What answers did you get; I got
1.a. 1,-1
b. (2,1) and (-1,2)
c. y=(1/2)x
2a. Show that
b. Circle radius 3 centre (1/2, -1)
c. a=0.5, c=-3 or 1
c. All the cross sections are circles so it must be a cone so it must be either a max or min not a saddle point. Since there is a cross section if z=42, s.p. at z=6 must be minimum
3. Find inverse of matrix using Cayley Hamilton, can't remember but I checked it with my calculator
4a. e=2 since e*e=e and a*a=2
b. 2*3=3, 1*2=1, 1*1=3*3=2 therefore 1*3 doesn't equal 1/2/3 therefore 1*3=4
c. basically a sudoku
d. Yes it is commutative since a*b=b*a from table
5a. a=alpha/100+1, b=-R
b. u(n)=a^nC+b((a^n-1)/(a-1))
c. show that. lim as n tends to infinity of un =(a^n)C+(b/(a-1))a^n (as a^n-1 tends to a^n)
therefore C+b/(a-1)<0 given answer follows immediately
d. can't remember exactly but 62,000 ish
6a. Show that. a/b=sqrt(7). Since a,b are rational, a/b is rational but root 7 is irrational so no integer solutions
b. Closure as (a+bsqrt7)(c+dsqrt7) belongs to G
associativity as multiplication is associative
identity is 1
inverse of (a+bsqrt7)=(a-bsqrt7)/(a^2-7b^2) this belongs to G
So a group
c. Not a group as (1+csqrt7)(1+dsqrt7)=1+7cd+(c+d)sqrt7, doesn't belong to H as 1+7cd doesn't equal 1 unless one of c,d=0
d. Counter example was <1,-1>. This is the only possible finite subgroup.
Some of those could well be wrong and I might have missed one