Algebra

Algebra is the largest block of the Mathematics paper (our estimate: about 8 of 35 questions). It has four distinct families: matrices and determinants, complex numbers with polynomial equations, sequences with counting, and the binomial / exponential / logarithmic series. Each family has two or three standard question shapes. Learn the shapes and the paper becomes predictable.

2.1 Matrices and determinants, types and properties, inverse of a matrix

A matrix is an array of m\times n numbers. Types: square, row, column, diagonal, scalar, identity I, zero, upper / lower triangular, symmetric (A^T=A), skew-symmetric (A^T=-A, diagonal zero), idempotent (A^2=A), involutory (A^2=I), nilpotent (A^k=0), orthogonal (AA^T=I).

Operations. Add / subtract same-size matrices; AB exists when columns of A = rows of B and is in general AB\ne BA. (AB)^T=B^TA^T, (AB)^{-1}=B^{-1}A^{-1}. Every square matrix is a sum of a symmetric and a skew-symmetric matrix, A=\tfrac{1}{2}(A+A^T)+\tfrac{1}{2}(A-A^T).

Determinant. For (det | a&b ; c&d |)=ad-bc; for 3×3 expand along a row with the signs +-+. Properties: - swapping two rows changes the sign; two equal (or proportional) rows give 0; - a common factor of a row comes out; so |kA|=k^n|A| for an n\times n matrix; - adding a multiple of one row to another changes nothing; - |AB|=|A||B|, |A^T|=|A|, |A^{-1}|=1/|A|, |\mathrm{adj}A|=|A|^{n-1}, A\,\mathrm{adj}A=|A|I.

Inverse. A^{-1}=\dfrac{\mathrm{adj}A}{|A|} exists iff |A|\ne0 (non-singular). For a 2×2 matrix A=[a\ b;\ c\ d] the inverse is \dfrac{1}{ad-bc}[d\ -b;\ -c\ a]: swap the diagonal, negate the other two. Elementary row operations reduce [A\,|\,I] to [I\,|\,A^{-1}].

Linear systems. Cramer's rule: x=D_x/D, y=D_y/D, z=D_z/D when D\ne0. If D=0 the system has no solution or infinitely many. A homogeneous system has non-trivial solutions iff D=0.

2.2 Complex numbers and polynomial equations

i^2=-1; i^{4k}=1, i^{4k+1}=i, i^{4k+2}=-1, i^{4k+3}=-i. For z=x+iy: conjugate \bar z=x-iy, modulus |z|=\sqrt{x^2+y^2}, z\bar z=|z|^2, \dfrac{1}{z}=\dfrac{\bar z}{|z|^2}. The argument \theta=\tan^{-1}(y/x) in the correct quadrant, principal value in (-\pi,\pi]. Polar form z=r(\cos\theta+i\sin\theta)=re^{i\theta}.

De Moivre: (\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta. The n roots of z^n=1 are e^{2\pi ik/n}, k=0,\dots,n-1. Cube roots of unity: 1,\omega,\omega^2 with \omega=\dfrac{-1+i\sqrt{3}}{2}; 1+\omega+\omega^2=0, \omega^3=1, \omega^2=\bar\omega. Useful: 1+\omega=-\omega^2, (1+\omega)^3=-1.

Handy: (1+i)^2=2i, \dfrac{1+i}{1-i}=i, |z_1z_2|=|z_1||z_2|, |z_1+z_2|\le|z_1|+|z_2|.

Polynomial equations

For ax^2+bx+c=0: roots \dfrac{-b\pm\sqrt{D}}{2a}, D=b^2-4ac; \alpha+\beta=-\dfrac{b}{a}, \alpha\beta=\dfrac{c}{a}. D>0 real distinct; D=0 equal; D<0 complex conjugate pair; for rational coefficients and D a perfect square the roots are rational, and irrational roots p+\sqrt{q} come with p-\sqrt{q}.

Symmetric functions: \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta; \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta); (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta. The equation with roots \alpha,\beta is x^2-(\alpha+\beta)x+\alpha\beta=0.

