Mathematics 1A

0 · Exam Blueprintread first

MATH1061 runs two parallel streams: this side is Calculus (limits → derivatives → integrals → series); flip for Linear Algebra. Assessment: weekly quizzes 8% · A1 5% · A2 10% · in-person Quiz A 15% · tutorials 2% · final exam 60%.

Most-tested moves: 0/0 limits (factor or L'Hôpital); differentiate with chain + product/quotient; classify critical points; evaluate a definite integral via FTC + a technique (sub / parts / partial fractions); write a Maclaurin series and use it.

Method marks: show the working — state the rule, then the substitution, then the answer. A dropped chain-rule factor or a missed +C is the standard mark-loss.

1 · Functions & LimitsWk 1–2

Function f:A→B, one output per input; range = image ⊆ codomain. Injective (1-1), surjective (onto, range = codomain), bijective = both ⇒ f⁻¹ exists. Composition (g∘f)(x)=g(f(x)).

Limit. lim_x→a f(x)=L: f(x) is forced arbitrarily close to L by taking x close to (≠) a. Two-sided limit exists iff both one-sided limits exist and agree.

Limit laws (if both limits exist)

algebra of limitslim(kf)=k·lim f · lim(f±g)=lim f ± lim g
lim(fg)=(lim f)(lim g)
lim(f/g)=lim f / lim g  only if lim g≠0

If f is continuous at a, lim_x→a f(x)=f(a) — so most limits are "plug in"; only the joints of piecewise functions need care. Never use the quotient law when the denominator limit is 0 — factor, rationalise or use L'Hôpital instead.

1b · Squeeze & Standard Limits★ memorise

squeeze (sandwich)g≤f≤h near a, lim g = lim h = L ⇒ lim f = L

standard limitslim_x→0 sin x / x = 1
lim_x→0 (1−cos x)/x = 0
lim_x→∞ (1+1/x)^x = e

Classic squeeze: −|x| ≤ x·sin(1/x) ≤ |x| ⇒ lim_x→0 x sin(1/x)=0.

1c · 0/0 Limits · Workedfactor first

lim_x→1 (x²−1)/(x²+x−2):

= (x−1)(x+1) / (x−1)(x+2)
= (x+1)/(x+2) → 2/3

Cancel the common factor causing the 0; the limit can exist even when f(1) is undefined.

Rationalise. lim_x→0 (√(1+x)−1)/x · (√(1+x)+1)/(√(1+x)+1) = 1/(√(1+x)+1) → 1/2.

At infinity. Divide by the highest power: lim_x→∞ (3x²+1)/(2x²−x) = 3/2 (horizontal asymptote).

2 · Continuity & IVTWk 3

f is continuous at a iff lim_x→a f(x)=f(a) (all three: limit exists, f(a) defined, equal). Polynomials, roots, e^x, ln, trig are continuous on their domains; sums/products/quotients/compositions stay continuous.

Intermediate Value Theoremf continuous on [a,b], N between f(a) & f(b)
⇒ ∃ c∈[a,b] with f(c)=N

Use IVT to show a root exists: f(a)<0<f(b) ⇒ some c with f(c)=0. Needs a closed interval; gives existence, not the value.

Inverse & hyperbolic. Restrict the domain so f is injective before inverting; f⁻¹ is the reflection of f in y=x. Key examples: cosh x=(e^x+e^−x)/2, sinh x=(e^x−e^−x)/2, with cosh²x − sinh²x = 1.

3 · The DerivativeWk 3–4

definitionf′(a)=lim_h→0 [f(a+h)−f(a)] / h
= slope of tangent at (a,f(a))

Tangent line: y = f(a) + f′(a)(x−a). Differentiable ⇒ continuous (not conversely — |x| has a corner at 0).

Rules

linearity / product / quotient / chain(kf)′=kf′ · (f±g)′=f′±g′
(fg)′ = f′g + fg′  (product)
(f/g)′ = (f′g − fg′)/g²  (quotient)
(g∘f)′(x) = g′(f(x))·f′(x)  (chain)

Implicit: differentiate F(x,y)=0 in x, chain-rule the y-terms, solve y′. e.g. x²+y²=1 ⇒ 2x+2yy′=0 ⇒ y′=−x/y.

Logarithmic: y=x^x ⇒ ln y = x ln x ⇒ y′/y = ln x + 1 ⇒ y′ = x^x(ln x + 1).

Quotient worked. d/dx (sin x / x) = (x cos x − sin x)/x². Chain worked. d/dx sin(x²) = cos(x²)·2x.

Higher derivatives f″, f‴ feed the second-derivative test and Taylor coefficients; for a product use the product rule repeatedly (or Leibniz's formula). Differentiable ⇒ continuous, but not the reverse (corners and cusps).

