Is ECON1003 statistics or maths?

Despite the 'Quantitative Methods' title, ECON1003 at the University of Sydney is mathematics for economics, NOT statistics. There is no descriptive statistics, probability, distributions, sampling, hypothesis testing or regression. The seven taught topics are straight lines, non-linear functions, financial mathematics, differentiation, functions of several variables, integration and linear algebra, all applied to demand/supply, cost/revenue, utility and production.

How is ECON1003 assessed?

ECON1003 is assessed by four online quizzes worth 10% in total (one week each, no extensions), a mid-semester exam worth 40% covering Weeks 1-4 up to the application of differentiation, and a cumulative final exam worth 50% covering all seven topics. A formula sheet is provided in both exams.

How do you use a Lagrangian in ECON1003?

To maximise f(x,y) subject to a constraint, write the Lagrangian L = f - lambda(constraint). Set the partial derivatives of L to zero, require all constraints to hold, keep the multipliers on inequality constraints non-negative, and apply complementary slackness (each inequality multiplier times its constraint equals zero). Solve the resulting cases, including the corner cases x=0 and y=0, then evaluate f at each candidate and pick the best. The multiplier lambda is the shadow price - how much the optimum changes if the constraint is relaxed by one unit.

What is the difference between compound and continuous interest?

Compounding m times a year grows a sum by P_t = P_0 (1+i/m)^(mt): you divide the annual rate by m and multiply the periods by m. Continuous compounding is the limit as m goes to infinity, giving P_t = P_0 e^(it). Continuous compounding yields slightly more for the same nominal rate. A common error is dividing the rate by m but forgetting to multiply the exponent by m.

When does the power rule for integration fail?

The integration power rule, integral of x^n dx = x^(n+1)/(n+1)+C, is valid for every n except n=-1, where it would divide by zero. The special case n=-1 integrates to the natural log: integral of x^(-1) dx = ln(x)+C. Always remember the +C, which matters when solving differential equations.

How do you find marginal revenue from a demand curve?

First write total revenue TR = P x Q with price P expressed in terms of Q, then differentiate: MR = dTR/dQ. Do not differentiate price directly. Revenue is maximised where MR=0 and profit where MR=MC.

What is the shadow price (the Lagrange multiplier)?

The value of a Lagrange multiplier lambda at the optimum is the shadow price: it tells you how much the optimal value of the objective changes if you relax that constraint by one unit. In a utility-maximisation problem it is the marginal utility of income; in a cost-minimisation problem it is the marginal cost of producing one more unit of output.

How do you solve a system of equations with the inverse matrix?

Write the system as AX=B, then X = A^(-1) B. The inverse exists only when det(A) is non-zero. For a 2x2 matrix, A^(-1) = (1/(ad-bc)) times [[d,-b],[-c,a]]. In general A^(-1) = C^T / det(A), where C is the matrix of cofactors - remember to transpose it and divide by the determinant.

ECON1003

Quantitative Methods in Economics

University of Sydney · School of Economics

Exam Revision
Sem 1 2026 · Side 1 of 2
Midterm 40% + Final 50% · formula sheet provided

SIDE 1/2 LINES → CALCULUS · Straight lines & elasticity · Quadratics/exponentials/logs · Financial maths · Differentiation & MR=MC maths, not stats Compiled by AskSia · mapped to the ECON1003 syllabus · asksia.ai/cheatsheet/usyd-econ1003

0 · How to Use Thisread first

ECON1003 is maths for economics — NOT statistics. No probability, distributions or hypothesis tests. Seven topics: lines · non-linear functions · financial maths · differentiation · several variables · integration · linear algebra (Bradley Chs 2,4,5,6,7,8,9).

Two exams: Midterm 40% (Wks 1–4, up to applications of differentiation — Side 1) · Final 50% (cumulative, all 7 — both sides). 4 quizzes = 10%. A formula sheet is provided; memorise the rest (power/product/chain rules, substitution, Lagrangian conditions, Cramer, inverse).

