cdot u3) u3
(e) proj_V = u1 u1^T + ... + u3 u3^T, if the ui's are column-vectors.
otherwise, u1^T u1 etc.
(f) should be crossed off.  We did not do this material
this year.

#10 a = 3,b = -2, y1,y2 are (2x1 - x2)/\sqrt{5}, (x1 + 2 x2)/\sqrt{5}

#11 use the fact that det(A - \lambda I) = det( (A - \lambda I)^T)

#12 use || v || = v^T v = (c_1 u_1 + ......... c_n u_n) (c_1 u_1 + 
............ c_n u_n)^T etc.