4 · Standard Derivativesd/dx · ★ table

cosh²x − sinh²x = 1. Combine with the chain rule for any composite, e.g. d/dx ln(f(x)) = f′(x)/f(x).

5 · L'Hôpital's RuleWk 5

indeterminate 0/0 or ∞/∞lim_x→a f/g = lim_x→a f′/g′
(when the RHS exists)

Check the form first. Other forms: 0·∞ → rewrite as 0/0 or ∞/∞; for 1^∞, 0⁰, ∞⁰ take logs, then apply.

Worked. lim_x→0 (e^x−1−x)/x² is 0/0 → (e^x−1)/2x still 0/0 → e^x/2 → 1/2.

0·∞ worked. lim_x→0⁺ x ln x = lim (ln x)/(1/x) = (∞/∞) → (1/x)/(−1/x²) = −x → 0.

Warning: never apply L'Hôpital to a determinate form (e.g. 3/0 or 5/2) — re-check the form after every step.

6 · Extrema & Curve SketchingWk 5–6

First derivative test

• f′>0 ⇒ increasing; f′<0 ⇒ decreasing
• Critical point f′(c)=0: necessary, not sufficient (x³ at 0)
• f′: −→+ at c ⇒ local min; +→− ⇒ local max

Second derivative test

• f″>0 ⇒ concave up; f″<0 ⇒ concave down
• f′(c)=0, f″(c)>0 ⇒ min; f″(c)<0 ⇒ max
• Inflection: concavity changes (f″=0 necessary, check sign change)

Sketch checklist: domain · intercepts · sign of f′ (incr/decr, crit pts) · sign of f″ (concavity, inflections) · end behaviour x→±∞.

6b · Optimisation · Workedmethod marks

Method: write the quantity & its domain → f′(x)=0 for critical points → classify (1st/2nd test) → compare with endpoints for the global optimum.

Eg. max area of a rectangle of perimeter 20: A=x(10−x), A′=10−2x=0 ⇒ x=5; A″=−2<0 ⇒ max. A=25 (a square).

Trap: f″(c)=0 does not force an inflection; never forget endpoint values on a closed interval.

7 · Riemann IntegralWk 9

definite integral = signed area∫_a^b f dx = lim_N→∞ Σ_k f(x_k*)Δx
Δx = (b−a)/N

Exists for continuous f. Lower/upper sums L_N ≤ ∫ ≤ U_N bound it (handy for monotonic f). Sub-intervals partition a=x₀<x₁<…<x_N=b; the sample point x_k* ∈ [x_k−1,x_k].

basic properties∫_a^a f = 0 · ∫_a^b f = −∫_b^a f
∫_a^b(αf+βg) = α∫f + β∫g (linearity)
∫_a^c f + ∫_c^b f = ∫_a^b f

Signed area: regions below the x-axis count negative — split at the zeros if you want true geometric area.

7b · Curve Sketch · Workedf=x³−3x

f′=3x²−3=0 ⇒ x=±1; f″=6x. x=−1: f″<0 ⇒ local max f=2; x=1: f″>0 ⇒ local min f=−2. Inflection at x=0. Odd, roots 0,±√3, ends ±∞. The sign charts of f′ and f″ give the whole shape. On a closed interval, also test endpoints for the global extremum; a critical point alone need not be one.

8 · Fundamental TheoremFTC · Wk 10

FTC I & III: d/dx ∫_c^x f(t) dt = f(x)
II: ∫_a^b F′(x) dx = F(b) − F(a)

variable limits (Leibniz)d/dx ∫_c^g(x) f(t) dt = f(g(x))·g′(x)
d/dx ∫_x^c f(t) dt = −f(x)

Don't drop g′(x) on a variable upper limit; the sign flips for a variable lower limit.

FTC worked. ∫₀^π/2 cos x dx = [sin x]₀^π/2 = 1 − 0 = 1. And d/dx ∫₀^x² sin t dt = sin(x²)·2x.

Part I → II. Part I says ∫_c^x f is an antiderivative of f; Part II evaluates any antiderivative at the endpoints. Continuity of f on [a,b] is the hypothesis both need; FTC is what turns "antiderivative" into a number.

9 · Antiderivatives∫ · ★ table

Always +C on an indefinite integral.

10 · Substitution & PartsWk 10–11

substitution (undo chain)∫ f(u(t))·u′(t) dt = ∫ f(s) ds,  s=u(t)
change the limits in the definite case

by parts (undo product)∫ u dv = uv − ∫ v du
choose u by LIATE

LIATE = Log · Inverse-trig · Algebraic · Trig · Exp — pick the earlier type as u (its derivative simplifies). Eg ∫ x e^x dx: u=x, dv=e^xdx ⇒ x e^x − e^x + C.

ln by parts: ∫ ln x dx: u=ln x, dv=dx ⇒ x ln x − ∫ x·(1/x) dx = x ln x − x + C.