Sia → Method marks matter — show every step; partial credit is given. Round only the FINAL answer to the stated decimal places.

1 · The Straight LineCh2

liney = mx + c m = slope, c = y-intercept
m = (y₂−y₁)/(x₂−x₁) = rise/run

Slope m: "x up 1 ⇒ y changes by m." Intercept c: y when x=0. Horizontal line y=k; vertical x=k (undefined slope).

worked · line thru (2,1), m=11 = 1·2 + c ⇒ c = −1 ⇒ y = x − 1

Trap → Non-standard form (e.g. 2y+4x−4=0) must be rearranged to y=mx+c before reading m,c: ⇒ y = −2x + 2.

2 · Demand & SupplyCh2

Demand Q=f(P), e.g. Q=200−2P (−2: a $1 rise drops Qd by 2). Inverse demand P=f(Q): P=100−0.5Q. The two slopes are reciprocals, not equal.

cost · revenue · profitTC = FC + VC (e.g. 800 + 1.5Q)
R = P·Q Profit = R − TC
Break-even: set Profit = 0, solve Q

worked · TC=800+1.5Q, P=3.5Profit = 3.5Q − (800+1.5Q) = 2Q−800
=0 ⇒ break-even Q = 400

Trap → Fixed cost is the intercept of TC, never part of marginal/variable cost. Read whether Q=f(P) or P=f(Q) is wanted.

3 · Budget LineCh2

two goods, income mm = p₁x₁ + p₂x₂
x₂ = m/p₂ − (p₁/p₂)x₁

Slope −p₁/p₂ = relative price; intercept m/p₂ = max of good 2. Price rise pivots (slope); income change shifts parallel (intercept only) — don't confuse.

worked · m=120, p₁=4, p₂=6x₂ = 20 − (2/3)x₁ (max 30 of good 1, 20 of good 2)

4 · Linear ElasticityCh2

point elasticityE = %ΔQ / %ΔP = (dQ/dP)·(P/Q)
demand P=a−bQ: Ed = (−1/b)·(P/Q)
supply P=c+dQ: Es = (1/d)·(P/Q)

Demand: elastic Ed<−1 · inelastic −1<Ed<0 · unit Ed=−1.

worked · P=2400−0.5Q at P=1800Q=1200 ⇒ Ed=(−1/0.5)(1800/1200)=−3
(1% price ↑ ⇒ 3% quantity ↓, elastic)

Trap → Elasticity is NOT the slope — it varies along a straight line: elastic at high P, inelastic at low P, unit at the midpoint.

5 · QuadraticsCh4

solve ax²+bx+c=0x = (−b ± √(b²−4ac)) / (2a)
discriminant Δ = b²−4ac

Δ>0 two real roots · Δ=0 one · Δ<0 no real roots (write "no real solution"). Parabola: one turning point (min if a>0, max if a<0); axis midway between roots; y-int = c.

worked · y=2x²−7x−9roots x=−1, 4.5 ⇒ axis x=1.75
sub back ⇒ turning-point y = −15.125

discriminant check · x²+x+1Δ = 1−4 = −3 < 0 ⇒ no real roots

Economic use: profit = quadratic in Q ⇒ the vertex gives the profit-maximising Q directly.

6 · TransformationsCh4

Move	Effect
f(x)+c	up c · f(x)−d down d
f(x−c)	RIGHT c · f(x+d) LEFT d
−f(x)	reflect in x-axis
f(−x)	reflect in y-axis

Combine in order: e.g. y = −(x−3)²+5 is x² reflected, shifted right 3, up 5 (vertex (3,5), opens down).

Trap → Horizontal shifts go the "wrong" way — (x−2)² moves right, not left.

