PAU Applied Mathematics for Social Sciences II - Canary Islands / ULPGC & ULL common mistakes
PAU Applied Mathematics for Social Sciences II - Canary Islands / ULPGC & ULL · Understand the traps and how to avoid them.
When applying Rouché–Capelli, the candidate computes the rank of the coefficient matrix A and the augmented matrix (A|b) but only reduces A to echelon form, assuming that the rank of the augmented matrix is automatically the same, then declares the system consistent even when a row of the form (0 0 … 0 | k) with k ≠ 0 appears in the augmented reduction.
Textbook presentations often treat the augmented column as an afterthought, and candidates memorise 'rank A = rank A|b means compatible' as a slogan without practising the augmented reduction. They also confuse the number of unknowns n with the number of equations, so they never check rank against n to distinguish determined from indeterminate cases.
What to do instead
Always reduce the full augmented matrix (A|b) in one pass and read both ranks off the same echelon form. Then apply the three-case rule explicitly: rank A ≠ rank A|b → incompatible; rank A = rank A|b = n → compatible determined; rank A = rank A|b < n → compatible indeterminate with n − rank free parameters. Write the three ranks/values in a labelled line before concluding, and state the number of free parameters when indeterminate.
The candidate solves a matrix equation such as AX = B or AXB = C by 'dividing by A', writing X = B/A, or multiplies both sides on the wrong side, e.g. from AX = B concludes X = BA⁻¹ instead of X = A⁻¹B.
Scalar arithmetic habits carry over: division is commutative for numbers, so candidates assume order does not matter. Many also never internalise that matrix multiplication is non-commutative and that a left factor must be cancelled by a left inverse.
What to do instead
Treat matrix equations as operations on both sides with the same side. For AX = B, left-multiply by A⁻¹: X = A⁻¹B. For XA = B, right-multiply: X = BA⁻¹. For AXB = C, X = A⁻¹CB⁻¹. Before computing, verify A (and B) are invertible by checking det ≠ 0, and always substitute the answer back into the original equation to confirm.
In a linear programming problem the candidate writes the constraints but optimises the objective function at an interior point or at the origin, or evaluates the objective only at the vertices that are easy to read, ignoring the vertex created by the intersection of two non-axis constraints, and so reports a suboptimal value.
Candidates trust the drawing more than the algebra: the intersection of two slanted lines is hard to read off a sketch, so it gets skipped. They also forget the fundamental theorem that a linear objective on a bounded feasible region attains its optimum at a vertex, and they fail to check whether the region is bounded at all.
What to do instead
List every constraint line, compute all pairwise intersection points algebraically, and keep only those satisfying all constraints (feasibility test by substitution). Build a table of vertex coordinates and objective value, then select max/min from the table. Explicitly state whether the region is bounded; if unbounded, check whether the objective is bounded on the region before claiming an optimum.
When computing a limit that gives 0/0 or ∞/∞, the candidate substitutes the value, gets an indeterminate form, and either declares 'the limit does not exist' or cancels terms incorrectly, for example cancelling x in (x² − 4)/(x − 2) to get x − 4 or cancelling across a sum.
Indeterminate forms look like arithmetic failure, so candidates stop rather than transform. Weak algebraic fluency (factorisation of quadratics, rationalising radicals, dividing by the highest power) means they reach for invalid cancellations instead of legitimate ones.
What to do instead
Classify the indeterminacy first (0/0, ∞/∞, ∞ − ∞, 0·∞, 1^∞). For 0/0 with polynomials, factor and cancel the common factor; with radicals, multiply by the conjugate. For limits at infinity, divide numerator and denominator by the highest power of x present. For 1^∞ use the exponential form or logarithms. Only cancel factors that appear in every term of both numerator and denominator.
The candidate computes a derivative of a product or quotient by differentiating each factor separately, e.g. writing (fg)′ = f′g′ or (f/g)′ = f′/g′, and mishandles the chain rule by omitting the inner derivative, e.g. differentiating sin(x²) as cos(x²).
The sum rule is the first rule learned and is the only one that distributes over the operation, so it becomes a default template. Chain-rule omission happens because the outer derivative is computed correctly and the candidate stops there, satisfied with a plausible-looking answer.
What to do instead
Before differentiating, name the structure out loud or in writing: 'this is a product / quotient / composition'. Apply the matching rule: (fg)′ = f′g + fg′; (f/g)′ = (f′g − fg′)/g²; (f∘g)′ = f′(g(x))·g′(x). For composites, always write the inner function and its derivative as a separate step before simplifying.