Sub worked: ∫ 2x e^x² dx, s=x², ds=2x dx ⇒ ∫ e^s ds = e^x² + C. For a definite integral, change the limits too (don't back-substitute).

Cyclic parts: ∫ e^x cos x dx — parts twice returns the original integral I; solve algebraically: I = ½e^x(sin x + cos x) + C.

Definite sub: ∫₀¹ 2x(x²+1)³ dx, s=x²+1 (1→2): ∫₁² s³ ds = [s⁴/4]₁² = 15/4.

x² by parts (twice): ∫ x² e^x dx = x²e^x − 2∫ x e^x dx = x²e^x − 2(xe^x − e^x) + C = e^x(x²−2x+2) + C. A polynomial × e^x/sin/cos: parts lowers the polynomial degree each pass.

11 · Partial Fractions & Trig Subrational + roots

distinct linear factors(a+bx)/[(x−λ)(x−μ)] = A/(x−λ) + B/(x−μ)

Solve A,B by cover-up / equating coefficients. Eg 1/[(x+1)(x−1)] = (−½)/(x+1) + (½)/(x−1). Reduce an improper fraction (divide) before splitting.

Trig integrals & substitution

∫ sin^mx cosⁿx dx: if a power is odd, peel one factor & use sin²+cos²=1; if both even, use sin²x=(1−cos2x)/2, cos²x=(1+cos2x)/2.

trig substitution√(a²−x²) ⇒ x=a sinθ
√(a²+x²) ⇒ x=a tanθ
√(x²−a²) ⇒ x=a secθ

PF integral worked. ∫ dx/[(x+1)(x−1)] = ∫(−½/(x+1) + ½/(x−1))dx = ½ ln|(x−1)/(x+1)| + C.

Trig-sub idea. ∫ dx/√(1−x²) with x=sinθ, dx=cosθ dθ ⇒ ∫ dθ = θ = sin⁻¹x + C (recovers the standard antiderivative). Repeated factors (x−λ)² need a B/(x−λ)² term too; an irreducible quadratic gets a (Cx+D)/(quadratic) term.

11b · Improper IntegralsWk 11–12

Defined as limits: ∫_a^∞ f = lim_b→∞ ∫_a^b f; singularity at c handled with a one-sided limit. Converges iff the limit is finite.

benchmarks∫₁^∞ dx/x^p converges iff p>1
∫₀¹ dx/x^p converges iff p<1

Worked. ∫₁^∞ dx/x² = lim_b→∞ [−1/x]₁^b = lim (1 − 1/b) = 1 (converges, p=2>1).

Trap: a singularity inside [a,b] must be split, and both pieces must converge for the whole integral to converge.

12 · Geometric ApplicationsWk 12

area between curves (f≥g)A = ∫_a^b [f(x) − g(x)] dx  (top − bottom)

arc length of y=f(x)L = ∫_a^b √(1 + (f′(x))²) dx

volume of revolutionabout x-axis (disc): V = ∫_a^b π f(x)² dx
about y-axis (shell): V = ∫_a^b 2π x f(x) dx

Trap: subtract in the right order; don't swap the disc (πf²) and shell (2πxf) formulas.

Area worked. Between y=x and y=x² on [0,1]: x≥x² there, so A = ∫₀¹(x−x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.

Volume worked. y=√x on [0,4] about the x-axis: V = ∫₀⁴ πx dx = π[x²/2]₀⁴ = 8π.

Arc length idea. For y=f(x), the element ds=√(1+(f′)²)dx comes from Pythagoras on (dx, dy); integrate it for total length.

Find the intersection first. For area between curves, set f(x)=g(x) to get the limits a,b, then integrate top−bottom; if they cross inside, split the integral at the crossing. Sketch first to see which curve is on top.

13 · Taylor & MaclaurinWk 7–8

Taylor polynomial about aT_n(x) = Σ_k=0ⁿ f^(k)(a)/k! · (x−a)^k
= f(a) + f′(a)(x−a) + f″(a)/2!·(x−a)² + …

a=0 ⇒ Maclaurin. T_n matches f and its first n derivatives at a.

remainder R_n = f − T_nLagrange: R_n = f⁽ⁿ⁺¹⁾(c)/(n+1)! · (x−a)ⁿ⁺¹
series equals f only where R_n→0

13b · Standard Series★ know these

Maclaurin seriese^x = Σ x^k/k!  (all x)
sin x = Σ (−1)^k x^2k+1/(2k+1)!
cos x = Σ (−1)^k x^2k/(2k)!
ln(1+x) = Σ (−1)^k−1 x^k/k  (|x|<1)
(1+x)^α = Σ C(α,k) x^k  (|x|<1)

geometric1+x+…+xⁿ = (1−xⁿ⁺¹)/(1−x)
Σ_k≥0 x^k = 1/(1−x)  (|x|<1)

Term-by-term trick: substitute / differentiate / integrate a known series rather than recompute derivatives.