7 · Exponentials & eCh4

index rules · y=aˣaᵐ·aⁿ = aᵐ⁺ⁿ aᵐ/aⁿ = aᵐ⁻ⁿ
(aᵐ)ᵏ = aᵐᵏ a⁰=1 a⁻ⁿ=1/aⁿ
e ≈ 2.71828

Solve by matching base, then equate exponents: 2ˣ = 1/16 = 2⁻⁴ ⇒ x = −4. Growth: P = P₀e^{rt}.

worked · 3^{2x} = 8181 = 3⁴ ⇒ 2x = 4 ⇒ x = 2

Trap → Can't match the base? Take logs of both sides and use log(aᵇ)=b·log a to drop the exponent down.

8 · LogarithmsCh4

log_a(b) = power on a giving b. log=log₁₀, ln=logₑ.

log ruleslog(a)+log(b) = log(ab)
log(a)−log(b) = log(a/b)
log(aᵇ) = b·log(a)
change base: log_c(a) = log(a)/log(c)

worked · 1750 = 753e^{0.03t}ln(1750/753) = 0.03t ⇒ t ≈ 28

Trap → ln(A+B) ≠ lnA + lnB. Rules apply only to ×/÷ inside one log; argument must be >0.

9 · Sequences & SeriesCh5

n-th termarithmetic: aₙ = a + (n−1)d
geometric: aₙ = a·rⁿ⁻¹

arithmetic sum (+d)Sₙ = (n/2)(2a + (n−1)d)

geometric sum (×r)Sₙ = a(1−rⁿ)/(1−r)
infinite (|r|<1): S = a/(1−r)

worked · 2+6+18+… (8 terms)a=2, r=3 ⇒ S₈ = 2(3⁸−1)/2 = 6560

Trap → Want the n-th term or the sum? And the infinite sum needs |r|<1.

10 · InterestCh5 · on sheet

growth of P₀simple: Pₜ = P₀(1+it)
compound: Pₜ = P₀(1+i)ᵗ
m/yr: Pₜ = P₀(1+i/m)^{mt}
continuous: Pₜ = P₀e^{it}
PV: P₀ = Pₜ/(1+i)ᵗ

worked · 10000→20000 in 6yr2=(1+i)⁶ ⇒ i = 2^{1/6}−1 ≈ 0.122

Trap → With m periods/yr you must divide rate (i/m) AND multiply periods (mt) — the classic slip drops the mt.

11 · Depreciation & NPVCh5

Straight-line: subtract a fixed amount/yr. Reducing-balance: Aₜ = A₀(1−i)ᵗ; total depr = A₀ − Aₜ.

net present valueNPV = Σ cashflow_t/(1+i)ᵗ
outlay at t=0 enters NEGATIVE

NPV > 0 ⇒ project beats the discount rate.

worked · $1000, quarterly, 8%, 2yrP = 1000(1+0.08/4)^{4·2}
= 1000(1.02)⁸ ≈ $1171.66

worked NPV · −1000 now, +600/yr ×2, i=10%−1000 + 600/1.1 + 600/1.21
≈ −1000 + 545.5 + 495.9 = +41.3 ⇒ accept

12 · Annuities & LoansCh5 · on sheet

future value (deposit A₀)Vₜ = A₀·[(1+i/m)^{tm} − 1]/(i/m)

present value / loan LL = A₀·[1 − (1+i/m)^{−mt}]/(i/m)
solve for A₀ = each payment

Total interest = (payment × no. payments) − L.

worked · loan $20000, 6%/yr, 5yr ann.L = A₀·[1−(1.06)⁻⁵]/0.06
20000 = A₀·4.2124 ⇒ A₀ ≈ $4747.93/yr

Trap → FV uses (…)^{tm} − 1; PV/loan uses 1 − (…)^{−tm} (negative exp). Match m to payment frequency (monthly m=12).

13 · The DerivativeCh6

definition (not examined to compute)f'(x) = lim_{k→0} [f(x+k)−f(x)]/k

= instantaneous rate of change = slope of the tangent. Notation: f'(x), dy/dx, df/dx all mean the same thing.