When finding extrema, the candidate sets f′(x) = 0, solves, and immediately declares each solution a maximum or minimum without checking the sign change of f′ or the sign of f″, and often forgets to check whether the critical points lie inside the domain of the function (e.g. x > 0 for a cost or production function).
The procedure is remembered as 'derivative equals zero gives the optimum', which is true for the location of stationary points but not for their classification or for endpoints. Economic contexts with restricted domains (quantity ≥ 0) are treated as if the domain were all reals.
What to do instead
Follow a fixed checklist: (1) state the domain; (2) find f′ and solve f′ = 0 within the domain; (3) classify each critical point by the sign change of f′ across it or by f″ at it; (4) evaluate f at endpoints of a closed domain and compare; (5) confirm the answer is a maximum or minimum as asked, and give the function value, not just x.
The candidate confuses concavity with the sign of f itself, or with the sign of f′, and reports an inflection point wherever f″ = 0 without checking that the concavity actually changes there (e.g. claiming x = 0 is an inflection point of f(x) = x⁴).
Three related quantities (f, f′, f″) carry three different meanings, and under exam pressure they collapse into 'the derivative tells you the shape'. The condition f″ = 0 is necessary but not sufficient for an inflection point, and that subtlety is easily lost.
What to do instead
Anchor the meanings: f′ > 0 increasing, f′ < 0 decreasing; f″ > 0 convex (concave up), f″ < 0 concave (concave down). For inflection points, require f″ = 0 (or undefined) AND a sign change of f″ on either side; test with a sign table of f″. Never infer concavity from the sign of f or f′.
When computing a definite integral with Barrow's rule, the candidate applies the chain rule in reverse incorrectly, e.g. writes ∫₀² e^{2x} dx = [e^{2x}]₀² instead of dividing by 2, or evaluates an area between curves as ∫(f − g) without checking which function is on top, obtaining a negative or wrong-signed area.
Antidifferentiation is the step with the least algorithmic feedback: an incorrect antiderivative still looks like an answer. The area formula ∫(upper − lower) is memorised as ∫(f − g), and the roles of f and g are not verified on the interval.
What to do instead
Differentiate your antiderivative as a check before applying Barrow. For linear inner functions use ∫f(ax + b)dx = F(ax + b)/a. For areas, first find the intersection points to get the limits, then determine on each subinterval which function is greater (test one interior point) and integrate (upper − lower); if the curves cross inside the interval, split the integral at the crossing point.
The candidate computes a conditional probability P(A|B) as P(B|A), or multiplies P(A)·P(B) for events that are not independent, and in Bayes' theorem problems uses the marginal P(B) as if it were P(B|A) or omits the total-probability denominator entirely.
The notation P(A|B) reads naturally as 'A given B' but the formula P(A∩B)/P(B) is easy to invert under time pressure, and the symmetry of the product rule for independent events is over-generalised to all events. Bayes problems require constructing a partition, which is a modelling step candidates skip.
What to do instead
Always write the definition P(A|B) = P(A∩B)/P(B) and identify numerator and denominator from the problem wording before substituting. Check independence by testing P(A∩B) = P(A)P(B), not by assumption. For Bayes, draw a tree with branches for the partition events, compute each path probability, and use P(A|B) = P(B|A)P(A) / Σ P(B|Aᵢ)P(Aᵢ), verifying the denominator is the total probability of the observed event.
When working with the normal distribution, the candidate uses the raw value x with the standard normal table instead of standardising to z = (x − μ)/σ, or mishandles the direction of the inequality (computing P(Z < z) when P(Z > z) is asked) and forgets the continuity correction when approximating a binomial by a normal.
The table gives Φ(z) for one specific form, and the transformation from an arbitrary normal to the standard one is an extra step that gets dropped. The symmetry P(Z > z) = 1 − Φ(z) and the ±0.5 correction are details that are learned once and rarely rehearsed.
What to do instead
Standardise first, always: write z = (x − μ)/σ with the numbers substituted. Sketch the standard normal curve and shade the required region, then translate the shading into Φ expressions using symmetry: P(Z > z) = 1 − Φ(z), P(a < Z < b) = Φ(b) − Φ(a). For binomial-to-normal approximation, apply the continuity correction (use x ± 0.5) and state the conditions np > 5 and n(1 − p) > 5.