From Σx^k=1/(1−x): differentiate ⇒ Σ k x^k−1 = 1/(1−x)²; integrate −x into 1/(1+x) ⇒ recovers ln(1+x).

Geometric series value: Σ_k≥0 ar^k = a/(1−r) for |r|<1. Eg Σ (1/2)^k = 1/(1−½) = 2; diverges for |r|≥1. The partial sum is a(1−rⁿ⁺¹)/(1−r).

13c · Taylor · Workeduse a known series

Maclaurin of cos x to x⁴: cos x ≈ 1 − x²/2 + x⁴/24.

So lim_x→0 (1−cos x)/x² = 1/2 (the x²/2 term dominates) — faster than two L'Hôpitals.

And e^x² = 1 + x² + x⁴/2 + … by substituting x² into e^x — no new derivatives needed. Integrate term-by-term ⇒ ∫ e^x² dx series (no elementary closed form). The first non-zero term controls the small-x behaviour.

About a≠0. ln x about a=1: f(1)=0, f′=1/x→1, f″=−1/x²→−1 ⇒ ln x ≈ (x−1) − (x−1)²/2 + (x−1)³/3 − … (matches ln(1+u), u=x−1). Build coefficients f^(k)(a)/k! one derivative at a time.

Radius / convergence. e^x, sin, cos converge for all x; ln(1+x) and (1+x)^α only for |x|<1; geometric for |x|<1. Always state the radius before equating a series to f. Off-by-one in the factorial/power is the standard slip.

Calculus Formula Beltside 1

(g∘f)′=g′(f)·f′ · (f/g)′=(f′g−fg′)/g²
∫ u dv = uv − ∫ v du · LIATE
FTC: ∫_a^bF′=F(b)−F(a) · d/dx∫_c^xf=f(x)
L'Hôpital: 0/0,∞/∞ ⇒ lim f′/g′
Σx^k=1/(1−x) (|x|<1)

14 · Complex NumbersWk 1–2

i²=−1; z=a+ib, Re(z)=a, Im(z)=b (both real). ℝ⊂ℂ.

arithmetic(a+ib)+(c+id) = (a+c)+i(b+d)
(a+ib)(c+id) = (ac−bd)+i(ad+bc)

Conjugate z̄=a−ib: z z̄ = a²+b² ∈ ℝ, and z̄w̄ = z̄·w̄, z+w‾ = z̄+w̄.

division — realise the denominatorw/z = w z̄ / (z z̄) = w z̄ / |z|²

Equality: a+ib = c+id ⟺ a=c and b=d.

Powers of i cycle: i, −1, −i, 1, i, … (period 4) — reduce the exponent mod 4.

Division worked. (3+i)/(1−2i) · (1+2i)/(1+2i) = (3+6i+i−2)/(1+4) = (1+7i)/5 = 1/5 + (7/5)i.

Quadratic over ℂ. z²+z+1=0 ⇒ z = (−1±√(−3))/2 = −1/2 ± (√3/2)i — a conjugate pair, the primitive cube roots of unity (≠1). The quadratic formula works over ℂ with √ of a negative; the discriminant being negative is what forces complex roots.

15 · Modulus, Polar & EulerArgand plane

Modulus |z|=√(a²+b²)=√(z z̄) = distance from 0; |z−w| = distance z↔w. |zw|=|z||w|, |z/w|=|z|/|w|, triangle ineq |z+w|≤|z|+|w|.

Argument arg z=θ with z=|z|(cosθ+i sinθ); principal Arg z∈(−π,π]. Read the quadrant from the diagram — not just tan⁻¹(b/a).

polar / exponential (Euler)e^iθ = cosθ + i sinθ  ⇒ z = r e^iθ
z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)}
z₁/z₂ = (r₁/r₂) e^{i(θ₁−θ₂)}

Euler's identity: e^iπ + 1 = 0. Multiplying = multiply moduli, add arguments.

Useful identities: Re(z)=(z+z̄)/2, Im(z)=(z−z̄)/2i, z⁻¹ = z̄/|z|² (|z|=1 ⇒ z⁻¹=z̄), and cosθ=(e^iθ+e^−iθ)/2, sinθ=(e^iθ−e^−iθ)/2i.

15b · de Moivre & RootsWk 2

de Moivre(cosθ + i sinθ)ⁿ = cos nθ + i sin nθ
(r e^iθ)ⁿ = rⁿ e^inθ

n-th roots of w = R e^iφz_k = R^1/n exp(i(φ+2kπ)/n)
k = 0,1,…,n−1  (spaced 2π/n apart)

Roots of unity (zⁿ=1): e^2πik/n — n points equally spaced on the unit circle. List all n roots (run k=0…n−1).