14 · Differentiation RulesCh6 · MEMORISE

core rulespower: xⁿ → n·xⁿ⁻¹ const → 0
[Kf]' = Kf' [f+g]' = f'+g'
aˣ → aˣ·ln a eˣ → eˣ
log_a x → 1/(x ln a) ln x → 1/x
chain: dy/dx = (dy/du)(du/dx)
product: [fg]' = f'g + g'f
quotient: [f/g]' = (f'g − g'f)/g²

rewrite-first worked · y = 3/x²= 3x⁻² ⇒ y' = −6x⁻³ = −6/x³

Trap → Rewrite roots/fractions as powers first (√x=x^{0.5}, 1/x=x⁻¹). Quotient order is f'g − g'f. Chain: ×inner derivative.

15 · Curve ShapeCh6

Sign	Meaning
f'>0 / <0 / =0	↑ / ↓ / stationary
f''>0	concave up (convex)
f''<0	concave down (concave)

Turning pt: f'=0 AND slope changes sign. Inflection: f''=0 AND concavity changes.

worked · y = x³ − 3x²f'=3x²−6x=0 ⇒ x=0,2 (turning pts)
f''=6x−6=0 ⇒ x=1 (inflection)
f''(0)=−6<0 MAX · f''(2)=6>0 MIN

Higher derivatives: f''' and beyond just differentiate again; f'' is the one with economic meaning (curvature, diminishing returns). On a TC curve f''>0 means MC is rising.

Trap → f'=0 ≠ turning point (e.g. y=x³ at 0); always check the sign actually changes.

16 · OptimisationCh6

2nd-derivative test (f'=0)f''<0 ⇒ local MAX
f''>0 ⇒ local MIN
f''=0 ⇒ inconclusive (use f' sign test)

Global: evaluate f at every stationary point + compare.

worked MIN · f=x²−6x+5f'=2x−6=0 ⇒ x=3 · f''=2>0 ⇒ MIN
f(3) = −4

worked MAX · f=−x²+4xf'=−2x+4=0 ⇒ x=2 · f''=−2<0 ⇒ MAX
f(2) = 4

17 · Economic ApplicationsCh6 · exam fave

marginalsMR = dTR/dQ MC = dTC/dQ
revenue max: MR = 0
profit max: MR = MC

Non-linear elasticity: Ed = (dQ/dP)·(P/Q) with calculus for dQ/dP.

worked · P=100−Q, TC=Q²TR=PQ=100Q−Q² ⇒ MR=100−2Q
MC=2Q · MR=MC ⇒ Q=25, P=75
check: revenue max (MR=0) at Q=50 — different!

Trap → To get MR, first write TR=P·Q with P in terms of Q, THEN differentiate. Never differentiate price directly.

17b · Worked DerivativesCh6 · the rules in action

chain · y = (3x²+1)⁵= 5(3x²+1)⁴·6x = 30x(3x²+1)⁴

product · y = x²eˣ= 2x·eˣ + x²·eˣ = eˣ(x²+2x)

quotient · y = (x+1)/(x−1)= [(x−1)−(x+1)]/(x−1)² = −2/(x−1)²

18 · Midterm Blueprint40% · 60 min

Held 19 Apr; covers Wks 1–4 to applications of differentiation: lines, elasticity, quadratics, logs/exponentials, financial maths, differentiation + MR/MC (Side 1). 60 min, formula sheet provided.

Drill: rearrange to y=mx+c · pick FV vs PV annuity · MR=MC profit · 2nd-deriv test. Show work; round at the end.

Bring nothing you don't understand — the formula sheet gives you the quadratic formula, interest/annuity/NPV, elasticity, the aˣ and log derivatives and the quotient rule, but not the power/product/chain rules. Those you must know cold.

