Topic 1 [145 marks] Markscheme. Examiners report. since. I is the identity A1 and. R is reflexive. it follows that B = AX 1 A1 and.

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1 Topic [ marks] a. since = I where I is the identity and det(i) =, R is reflexive RB = BX where det(x) = M it follows that B = X and det( X ) = det(x ) = R is symmetric RB and BRC = BX and B = CY where det(x) = det(y ) = M it follows that = CY X det(y X) = det(y )det(x) = R is transitive hence R is an equivalence relation G [ marks] [ marks] This question was not well done in general, again illustrating that questions involving both matrices and equivalence relations tend to cause problems for candidates. common error was to assume, incorrectly, that RB and BRC = BX and B = CX, not realizing that a different " x" is required each time. In proving that R is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this. b. for reflexivity, we require R so that = I (for all S ) M since det(i) = and we require I S the only possibility is n = [ marks] [ marks]

2 In proving that R is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this. a. recognizing that the function needs to be injective and surjective R Note: ward R if this is seen anywhere in the solution. injective: let U,V U = V be -D column vectors such that M U = V M U = V this shows that f is injective surjective: let W M then there exists Z = W such that Z = W M this shows that f is surjective therefore f is a bijection G This proved to be a difficult question for some candidates. Most candidates realised that they had to show that the function was both injective and surjective but many failed to give convincing proofs. Some candidates stated, incorrectly, that f was injective because X is uniquely defined, not realising that they had to show that X = Y X = Y.

3 b. (i) the relationship is ad = bc [ marks] (ii) it follows that c d = = λ so that a b (c,d) = λ(a,b) (iii) EITHER let p W = [ ] be a -D vector q then a b p W = [ ][ ] λa λb q M the image always satisfies y = λx so f is not surjective and therefore not a bijection OR consider ap + bq = [ ] λ(ap + bq) a b b ab [ ][ ] = [ ] λa λb λab a b ab [ ][ ] = [ ] λa λb a λab this shows that f is not injective and therefore not a bijection R R [ marks] Solutions to (b) were disappointing with many candidates failing to realise that they had either to show that X was confined to a subset of R R or that two distinct vectors had the same image under f.

4 c. the identity element is R consider, for r m, using as a generator M combined with itself r times gives r and as r increases from to m, the group is generated ending with when r = m it is therefore cyclic G [ marks] [ marks] This question was well answered in general with solutions to (c) being the least successful. d. (i) by Lagrange the order of each element must be a factor of m and if m is prime, its only factors are and m R since is the only element of order, all other elements are of order m and are therefore generators R (ii) since x (m x) = + m the inverse of x is (m x) (M) (iii) consider M there are (m ) inverse pairs N Note: ward M for an attempt to list the inverse pairs, for completing it correctly and for the final answer.

5 This question was well answered in general with solutions to (c) being the least successful. e. since a, b are unequal primes the only factors of m are a and b there are therefore only subgroups of order a and b R they are {,a,a,,(b )a} {,b,b,,(a )b} [ marks] [ marks] This question was well answered in general with solutions to (c) being the least successful. a. let = and consider det() = (M) the vectors form a basis because the determinant is non-zero (or because the matrix is non-singular) [ marks] R [ marks]

6 b. let = λ + μ + v 6 M [ marks] so that EITHER λ + μ + v = λ + μ + v = λ + μ + v = 6 M OR λ μ = v 6 M THEN giving λ =, μ =, v = () hence = [ marks] a. reducing to row echelon form [ marks] k (M)() k (i) this shows that the rank of the matrix is (ii) the equations can be solved if k = [ marks]

7 b. let z = λ [ marks] then y = λ and λ x = ( λ + =) λ Note: ccept equivalent expressions. [ marks] a. (i) = 6 [6 marks] = 6 = 6 M = 6 = G (ii) = = ( ) = M G = ( ) = Note: ccept alternative solutions that include correct calculation of both sides of the expression. [6 marks]

8 b. (i) conjecture: n = (n ) (n ) [ marks] (ii) first check that the result is true for n = the formula gives = which is correct assume the result for n = k, i.e. M k = (k ) (k ) so k+ = [(k ) (k )] = (k ) (k ) so true for n = k true for n = k + and since true for n =, = (k )( ) (k ) = k (k ) the result is proved by induction R M M Note: Only award the R mark if a reasonable attempt at a proof by induction has been made. [ marks] 6a. a b a c b a = c b c M [ marks] = this shows that each matrix is self-inverse [ marks]