For ax^3+bx^2+cx+d=0: \sum\alpha=-\dfrac{b}{a}, \sum\alpha\beta=\dfrac{c}{a}, \alpha\beta\gamma=-\dfrac{d}{a}. A polynomial with real coefficients has complex roots in conjugate pairs (so an odd-degree one has at least one real root). Factor theorem: (x-a) divides f iff f(a)=0. Rational roots p/q have p\mid constant term, q\mid leading coefficient.

2.3 Sequence and series, permutation and combination

Progressions

	AP	GP	HP
n-th term	a+(n-1)d	ar^{n-1}	reciprocals form an AP
sum of n terms	\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)	\dfrac{a(r^n-1)}{r-1}	–
sum to infinity	–	\dfrac{a}{1-r}, |r|<1	–
mean of a,b	\frac{a+b}{2}	\sqrt{ab}	\frac{2ab}{a+b}

\text{AM}\ge\text{GM}\ge\text{HM}, and \text{GM}^2=\text{AM}\cdot\text{HM} for two numbers. Three numbers in AP: a-d,a,a+d; in GP: a/r,a,ar. If S_n is known, t_n=S_n-S_{n-1}. Standard sums: \sum n=\dfrac{n(n+1)}{2}, \sum n^2=\dfrac{n(n+1)(2n+1)}{6}, \sum n^3=\Big[\dfrac{n(n+1)}{2}\Big]^2. Arithmetico-geometric series: multiply by r, subtract. Method of differences: write t_r as f(r)-f(r+1) so the sum telescopes.

Counting

Fundamental principle: independent stages multiply, alternatives add.

• Arrangements of n things with repeats p,q,r: \dfrac{n!}{p!\,q!\,r!} (BANANA: \dfrac{6!}{3!\,2!}=60).
• Circular arrangements of n: (n-1)!; of a necklace (flip allowed): \tfrac{1}{2}(n-1)!.
• Words using all letters where certain letters stay together: glue them into one block, then multiply by the block's internal arrangements.
• No two together: arrange the others first, then place the restricted ones in the gaps: n!\times\,^{(n+1)}P_r.
• Digits: the first digit cannot be 0 (four-digit numbers with distinct digits: 9\times9\times8\times7=4536).
• At least / at most selections: add cases or subtract the complement from ^nC_r total.
• Diagonals of an n-gon: \dfrac{n(n-3)}{2}=\,^nC_2-n. Triangles from n points with k collinear: ^nC_3-\,^kC_3.
• Distribution of n identical items into r boxes (empty allowed): ^{n+r-1}C_{r-1}.

2.4 Binomial theorem, exponential and logarithmic series

(a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r has n+1 terms. General term T_{r+1}={}^nC_r\,a^{n-r}b^r. Middle term: T_{n/2+1} if n even; two middle terms T_{(n+1)/2},T_{(n+3)/2} if n odd. Constant term: set the power of x to zero ((x+1/x)^6: r=3, ^6C_3=20). Sum of coefficients =2^n (put x=1); even-place sum = odd-place sum =2^{n-1}. Greatest coefficient is ^nC_{n/2} for even n. Ratio of consecutive terms: \dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac{b}{a}.

For any real n and |x|<1:

For small x: (1+x)^n\approx1+nx (better: add the x^2 term). (1-x)^{-1}=1+x+x^2+\cdots.

Also \ln 2=1-\frac{1}{2}+\frac{1}{3}-\cdots and e^{-1}=\sum\frac{(-1)^n}{n!}.