Worked. Cube roots of 8: 8=8eⁱ⁰, R^1/3=2, angles 0, 2π/3, 4π/3 ⇒ 2, −1+i√3, −1−i√3.

Polar worked. z=1+i: r=√2, θ=π/4 ⇒ z=√2 e^iπ/4; z⁸ = (√2)⁸ e^i2π = 16.

Argand regions. |z|=2 is a circle radius 2; |z−i|≤2 is a disc centred at i; Re(z)>−1 is a half-plane. Read loci as distances/angles, not coordinates.

de Moivre for identities: expand (cosθ+i sinθ)³ and match real/imaginary parts ⇒ cos3θ = 4cos³θ − 3cosθ.

Modulus worked. z=3−4i: |z|=√(9+16)=5; arg in the 4th quadrant, θ=tan⁻¹(−4/3)≈−53°. z̄=3+4i, z z̄=25.

The n roots of unity sum to 0 (for n≥2) and are the vertices of a regular n-gon on the unit circle — a quick sketch checks your answer. One root is always z=1; the rest are its powers ω^k.

16 · Polynomials over ℂroots

Complex exponential. z=a+ib ⇒ e^z=e^a(cos b + i sin b); |e^z|=e^{Re z}, arg(e^z)=Im z. Solving e^z=w gives ∞-many solutions (b mod 2π).

Fundamental Theorem of Algebra. Every non-constant p(z) has a root in ℂ; counted with multiplicity, degree n ⇒ exactly n roots.

Conjugate root theorem. p with real coefficients, α a root ⇒ ᾱ a root. Then (z−α)(z−ᾱ) = z² − 2Re(α)z + |α|² divides p; divide out to find the rest.

Worked. z⁴+1=0: roots are the 4th roots of −1 = e^iπ, i.e. e^iπ/4, e^i3π/4, e^i5π/4, e^i7π/4 — two conjugate pairs, so z⁴+1 = (z²−√2z+1)(z²+√2z+1) over ℝ.

17 · Vectors in ℝⁿWk 3–4

AB⃗ = OB⃗ − OA⃗ = [b₁−a₁,…] (head − tail). a+b head-to-tail; ca scales.

length / unit vector‖a‖ = √(a₁²+…+aₙ²) · ‖ca‖=|c|‖a‖
â = a/‖a‖

Dot product (scalar)

u·v & angleu·v = u₁v₁+…+uₙvₙ · ‖u‖=√(u·u)
u·v = ‖u‖‖v‖cosθ ⇒ cosθ = u·v / (‖u‖‖v‖)

Orthogonal ⟺ u·v=0. Cauchy–Schwarz |u·v|≤‖u‖‖v‖.

17b · ProjectionWk 4

project v onto uproj_u(v) = (u·v / ‖u‖²) u = (û·v) û

Decompose v = a + b: a ∥ u, b = v − proj_u(v) ⊥ u. Distance point P → line (dir u, A on line): ‖AP⃗ − proj_u(AP⃗)‖.

Trap: divide by ‖u‖², not ‖u‖; u·v is a scalar, not a vector.

Worked. v=[3,4] onto u=[1,0]: u·v=3, ‖u‖²=1 ⇒ proj=[3,0]; orthogonal part [0,4]; so dist from (3,4) to the x-axis = 4.

18 · Cross Productℝ³ only · Wk 5

u×v= [u₂v₃−u₃v₂, u₃v₁−u₁v₃, u₁v₂−u₂v₁]
= det[ i j k ; u₁ u₂ u₃ ; v₁ v₂ v₃ ]

Orthogonal to both u,v (right-hand rule). Anti-commutative u×v=−v×u; u×u=0; not associative. ‖u×v‖ = ‖u‖‖v‖ sinθ = parallelogram area.

Worked. u=[1,0,0], v=[0,1,0]: u×v = [0,0,1] = k (right-hand rule), area = 1. And v×u = [0,0,−1] = −(u×v).

Scalar triple u·(v×w) = det[u;v;w] = volume of the parallelepiped; zero ⇒ the three vectors are coplanar.

Triangle area = ½‖AB⃗ × AC⃗‖. Eg A,B,C with AB⃗=[1,0,0], AC⃗=[0,2,0] ⇒ ‖[0,0,2]‖/2 = 1.

Parallel test: u ∥ v iff u×v=0 (or v=cu). Perpendicular test: u ⊥ v iff u·v=0. Two quick checks for line/plane relations.

Dot worked. u=[1,2,2], v=[2,0,1]: u·v = 2+0+2 = 4, ‖u‖=3, ‖v‖=√5 ⇒ cosθ = 4/(3√5) ≈ 0.596 ⇒ θ ≈ 53°.