★ Formula Belt · Side 1memorise cold

lines & elasticityy = mx + c · m = Δy/Δx
Profit = PQ − TC; break-even Profit=0
budget x₂ = m/p₂ − (p₁/p₂)x₁
Ed = (dQ/dP)(P/Q)

non-linearx = (−b±√(b²−4ac))/(2a)
Δ = b²−4ac
log(ab)=log a+log b · log(aᵇ)=b log a

financePₜ = P₀(1+i)ᵗ · = P₀(1+i/m)^{mt}
continuous Pₜ = P₀e^{it}
NPV = Σ CFₜ/(1+i)ᵗ
annuity PV ∝ 1−(1+i/m)^{−tm}

calculusxⁿ→nxⁿ⁻¹ · eˣ→eˣ · ln x→1/x
chain (dy/du)(du/dx) · aˣ→aˣln a
[fg]'=f'g+g'f · [f/g]'=(f'g−g'f)/g²
2nd-deriv: f''<0 max, f''>0 min
MR=MC at profit max · MR=0 at rev max

★ Exam Disciplinedon't lose easy marks

Rearrange first — get y=mx+c or P=f(Q) before reading anything off
Units & signs — Ed is negative; FC sits in the intercept
m-periods — divide rate, multiply exponent, both
Δ<0 — say "no real solution," don't force it
+C habit even on Side 1 (it matters on Side 2)
Round last — to the stated decimals only

Sia → The single most-tested move on this side is profit maximisation: build TR(Q), differentiate to MR, set MR=MC, solve, back-out P. Drill it until automatic.

★ Quick Self-Checkcan you do these?

Rearrange 3y−6x+9=0 → y=2x−3
Ed of P=50−2Q at Q=10 (P=30) = −1.5
2ˣ = 32 ⇒ x = 5
$5000 at 8% cts for 3yr = 5000e^{0.24} ≈ $6356
d/dx[x²ln x] = 2x ln x + x
Roots of x²−5x+6 = 0 → x = 2, 3
Break-even: TC=100+4Q, P=9 → Q = 20
Turning pt of y=x²−8x+1 → (4, −15)
∫(6x²+2) dx → 2x³ + 2x + C

★ Quiz & Logistics10% · don't lose it

4 online quizzes (10% total), 1 week each, NO extensions — do them early. Q&A via Ed, not email. Textbook: Bradley, Essential Mathematics for Economics & Business, 4th ed. (Chs 2,4,5,6,7,8,9). The unit is maths in the service of economics — every technique gets an economic reading (slope = marginal effect, λ = shadow price, integral = surplus).

ECON1003

Quantitative Methods in Economics

University of Sydney · School of Economics

Exam Revision
Sem 1 2026 · Side 2 of 2
Midterm 40% + Final 50% · formula sheet provided

SIDE 2/2 SEVERAL VARIABLES → MATRICES · Partials · MPL/MPK · MRTS · MRS · the LAGRANGIAN · Integration & surplus · Linear algebra · Cramer · inverse maths, not stats Compiled by AskSia · mapped to the ECON1003 syllabus · asksia.ai/cheatsheet/usyd-econ1003

19 · Partial DerivativesCh7

Method: for ∂f/∂x, treat every other variable as a constant. 2nd-order: f_xx, f_yy; cross-partials f_xy = f_yx (mixed partials equal).

worked · z = x²y + 2y + 4∂z/∂x = 2xy ∂z/∂y = x² + 2

total differentialdz = (∂z/∂x)dx + (∂z/∂y)dy

Use the differential to approximate a small change: if x rises by dx and y by dy, z changes by about dz — handy for "% change in output" questions.

Trap → Track which variable is "the constant" — lone variables & constant-multiple terms vanish or simplify depending on it.

20 · Production · MPL/MPKCh7

Q = f(L,K). MPL = ∂Q/∂L, MPK = ∂Q/∂K. Diminishing returns: Q_LL < 0 (MPL falls as L↑). Cobb–Douglas e.g. Q = 10L^{0.7}K^{0.3}.

worked · Q = 10L^0.7 K^0.3MPL = 7L⁻⁰·³K⁰·³ MPK = 3L⁰·⁷K⁻⁰·⁷

isoquant & MRTSisoquant = all (L,K) with same Q
MRTS = dK/dL = −MPL/MPK

21 · Utility · MRSCh7

U(x,y); MUx = ∂U/∂x, MUy = ∂U/∂y. Indifference curve = all (x,y) with same U.

