MATH40003 Revised Historical Coverage Practice Paper 5

Question 1 Historical basis: 2021 Q5, 2020 Q5 and 2025 Q4(b). It directly covers the previously under-trained high-mark determinant/minor ability chain. 30 minutes. Recognition signal: sparse matrices, eigenspace dimensions and the largest nonzero minor. First key step: For the monomial matrix, permute rows to a diagonal matrix. For rank, prove “rank at least k iff some k-minor is non-zero”. Permute the rows of A so that the non-zero entries lie on the diagonal. Row permutations change determinant only by a sign, so det A = (cid:6)x (cid:1) (cid:1) (cid:1) x . 1 n Suppose the new entry is in row i, column j, and the original non-zero entry in column j is x in row r 6= i. Subtract (λ/x) times row r from row i. This restores A without changing determinant, so det B = det A. Choose a basis v , v of E and extend it to a basis of F 4. In this basis the matrix of C has 1 2 µ block upper-triangular form ( ) µI (cid:3) 2 . 0 D Therefore χ (t) = (t (cid:0) µ)2 det(tI (cid:0) D). C 2 If a k-minor is non-zero, the selected columns are independent, so rank is at least k. Con- versely, choose k independent columns from a rank-k matrix; within them choose k in- dependent rows. Their intersection is a non-singular k-minor. Taking the largest such k proves the result. Marking points: 4 monomial determinant; 4 added entry; 5 characteristic factor; 7 rank theorem. Common errors: Expanding a huge determinant instead of using the structure; assuming geomet- ric multiplicity automatically implies algebraic multiplicity without the basis argument; proving only one direction of the rank theorem. Final self-check: Test part (b) on a 2 (cid:2) 2 example; verify the block matrix has µI in the first two 2 columns; identify the selected rows and columns in the rank proof. 2/5

Question 2 Historical basis: 2024 Q4(b). Directly restores the tridiagonal determinant recurrence and the 2x2 diagonalisation method. 30 minutes. Recognition signal: a tridiagonal matrix determinant d and a request for a closed form. n First key step: Expand along the last row, then the last column of the remaining minor, to obtain a second-order recurrence. Laplace expansion gives d n = (b + c)d n−1 (cid:0) bc d n−2 . Thus ( ) ( ) ( ) d n b + c (cid:0)bc d n−1 = . d n−1 1 0 d n−2 The recurrence matrix has characteristic polynomial t2 (cid:0) (b + c)t + bc = (t (cid:0) b)(t (cid:0) c), with eigenvectors (b, 1)T and (c, 1)T . Since ( ) (( ) ( )) 1 1 b c = (cid:0) , 0 b (cid:0) c 1 1 applying the nth power gives bn+1 (cid:0) cn+1 d = . n b (cid:0) c For b = 2, c = 1, the recurrence starts d = 1, d = 3 and gives d = 7, d = 15, d = 31. The 0 1 2 3 4 formula gives (25 (cid:0) 1)/(2 (cid:0) 1) = 31. Marking points: 5 recurrence; 3 matrix form; 8 diagonalisation/closed form; 4 numerical check. Common errors: Using b + c as both the diagonal and an eigenvalue; omitting d = 1; shifting the 0 power by one. Final self-check: Check n = 0 and n = 1 in the closed formula; substitute it into the recurrence algebraically. 3/5

Question 3 Historical basis: 2025 Q5 direct A/B coverage and 2020 Q1(f). It combines the 2025 QR/integers question with a concise O2 classification, while retaining a two-hour workload. 30 minutes. Recognition signal: apply Gram-Schmidt to three given columns, then classify a matrix group using the integer constraints on orthogonal columns. First key step: Compute q , subtract its projection from v , then subtract both projections 1 2 from v ; the R entries are q (cid:1) v . 3 i j Gram–Schmidt gives 1 1 1 q = p (1, 1, 0)T , q = p (1, (cid:0)1, 2)T , q = p ((cid:0)1, 1, 1)T . 1 2 3 2 6 3 Thus Q = (q 1 jq 2 jq 3 ) and p p p  2 1p/ 2 1/p2   R = 0 6/2 1/p6 . 0 0 2/ 3 Direct multiplication verifies A = QR. ( ) a c If X = 2 O (R), then det X = (cid:6)1. Orthogonality determines the second column from b d 2 the first. For determinant 1, ( ) cos θ (cid:0) sin θ X = ; sin θ cos θ for determinant (cid:0)1, ( ) cos θ sin θ X = . sin θ (cid:0) cos θ For X 2 O (Z), each integer column has squared norm 1, so it has exactly one non-zero en- n try, equal to (cid:6)1. Orthogonality puts these entries in different rows. Thus the matrices are precisely signed permutation matrices. They are closed under multiplication and inverse, and there are n! position choices and 2n sign choices. Marking points: 8 QR; 5 O2 classification; 7 signed permutation classification/group/order. Common errors: Forgetting to square the norm in projection denominators; allowing a zero or two non-zero integer entries in a unit column; proving only that permutation matrices are included, not classifying all of O (Z). n Final self-check: Check QT Q = I and QR = A; count choices column by column; verify the two O2 forms have determinant (cid:6)1. 4/5

Question 4 Historical basis: 2021 Q6, 2023 Q4(d), 2022 Q6(d) and the finite-group counting result. This gap paper restores several rotation abilities but keeps each proof short and independent. 30 minutes. Recognition signal: the set of squares, power maps on roots of unity, permutation cycle types, counting elements of prime order, and the Cayley action. First key step: Use the subgroup test for squares in an abelian group; use element orders for the power map; count order-p elements by their unique cyclic subgroups. If G is abelian, g2h2 = (gh)2 and (g2) −1 = (g −1)2, so the squares form a subgroup. If jGj is odd and jgj = 2k (cid:0) 1, then g = (gk)2, so every element is a square. For the pth roots of unity, (zq)p = 1 and the map is a homomorphism because the group is abelian. Its kernel is trivial: an element in the kernel has order dividing both p and q. Hence the map is injective and, on a finite set, surjective. Order-three permutations in S have shapes [3, 15] or [3, 3, 1, 1]. The counts are 8 ( ) ( )( ) 8 22 8 5 2 = 112, = 1120, 3 2! 3 3 for a total of 1232. Each subgroup of order p contains exactly p (cid:0) 1 elements of order p, and distinct such subgroups intersect only in the identity. Hence the total is divisible by p (cid:0) 1. For Cayley, let G act on itself by left multiplication: λ (x) = gx. Each λ is a permutation, g g g 7! λ is a homomorphism, and it is injective because λ (e) = g. Thus G is isomorphic to a g g subgroup of S |G|. Marking points: 6 squares; 4 unit roots; 4 cycle count; 3 divisibility; 3 Cayley. Common errors: Assuming (gh)2 = g2h2 in a non-abelian group; failing to divide by 2! when counting two 3-cycles; not proving distinct order-p subgroups meet trivially. Final self-check: For the power map, check kernel via orders; sum the two cycle-shape counts; evaluate the Cayley embedding at the identity. 5/5