Orthogonality solve. [1,k,2]·[3,−1,1]=0 ⇒ 3−k+2=0 ⇒ k=5 makes them perpendicular.

The cross product lives only in ℝ³; in ℝ² or ℝ⁴ use dot product, projection and determinants instead. Distance from P to a line = length of the orthogonal component; the sign of u·v shows acute (+) vs obtuse (−) angle, and u·v=0 is a right angle.

Cauchy–Schwarz |u·v|≤‖u‖‖v‖ guarantees cosθ∈[−1,1], so the angle formula always makes sense.

19 · Lines & PlanesWk 5–6

line — vector / parametricx = p + t d,  t∈ℝ  (d≠0 direction)
x_i = p_i + t d_i · through A,B: d = AB⃗

plane in ℝ³ — normal formn·(x − p) = 0 ⟺ ax+by+cz = d
n = [a,b,c]

A plane = point + normal (or point + two non-parallel directions, then n = d₁×d₂). Line in ℝ²: n·x = n·p ⟺ ax+by=c with n·d=0.

Lines in ℝ³ have no normal/general form. Relations: coincide / intersect / parallel; in ℝ³ also skew. Trap: don't conclude intersection after matching only 2 of 3 coordinates.

20 · Linear SystemsWk 7

Linear eqn a₁x₁+…+aₙxₙ=b (no products/powers). Augmented matrix [A|b]; homogeneous if all b_i=0.

Elementary row operations (preserve solutions)

1. Swap R_i↔R_j
2. Scale R_i→cR_i (c≠0)
3. R_i→R_i+cR_j

REF: zero rows at bottom, each leading entry right of the one above. RREF: + leading 1s, zeros elsewhere in their column (RREF is unique). Gaussian = REF + back-sub; Gauss–Jordan = RREF.

Outcomes: a row [0 … 0 | c], c≠0 ⇒ inconsistent. Else non-leading columns = free variables ⇒ unique (none) or ∞-many (≥1).

Worked. x+y=3, 2x−y=0: R2−2R1 ⇒ −3y=−6 ⇒ y=2, x=1. As [A|b]: [1 1|3; 2 −1|0] → [1 1|3; 0 −3|−6] → x=1, y=2.

21 · Matrix AlgebraWk 8

A=[a_ij], size m×n. Add (same size) & scale entrywise.

multiplication (inner dims match)[AB]_ij = Σ_k a_ikb_kj = (row i of A)·(col j of B)
A is p×q, B is q×r ⇒ AB is p×r

AB ≠ BA in general; associative; I A = A. AB=O does not imply A=O or B=O. System: Ax=b.

Worked. [1 2; 3 4][5; 6] = [1·5+2·6; 3·5+4·6] = [17; 39]. Powers A^k need A square.

Geometry of solutions (3 unknowns): planes meeting in a point (unique) / in a line (∞-many, 1 free var) / not at all (inconsistent).

Plane worked. Through P=(1,0,2) with normal n=[2,−1,3]: 2(x−1)−(y)+3(z−2)=0 ⇒ 2x − y + 3z = 8. Two planes are parallel iff their normals are; the angle between planes equals the angle between their normals.

Homogeneous Ax=0 is always consistent (x=0 works); it has a nonzero solution iff there is a free variable iff det A=0 (square A).

Rank. rank(A) = number of leading entries in REF = number of pivots. For an n-variable system: unique solution iff rank = n (no free variables). More unknowns than equations ⇒ at least one free variable.

Inconsistent example. x+y=1, x+y=2 reduces to [1 1|1; 0 0|1] — the row [0 0|1] is the contradiction 0=1 ⇒ no solution (parallel lines).

EROs never change the solution set, so the REF/RREF you reach is logically the same system — just easier to read off. RREF is unique; REF is not. Set each non-leading variable as a free parameter, then write the leading ones in terms of them.

22 · Transpose & InverseWk 9–10

transpose(Aᵀ)ᵀ=A · (A+B)ᵀ=Aᵀ+Bᵀ
(AB)ᵀ = BᵀAᵀ  (order reverses!)

Inverse: AB=BA=I ⇒ B=A⁻¹ (unique). (A⁻¹)⁻¹=A; (AB)⁻¹ = B⁻¹A⁻¹ (order reverses); (Aᵀ)⁻¹=(A⁻¹)ᵀ.

2×2 inverseA=[a b; c d] invertible ⟺ ad−bc≠0
A⁻¹ = 1/(ad−bc) · [d −b; −c a]

n×n via row reduction: [A | I] → [R | B]. R=I ⇒ A⁻¹=B; zero row in R ⇒ not invertible. Then x = A⁻¹b solves Ax=b.