marginal rate of substitutionMRS = dy/dx = −MUy/MUx

worked MRS · U = x^0.5 y^0.5MUx = 0.5x⁻⁰·⁵y⁰·⁵, MUy = 0.5x⁰·⁵y⁻⁰·⁵
MRS = −MUy/MUx = −x/y

partial elasticitiesprice Ed = (∂Q/∂P_A)(P_A/Q)
income E_Y = (∂Q/∂Y)(Y/Q)
cross Ec = (∂Q/∂P_B)(P_B/Q)

Signs read the goods: Ec>0 substitutes, Ec<0 complements; E_Y>0 normal, E_Y<0 inferior.

22 · Unconstrained Optim.Ch7

2 variablesset ∂f/∂x = 0 AND ∂f/∂y = 0
solve simultaneously, evaluate f

(Full 2nd-order test mentioned, not examined — assume the extremum exists if asked.)

worked · f = x² + y² − 4x − 6y2x−4=0 ⇒ x=2 · 2y−6=0 ⇒ y=3
min at (2,3), f = −13

Trap → The two first-order conditions are simultaneous — substitute one into the other if they're coupled.

23 · The LagrangianCh7 · LECTURE NOTES

Max f(x,y) s.t. equality h(x,y)−c=0 & inequality g(x,y)−b ≤ 0 constraints. Each constraint gets its own multiplier λ.

build it (minus each constraint)L = f − λ₁(h₁−c₁) − λ₂(g₁−b₁) − …

conditions — ALL must hold1. ∂L/∂x = 0, ∂L/∂y = 0
2. all constraints hold
3. inequality multipliers λ ≥ 0
4. compl. slackness: λⱼ(gⱼ−bⱼ) = 0

Method: write L → list conditions → solve case-by-case (x=0? y=0? both >0?) → evaluate f at each → pick best. Min f = max(−f).

case logic for one inequality g−b≤0Case A: λ=0 (slack) → ignore constraint, check g≤b
Case B: g=b (binds) → solve with λ≥0

Trap → Complementary slackness spawns cases — students solve only the interior & forget the x=0, y=0 corners. Reject any candidate forcing an inequality λ negative.

23b · Worked · Utility Maxthe whole move

max U=xy s.t. 2x+y=10L = xy − λ(2x+y−10)
L_x: y − 2λ = 0 ⇒ y = 2λ
L_y: x − λ = 0 ⇒ x = λ
⇒ y = 2x; sub: 2x+2x=10 ⇒ x=2.5
x=2.5, y=5, λ=2.5, U=12.5

At the optimum MRS = price ratio: MUx/MUy = y/x = 2 = p₁/p₂. ✓

Cost-min mirrors it: min C s.t. Q̄ = f(L,K) gives the tangency MRTS = w/r (input-price ratio).

24 · The Multiplier λCh7 · shadow price

λ = shadow price: how much the optimal objective changes if you relax that constraint by one unit.

In utility-max s.t. budget: λ = marginal utility of income (above, λ=2.5 ⇒ +$1 income raises U by ≈2.5). In cost-min s.t. output: λ = marginal cost of one more unit. Uses: utility maximisation & cost minimisation.

Sia → If asked "what does λ mean?", answer in units of the objective per unit of the constraint — that one sentence is worth marks.

24b · Lagrangian Checklistbefore you stop

All ∂L/∂(var) = 0; constraints checked
Inequality λ's all ≥ 0
Complementary slackness on each ≤
Corner cases (x=0, y=0) evaluated
f compared across ALL candidates

25 · The IntegralCh8

anti-derivative∫f(x)dx = g(x) + C, where g'(x)=f(x)
ALWAYS add + C (indefinite)

rules∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫x⁻¹ dx = ln(x) + C (the n=−1 case)
∫K dx = Kx + C · ∫eˣ dx = eˣ + C

Trap → Power rule breaks at n=−1 (÷0) — that case is the log. Dropping +C is fatal once you solve for C in an ODE.