Formula and fact sheet

Topic	Result
|kA|, |\mathrm{adj}A|	k^n|A|, |A|^{n-1}
2×2 inverse	\frac{1}{ad-bc}[d\ -b;\,-c\ a]
\omega	1+\omega+\omega^2=0, \omega^3=1
quadratic	\alpha+\beta=-b/a, \alpha\beta=c/a, D=b^2-4ac
AP sum / GP sum	\frac{n}{2}[2a+(n-1)d] / a\frac{r^n-1}{r-1}
\sum n^2, \sum n^3	\frac{n(n+1)(2n+1)}{6}, [\frac{n(n+1)}{2}]^2
^nC_r, ^nP_r	\frac{n!}{r!(n-r)!}, \frac{n!}{(n-r)!}
binomial	T_{r+1}={}^nC_ra^{n-r}b^r
e^x, \ln(1+x)	\sum x^n/n!, x-x^2/2+x^3/3-\cdots

Chapter summary

• Determinants scale as k^n; rows in AP, or two proportional rows, give zero.
• A^{-1} needs |A|\ne0; for 2×2 swap the diagonal and negate the rest.
• Reduce powers of i and \omega by their cycles (4 and 3).
• For any quadratic question first write the sum and product of the roots.
• t_n=S_n-S_{n-1} and the infinite GP needs |r|<1.
• "No two together" = arrange the others, then fill gaps; "at least" = cases or complement.
• Constant term and coefficient questions are solved by the general term T_{r+1}.

Practice questions

Set A — Recall (16 questions)

1. The determinant of the matrix [2\ 3;\ 4\ 5] isa22b-2c2d-22
2. If A is a 3\times3 matrix with |A|=4, then |2A| equalsa8b16c32d64
3. For A=[a\ b;\ c\ d] with ad-bc\ne0, A^{-1} equals \dfrac{1}{ad-bc} timesa[d\ -b;\ -c\ a]b[d\ b;\ c\ a]c[a\ -b;\ -c\ d]d[-d\ b;\ c\ -a]
4. The value of i^{27} isaib-ic1d-1
5. The principal argument of -1+i isa\pi/4b-\pi/4c3\pi/4d-3\pi/4
6. If \omega is a complex cube root of unity, (1+\omega)^3 equalsa-1b1c0d8
7. The product of the roots of 2x^2-7x+3=0 isa-7/2b7/2c-3/2d3/2
8. The 20th term of the AP 3,7,11,\dots isa76b80c83d79
9. The sum to infinity of 9+3+1+\cdots isa12b13.5c27d13
10. The value of ^8C_3 isa56b336c24d112
11. The number of distinct arrangements of the letters of BANANA isa720b120c60d30
12. The number of terms in the expansion of (x+y)^{10} isa10b11c12d9
13. The sum of the binomial coefficients in (1+x)^7 isa64b49c256d128
14. The coefficient of x^3 in e^{2x} isa8/3b4/3c2/3d1/3
15. The sum of the series 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots isa\ln2b1ced\pi/4
16. A square matrix A with A^2=I is calledanilpotentbidempotentcinvolutorydorthogonal

Answers and explanations — Set A

1. (b) — 2\cdot5-3\cdot4=-2.

2. (c) — |2A|=2^3|A|=32. (b) wrongly uses 2^2.

3. (a) — the adjoint of a 2×2 matrix swaps a,d and negates b,c.

4. (b) — 27=4\cdot6+3, so i^{27}=i^3=-i.

5. (c) — the point (-1,1) is in the second quadrant: \pi-\pi/4.

6. (a) — 1+\omega=-\omega^2, so (1+\omega)^3=-\omega^6=-1.

7. (d) — c/a=3/2. (b) is the sum.

8. (d) — 3+19\times4=79.

9. (b) — a=9, r=\frac{1}{3}: 9/(1-\frac{1}{3})=13.5.

10. (a) — \frac{8\cdot7\cdot6}{3!}=56; 336 is ^8P_3.

11. (c) — \frac{6!}{3!\,2!}=60.

12. (b) — n+1=11.

13. (d) — 2^7=128.

14. (b) — \frac{(2x)^3}{3!}=\frac{8}{6}x^3=\frac{4}{3}x^3.

15. (a) — it is \ln(1+x) at x=1.

16. (c) — involutory. Idempotent is A^2=A.