2×2 worked. A=[1 2; 3 4], det=4−6=−2 ⇒ A⁻¹ = (−1/2)[4 −2; −3 1] = [−2 1; 3/2 −1/2].

Check: A A⁻¹ should give I — always verify one product. Symmetric A=Aᵀ; the diagonal of A and Aᵀ agree, off-diagonals reflect.

Solve via inverse. A=[1 2; 3 4], b=[5; 6]: x = A⁻¹b = [−2 1; 3/2 −1/2][5;6] = [−4; 9/2].

Order reverses, worked. (AB)ᵀ = BᵀAᵀ and (AB)⁻¹ = B⁻¹A⁻¹ — write them in reverse to keep the inner dimensions matching; getting the order wrong is a classic exam slip.

Only square matrices have inverses/determinants. A non-square or det-0 matrix is singular: Ax=b then has 0 or ∞-many solutions, never a unique one. (A⁻¹)ᵀ = (Aᵀ)⁻¹, so transpose and inverse commute. The 6 invertibility statements opposite are equivalent — prove one, get all.

22b · Invertibility Equivalences★ the big list

For square A, the following are equivalent:

• A is invertible
• Ax=b has a unique solution for every b
• Ax=0 has only x=0
• RREF(A) = I
• det A ≠ 0
• 0 is not an eigenvalue of A

Corollary: for square A, BA=I alone forces A⁻¹=B. Trap: never write 1/A or "divide" by a matrix.

23 · DeterminantsWk 10–11

cofactor (Laplace) expansiondet A = Σ_j (−1)^1+j a_1j det(A_1j)
2×2: ad − bc

3×3: a₁₁(a₂₂a₃₃−a₂₃a₃₂) − a₁₂(a₂₁a₃₃−a₂₃a₃₁) + a₁₃(a₂₁a₃₂−a₂₂a₃₁). Expand along any row/col (checkerboard signs (−1)^i+j) — pick one with many zeros.

Triangular shortcut: det = product of diagonal.

3×3 worked. det[2 0 1; 1 3 2; 0 1 1] = 2(3·1−2·1) − 0 + 1(1·1−3·0) = 2(1)+1 = 3 ⇒ invertible (expanded along row 1; row 3 has zeros too).

Product rule for det: det(AB)=det A·det B lets you read off det(A^k)=(det A)^k and det(A⁻¹)=1/det A without computing the matrices.

Vandermonde: det of the rows [1, x_i, …, x_iⁿ⁻¹] = ∏_i<j(x_j−x_i) — nonzero iff the x_i are distinct.

Cofactor & adjugate: C_ij=(−1)^i+jdet(A_ij); A⁻¹ = (1/det A)·adj(A), where adj(A)=[C_ij]ᵀ — the general inverse formula behind the 2×2 case. Since det(Aᵀ)=det A, you may expand along a column as easily as a row, and triangular matrices read off in one step (product of the diagonal).

23b · Determinant PropertiesEROs & rules

Effect of row operations

• Swap two rows: det → −det
• Scale a row by c: det → c·det
• Add a multiple of a row: det unchanged

further propertiesdet(cA) = cⁿ det A  (not c det A)
det(AB) = det A · det B
det(Aᵀ) = det A · det(A⁻¹) = 1/det A

A zero row/column ⇒ det=0. A invertible ⟺ det A ≠ 0. Note det(A+B) ≠ det A + det B. Reduce to triangular, tracking the row-op factors, for big matrices.

24 · Eigenvalues & EigenvectorsWk 11

For n×n A: scalar λ and nonzero v with Av = λv. λ = eigenvalue, v = λ-eigenvector.

characteristic equationAv=λv ⟺ (A−λI)v=0 has v≠0
⟺ det(A − λI) = 0

det(A−λI) = characteristic polynomial (degree n) ⇒ ≤ n eigenvalues (possibly complex). Eigenspace E_λ = solutions of (A−λI)v=0 (row-reduce [A−λI | 0]).

Multiplicities

Algebraic = multiplicity as a root of char. poly. Geometric = dim E_λ (free parameters). Always 1 ≤ geom ≤ alg. A invertible ⟺ 0 not an eigenvalue.

Sum of algebraic multiplicities = n (counting complex roots). If all are simple (n distinct λ), every geom mult is 1 and the matrix diagonalises automatically — no eigenspace check needed.

24b · Eigen · Worked2×2

A = [2 1; 1 2]. det(A−λI) = (2−λ)²−1 = λ²−4λ+3 = 0 ⇒ λ = 1, 3.

λ=3: (A−3I)v=0 ⇒ [−1 1; 1 −1]v=0 ⇒ v=[1,1]. λ=1: v=[1,−1]. Two distinct λ ⇒ independent eigenvectors ⇒ diagonalisable.