9 6b. closure: a b = a where each of a, b, c can only be ± this proves closure identity: the identity matrix is the group identity inverse: as shown above, every element is self-inverse M associativity: this follows because matrix multiplication is associative S is therefore a group belian: a a b b c c c b a c a a b G b b c c c = = = a a a a a a b b b b b b c c c c c c Note: Second line may have been shown whilst proving closure, however a reference to it must be made here. [9 marks] we see that the same result is obtained either way which proves commutativity so that the group is belian [9 marks] R 6c. since all elements (except the identity) are of order, the group is not cyclic (since S contains elements) R [ mark] [ mark]

10 a. (i) let a b M = ( ) b c (M) the eigenvalues satisfy det(m λi) = (M) (a λ)(c λ) b = λ λ(a + c) + ac b = discriminant = (a + c) (ac b ) = (a c ) + b () M this shows that the eigenvalues are real G [ marks] (ii) let the distinct eigenvalues be λ, λ, with eigenvectors, X X then λ X = MX λ X = MX and M transpose the first equation and postmultiply by X to give λ X T X = X T MX premultiply the second equation by X T λ X T X = X T MX it follows that (λ λ ) X T X = since λ λ, it follows that X T X = so that the eigenvectors are orthogonal R [ marks]

11 b. the eigenvalues satisfy λ = M 9 λ λ λ + 96 = λ =, first eigenvector satisfies x ( ) ( ) = ( ) y x ( ) = (any multiple of) y ( ) second eigenvector satisfies x ( ) ( ) = ( ) y x ( ) = y ( ) (any multiple of ) M

12 c. (i) consider the rotation in which (x,y) is transformed onto ( x, y ) defined by x ( x ) = ( ) so that y y the ellipse E becomes x x ( ) = ( ) y y G M ( x y x ) ( ) ( ) = 9 y ( x y x )( ) ( ) = y ( x ) + ( y ) = 6 M (ii) the angle of rotation is given by cosθ =,sin θ = M since a rotational matrix has the form cosθ sin θ ( ) sin θ cosθ so θ = 6 (anticlockwise)

13 . (a) the eigenvalues satisfy a λ b = (M) c d λ λ (a + d)λ + ad bc = using the sum and product properties of the roots of a quadratic equation λ + λ = a + d, λ λ = ad bc = det [ marks] (M) G R [ marks] (b) let f(λ) = λ (a + d)λ + ad bc putting b = a and d = c, consider M f() = a + c + a ac c + ac = therefore λ = is an eigenvalue G [ marks] Note: llow substitution for b, c into the quadratic equation for λ followed by solution of this equation. (c) using any valid method (M) the eigenvalues are and an eigenvector corresponding to λ = satisfies x x ( ) ( ) = ( ) or y y x ( ) ( ) = ( ) M y x ( ) = ( ) or any multiple y an eigenvector corresponding to λ = satisfies x x ( ) ( ) = ( ) or y y x ( ) ( ) = ( ) M y x ( ) = ( ) or any multiple y Note: ward M for calculating the first eigenvector and M for the second irrespective of the order in which they are calculated.

14 9. (a) = 6 = successive powers of are given by 9 (M) it follows, considering elements in the first rows, that solving, a + b + c = a + b = 6a + b = (M) M (a, b, c) = (,, ) [ marks] Note: ccept any other three correct equations. Note: ccept the use of the Cayley Hamilton Theorem. (b) (i) it has been shown that = + +I multiplying by, M = +I + whence =...I (ii) substituting powers of, 6 = = M Note: Follow through their equation in (b)(i).

15 Note: Line (ii) of (ii) must be seen. [ marks]

16 . [ marks] (a) (i) using row reduction, M () for consistency, (M) put M (ii) the rank of a matrix is the number of independent rows (or columns) [ marks] (b) (i) (M) since, the vectors form a basis R (ii) let M it follows that therefore [ marks] 6 μ μ μ = μ = z = α, t = β y = α + β; x = α β rank ( )= det( )= det( ) = a + b + c + d 6 6 = 6 a b c d = a b c d 6 6 = a = b = c = d = =

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