26 · SubstitutionCh8

Reverse of the chain rule: set u = inner fn, find du, rewrite all in u, integrate, sub back.

worked · ∫(5x−2)¹⁰ dxu=5x−2, dx=du/5
⇒ (5x−2)¹¹/55 + C

worked · ∫2x·e^{x²} dxu=x² ⇒ du=2x dx ⇒ ∫eᵘ du = e^{x²}+C

Trap → The leftover x's must all cancel via du — if any remain, substitution isn't the right tool.

27 · By PartsCh8 · on sheet

lecturer's form∫f g dx = f·∫g dx − ∫f'·(∫g dx) dx

worked · ∫ln x dxf=ln x, g=1 ⇒ x ln x − x + C

worked · ∫x eˣ dxf=x, g=eˣ ⇒ x eˣ − ∫eˣ dx = eˣ(x−1)+C

Trap → Pick f (to differentiate) so f' simplifies — logs & powers of x are good f.

28 · Definite Integral & AreaCh8

net signed area∫ₐᵇ f dx = F(b) − F(a)
between curves: ∫ₐᵇ [f − g] dx (f above g)

worked · ∫₁³ (2x+1) dx= [x²+x]₁³ = (9+3) − (1+1) = 10

Trap → Gives NET area — below-axis counts negative (∫₋₃³ x dx = 0). For "total area" split at crossings.

28b · TC from MCCh8 · econ use

recover total costTC = ∫ MC dQ + FC
the constant of integration = fixed cost

worked · MC = 3Q²+2, FC=50TC = Q³ + 2Q + 50

29 · Surplus & ODEsCh8

consumer / producer surplusCS = ∫₀^{Q*} [demand(Q) − P*] dQ
PS = ∫₀^{Q*} [P* − supply(Q)] dQ

Find equilibrium Q* (from P*) first — it's the upper limit.

worked CS · D: P=20−Q, P*=8Q*=12 ⇒ CS = ∫₀¹² (20−Q−8) dQ
= [12Q − Q²/2]₀¹² = 144 − 72 = 72

worked PS · S: P=2+Q, P*=8Q*=6 ⇒ PS = ∫₀⁶ (8−2−Q) dQ
= [6Q − Q²/2]₀⁶ = 36 − 18 = 18

differential equationsdy/dx=f(x): integrate, use a point to fix C
separable dy/dx=f(x)g(y):
collect y left, x right, integrate both

Economic uses: TC = ∫MC dQ; accumulated quantity from a rate.

worked separable · dy/dx = xy∫(1/y)dy = ∫x dx
ln y = x²/2 + C ⇒ y = Ae^{x²/2}

The classic econ ODE dP/dt = kP separates to P = P₀e^{kt} — the same exponential growth seen in continuous interest.

Trap → Keep + C until the initial condition is applied; on separable eqns remember the ln/exp step when re-isolating y.

30 · Matrices · BasicsCh9

Dimension = rows × cols; equal iff same dim & all entries match. Transpose Aᵀ: swap rows↔cols (a×b → b×a). Special: null, square, identity I (1's on diagonal).

arithmeticadd/subtract: element-wise, SAME dim only
scalar: × every entry
A(m×n)·B(n×k) = m×k; cols A = rows B

worked · product (entry = row·col)[[1,2],[3,4]]·[[5],[6]] = [[17],[39]]
(1·5+2·6=17 · 3·5+4·6=39)

Identity I acts like 1: AI = IA = A. Transposing reverses a product: (AB)ᵀ = BᵀAᵀ.

Trap → Multiplication is not commutative — AB ≠ BA (one may not even be defined). Check dims before multiplying.

30b · Economic UseCh9 · why bother

A market or input–output model is a system of linear equations; in matrix form AX = B. Solving it (Gaussian, Cramer, or the inverse) gives equilibrium prices/quantities all at once. The determinant flags whether a unique equilibrium even exists.