Any nonzero scalar multiple of an eigenvector is also an eigenvector (same λ) — pick the simplest representative (set a free parameter to 1).

Repeated λ can fail: [1 1; 0 1] has λ=1 (alg mult 2) but only a 1-D eigenspace (geom mult 1) ⇒ not diagonalisable. Complex λ are allowed: [0 −1; 1 0] has λ=±i.

Trace & det check: Σλ_i = trace(A), ∏λ_i = det(A) — a fast sanity check on your eigenvalues.

For A=[2 1; 1 2]: trace 4 = 1+3 ✓, det 3 = 1·3 ✓. Triangular matrices give eigenvalues straight off the diagonal.

Multiplicities worked. A=[2 0; 0 2]=2I: char poly (2−λ)², alg mult 2; E₂ is all of ℝ² ⇒ geom mult 2 = alg ⇒ diagonalisable (already diagonal). Contrast the [1 1;0 1] shear, geom 1 < alg 2.

Complex eigenvalues. A=[0 −1; 1 0] (rotation by 90°): det(A−λI)=λ²+1=0 ⇒ λ=±i. Real matrices can have complex eigenvalues in conjugate pairs — no real eigenvector here.

To verify a candidate v: just compute Av and check it equals λv — faster than re-solving the system. Eigenvectors for distinct eigenvalues are automatically independent, so they always give an invertible P. Real symmetric matrices are always diagonalisable, with mutually orthogonal eigenvectors and only real eigenvalues — a useful shortcut whenever A=Aᵀ in an exam question.

25 · DiagonalisationWk 12

A is diagonalisable if A = PDP⁻¹ (P invertible, D diagonal), i.e. P⁻¹AP = D.

theoremP = [v₁ … vₙ], D = diag(λ₁,…,λₙ)
iff Av_i=λ_iv_i and the v_i independent

Criterion: diagonalisable ⟺ for every eigenvalue, geometric mult = algebraic mult. n distinct eigenvalues ⇒ diagonalisable.

payoff — matrix powersA^k = P D^k P⁻¹, D^k = diag(λ₁^k,…,λₙ^k)

Worked. A=[2 1; 1 2]: P=[1 1; 1 −1], D=diag(3,1) ⇒ A^k = P diag(3^k,1) P⁻¹, giving each entry = (3^k±1)/2 — no repeated multiplication needed.

Why it works: in the eigenbasis A just scales each axis by its λ; P changes coordinates into that basis and P⁻¹ back. This is the engine behind Markov-chain long-run states and linear recurrences — also how A¹⁰⁰ becomes a one-line answer instead of 99 multiplications.

25b · Method Recipesdo this in order

• Eigen/diag: det(A−λI)=0 → λ → eigenspaces → if any geom≠alg, stop (not diag.) → else assemble P (columns), D (matching order)
• Solve Ax=b: row-reduce [A|b] → REF → free vars → back-sub
• Invert A: [A|I] → RREF → [I|A⁻¹]
• Plane from 3 pts: two directions → n=d₁×d₂ → n·(x−p)=0
• Roots of zⁿ=w: polar form → R^1/n, add 2π/n each root
• Angle / distance: cosθ=u·v/(‖u‖‖v‖); point→line via the orthogonal component; point→plane via |n·(P−Q)|/‖n‖
• Is it diagonalisable? distinct λ ⇒ yes; else check geom = alg for each repeated λ
• Sanity: Σλ = trace, ∏λ = det — check before assembling P, D

26 · High-Yield Trapsdon't lose marks

• L'Hôpital only on genuine 0/0 or ∞/∞ — verify first
• Keep the chain-rule inner factor; quotient is f′g−fg′
• Integration: always +C; convert limits in substitution; reduce improper fractions before partial fractions
• FTC variable limit: ×g′(x); sign flips for a lower limit
• Taylor: check R_n→0 / radius before claiming series = f
• Complex: arg from the correct quadrant; list all n roots; conjugate-root needs real coefficients
• Projection uses ‖u‖²; cross product is ℝ³-only & order-sensitive
• (AB)⁻¹, (AB)ᵀ reverse order; AB≠BA; no matrix division
• det(cA)=cⁿdet A; eigenvectors ≠ 0; match P & D column order
• RREF is unique, REF is not; only square matrices have det / inverse

Algebra Formula Beltside 2

e^iθ=cosθ+i sinθ · (re^iθ)ⁿ=rⁿe^inθ
n-th roots: R^1/nexp(i(φ+2kπ)/n)
proj_u(v)=(u·v/‖u‖²)u · ‖u×v‖=‖u‖‖v‖sinθ
2×2⁻¹ = (1/(ad−bc))[d −b; −c a]
det(A−λI)=0 · A=PDP⁻¹ ⇒ A^k=PD^kP⁻¹
Σλ=tr A · ∏λ=det A