Set B — Understanding (10 questions)

1. If A is 3\times3 with |A|=-3, then |\mathrm{adj}A| isa-9b9c-3d27
2. The value of (det | 1&2&3 ; 4&5&6 ; 7&8&9 |) isa0b1c-1d45
3. The value of (1+i)^8 isa8b16ic16d-16
4. If z=\dfrac{1+i}{1-i}, then z^{100} equalsaib-1c-id1
5. If \alpha,\beta are the roots of 2x^2-5x+1=0, then \alpha^2+\beta^2 isa25/4b21/4c23/4d19/4
6. The equation x^2+kx+9=0 has equal roots forak=\pm6bk=6 onlyck=3dk=\pm3
7. If S_n=3n^2+2n for an AP, its 10th term isa61b57c320d59
8. Three positive numbers are in GP and their product is 64. The middle number isa8b2c4d16
9. Five boys and three girls are seated in a row so that no two girls sit together. The number of ways isa4320b14400c720d40320
10. The constant term in the expansion of \big(x+\frac{1}{x}\big)^6 isa15b6c20d30

Answers and explanations — Set B

1. (b) — |\mathrm{adj}A|=|A|^{n-1}=(-3)^2=9.

2. (a) — rows in AP: R_3-R_2=R_2-R_1, so after operations a row of zeros appears.

3. (c) — (1+i)^2=2i, so (1+i)^8=(2i)^4=16.

4. (d) — z=i; i^{100}=(i^4)^{25}=1.

5. (b) — \alpha+\beta=\frac{5}{2}, \alpha\beta=\frac{1}{2}: \frac{25}{4}-1=\frac{21}{4}.

6. (a) — D=k^2-36=0 gives k=\pm6.

7. (d) — S_{10}-S_9=320-261=59 (general term 6n-1).

8. (c) — the terms are a/r,a,ar; the product is a^3=64, so a=4.

9. (b) — boys: 5!=120; the girls take 3 of the 6 gaps in order: ^6P_3=120; 120\times120=14400.

10. (c) — T_{r+1}={}^6C_r x^{6-2r}; constant when r=3: ^6C_3=20.

Set C — Application (8 questions)

1. If \omega is a complex cube root of unity, (1-\omega+\omega^2)(1+\omega-\omega^2) equalsa0b2c4d8
2. The roots of x^2-6x+k=0 are in the ratio 1:2. Then k equalsa8b9c12d6
3. The sum of all integers from 1 to 100 that are divisible by 3 or 5 isa2733b2318c2468d2418
4. The number of four-digit numbers with all digits distinct isa5040b4536c3024d4500
5. The number of diagonals of a decagon isa45b40c35d25
6. A committee of 4 is chosen from 6 men and 4 women with at least 2 women. The number of ways isa115b90c210d125
7. In (1+x)^n the coefficients of the 5th, 6th and 7th terms are in AP. The sum of all possible values of n isa21b14c7d28
8. Using the first three terms of the binomial series, (1.02)^5 is approximatelya1.100b1.102c1.110d1.104

Answers and explanations — Set C

1. (c) — 1+\omega^2=-\omega, so the factors are -2\omega and -2\omega^2; product 4\omega^3=4.

2. (a) — roots \alpha,2\alpha: 3\alpha=6, \alpha=2, k=2\alpha^2=8.

3. (d) — multiples of 3: 3\cdot\frac{33\cdot34}{2}=1683; of 5: 5\cdot\frac{20\cdot21}{2}=1050; of 15: 15\cdot\frac{6\cdot7}{2}=315; 1683+1050-315=2418.

4. (b) — 9\times9\times8\times7; the first digit cannot be 0 (5040 is 10\times9\times8\times7).

5. (c) — \frac{10\cdot7}{2}=35 (^{10}C_2-10).

6. (a) — ^4C_2\,^6C_2+{}^4C_3\,^6C_1+{}^4C_4=90+24+1=115.

7. (a) — ^nC_4+{}^nC_6=2\,^nC_5 reduces to n^2-21n+98=0: n=7,14; sum 21. Check n=7: 35+7=2\times21.

8. (d) — 1+5(0.02)+10(0.02)^2=1+0.1+0.004=1.104 (exact 1.10408).