31 · EliminationCh9

Strip variables → augmented matrix [A|b]. 3 row ops (none change the solution): swap rows · ×non-zero · add a multiple of one row to another.

Gaussian → row-echelon (more leading zeros each row down), then back-substitute. Gauss–Jordan → reduced row-echelon (1's on diagonal, 0's elsewhere) ⇒ read the solution off directly.

outcomes from the echelon formfull diagonal → unique solution
0 = 0 row → infinitely many
0 = nonzero row → no solution

Trap → Arithmetic slips in row ops are the #1 error — keep the RHS column attached and operate on the whole row.

32 · DeterminantsCh9 · 3×3 on sheet

2×2det[[a,b],[c,d]] = ad − bc

worked 3×3 · diag-heavydet[[1,2,0],[0,3,1],[0,0,4]] (triangular)
= product of diagonal = 1·3·4 = 12

Trap → The alternating + − + signs on the 3×3 are routinely forgotten. (Triangular det = product of the diagonal — a fast check.)

33 · Cramer's RuleCh9 · MEMORISE

solve Ax = bx = det(Bₓ)/det(A) · y = det(B_y)/det(A)
Bᵥₐᵣ = A with that column replaced by b
requires det(A) ≠ 0

worked · 2x+y=5, x+3y=10det A = 2·3−1·1 = 5
x = det[[5,1],[10,3]]/5 = (15−10)/5 = 1
y = det[[2,5],[1,10]]/5 = (20−5)/5 = 3
check: 2(1)+3=5 ✓

Trap → Replace the correct column per variable; the method fails entirely when det(A)=0 (no unique solution).

33b · Which Method?Ch9 · pick fast

One unknown wanted → Cramer (fewer dets)
Same A, many b's → find A⁻¹ once
Big / numeric system → Gauss–Jordan

34 · The Inverse MatrixCh9 · MEMORISE

defn — exists iff det(A) ≠ 0A·A⁻¹ = A⁻¹·A = I

cofactor methodA⁻¹ = Cᵀ / det(A)
C = cofactors (minor × +−+ sign)

2×2 shortcutA⁻¹ = (1/(ad−bc))·[[d,−b],[−c,a]]

worked · A = [[2,1],[1,1]]det = 2−1 = 1 ⇒ A⁻¹ = [[1,−1],[−1,2]]

solve a systemAX = B ⇒ X = A⁻¹B

Trap → Cofactor method: don't forget to TRANSPOSE (Cᵀ) and divide by det(A). If det(A)=0 there is no inverse.

★ Formula Belt · Side 2memorise cold

several variablesMPL=∂Q/∂L · MPK=∂Q/∂K
MRTS=−MPL/MPK · MRS=−MUy/MUx
L = f − λ(constraint) · λ = shadow price
compl. slackness λ(g−b)=0

integration∫xⁿ=xⁿ⁺¹/(n+1)+C (n≠−1)·∫x⁻¹=ln x+C
by parts: f∫g − ∫f'(∫g)
CS=∫₀^{Q*}(D−P*) · PS=∫₀^{Q*}(P*−S)

linear algebradet 2×2 = ad−bc
Cramer xᵢ = det(Bᵢ)/det(A)
A⁻¹ = Cᵀ/det(A) · X = A⁻¹B

On the sheet	You memorise
3×3 determinant	Cramer's rule
aˣ, log_a, quotient rule	inverse methods
by-parts formula	Lagrangian conditions

★ Final Exam Disciplineall 7 topics

Cumulative — Side 1 maths returns; redo profit-max & finance
Lagrangian — list ALL 4 conditions; check corners
Integration — n=−1 → ln; +C; find Q* before surplus
Matrices — dims first; det≠0 before Cramer/inverse; transpose cofactors
Surplus — equilibrium Q* is the upper limit, not P*
Show work — partial credit; round only at the end

Sia → The big Side-2 earners are the Lagrangian (with shadow-price interpretation) and a full matrix solve (Cramer or inverse) — rehearse one of each end-to-end the night before.