2026
MATHEMATICS II
SUMMARY SOLUTIONS MANUAL
KOEN HANEGREEFS
VUB
,Table of contents
Table of contents ..................................................................................................................................... 1
Chapter 1 - Linear Geometry ..................................................................................................................... 5
E1.1 - vector arithmetic, length .......................................................................................................... 5
E1.2 - dot product, angle ................................................................................................................... 6
E1.3 - perpendicularity / parameter .................................................................................................... 7
E1.4 - line through points, membership .............................................................................................. 9
E1.5 - plane through three points ..................................................................................................... 10
E1.6 - hyperplane equation to parametrise ....................................................................................... 12
E1.7 - degenerate parametrisation ................................................................................................... 13
E1.8 - hyperplane through points + line ............................................................................................ 15
Chapter 2 - Vector Spaces ...................................................................................................................... 17
E2.1 - subspace test........................................................................................................................ 17
E2.2 - subspace test, fails................................................................................................................ 18
E2.3 - linear independence in R^3 .................................................................................................... 20
E2.4 - basis for solution space ......................................................................................................... 21
E2.5 - basis of P_2 ........................................................................................................................... 23
E2.6 - subspace check in P_3........................................................................................................... 24
E2.7 - spanning set vs basis ............................................................................................................. 26
E2.8 - coordinate representation ..................................................................................................... 27
Chapter 3 - Maps Between Spaces.......................................................................................................... 29
E3.1 - linearity check ....................................................................................................................... 29
E3.2 - linearity fails .......................................................................................................................... 30
E3.3 - rank, nullity ........................................................................................................................... 32
E3.4 - kernel & range ....................................................................................................................... 33
E3.5 - matrix representation ............................................................................................................ 34
E3.6 - change of basis ..................................................................................................................... 36
E3.7 - homomorphism check ........................................................................................................... 38
E3.8 - construct homomorphism ..................................................................................................... 39
Chapter 4 - Matrix Operations ................................................................................................................. 42
E4.1 - matrix arithmetic ................................................................................................................... 42
1
, E4.2 - matrix product ....................................................................................................................... 43
E4.3 - non-commutativity ................................................................................................................ 45
E4.4 - 2x2 inverse ............................................................................................................................ 46
E4.5 - 3x3 inverse via Gauss-Jordan ................................................................................................. 48
E4.6 - elementary matrices .............................................................................................................. 50
E4.7 - matrix-product dimensions .................................................................................................... 52
E4.8 - composition = product ........................................................................................................... 53
Chapter 5 - Linear Systems ..................................................................................................................... 56
E5.1 - 2x2 Gauss ............................................................................................................................. 56
E5.2 - Gauss with free variable ......................................................................................................... 58
E5.3 - inconsistent system............................................................................................................... 60
E5.4 - particular + homogeneous ..................................................................................................... 61
E5.5 - Gauss-Jordan ........................................................................................................................ 63
E5.6 - kernel as solution set ............................................................................................................. 65
E5.7 - back-substitution mistake ..................................................................................................... 66
E5.8 - full Gauss workflow ............................................................................................................... 68
Chapter 6 - Determinants ....................................................................................................................... 70
E6.1 - 2x2 determinant .................................................................................................................... 70
E6.2 - 3x3 Sarrus ............................................................................................................................. 71
E6.3 - Laplace expansion ................................................................................................................. 72
E6.4 - triangular determinant ........................................................................................................... 74
E6.5 - row reduction for determinant ................................................................................................ 75
E6.6 - inverse via adjugate ............................................................................................................... 77
E6.7 - cofactor sign ......................................................................................................................... 79
E6.8 - determinant + Cramer + interpretation.................................................................................... 80
Chapter 7 - Eigen-Theory ........................................................................................................................ 83
E7.1 - 2x2 eigenvalues ..................................................................................................................... 83
E7.2 - eigenvectors.......................................................................................................................... 85
E7.3 - characteristic polynomial ...................................................................................................... 87
E7.4 - diagonalisation ...................................................................................................................... 89
E7.5 - 3x3 eigenvalues ..................................................................................................................... 91
E7.6 - non-diagonalisable ................................................................................................................ 93
2
, E7.7 - characteristic polynomial mistake .......................................................................................... 95
E7.8 - diagonalise + compute power ................................................................................................ 97
E7.9 - transition matrix & coordinate change .................................................................................... 99
E7.10 - diagonalisation via eigenbasis ............................................................................................ 101
Chapter 8 - Multivariable Calculus ........................................................................................................ 104
E8.1 - partial derivatives ................................................................................................................ 104
E8.2 - gradient .............................................................................................................................. 106
E8.3 - homogeneity ....................................................................................................................... 108
E8.4 - critical points + Hessian ....................................................................................................... 109
E8.5 - directional derivative ........................................................................................................... 111
E8.6 - chain rule ............................................................................................................................ 113
E8.7 - Hessian sign rule ................................................................................................................. 115
E8.8 - contour-plot signs ............................................................................................................... 117
E8.9 - multivariable limit non-existence ......................................................................................... 119
E8.10 - chain rule in polar coordinates ........................................................................................... 120
Chapter 9 - Optimisation ...................................................................................................................... 123
E9.1 - unconstrained min .............................................................................................................. 123
E9.2 - saddle point ........................................................................................................................ 124
E9.3 - Lagrange - single constraint ................................................................................................. 126
E9.4 - constrained - substitution & Lagrange .................................................................................. 127
E9.5 - open container cost min ...................................................................................................... 129
E9.6 - Hessian classification with leading minors ........................................................................... 131
E9.7 - Lagrange - missing condition ................................................................................................ 132
E9.8 - marketing-budget Lagrangian ............................................................................................... 134
Chapter 10 - Multiple Integrals .............................................................................................................. 136
E10.1 - rectangular double integral ................................................................................................ 136
E10.2 - iterated integral ................................................................................................................. 137
E10.3 - polar conversion ................................................................................................................ 138
E10.4 - polar - circular region ......................................................................................................... 140
E10.5 - Jacobian - linear substitution.............................................................................................. 141
E10.6 - polar - variable bounds ...................................................................................................... 143
E10.7 - missing Jacobian ............................................................................................................... 145
3
, E10.8 - iterated then sub ............................................................................................................... 146
E10.9 - reverse the order of integration ........................................................................................... 148
Chapter 11 - Ordinary Differential Equations ......................................................................................... 151
E11.1 - separable ODE .................................................................................................................. 151
E11.2 - 1st-order linear .................................................................................................................. 153
E11.3 - 2nd-order homogeneous, distinct roots .............................................................................. 155
E11.4 - Bernoulli ODE .................................................................................................................... 156
E11.5 - homogeneous-substitution ODE ........................................................................................ 158
E11.6 - 2nd-order - complex roots .................................................................................................. 160
E11.7 - ODE classification ............................................................................................................. 161
E11.8 - 1st-order linear with finance setup ..................................................................................... 163
E11.9 - debt with constant repayment ............................................................................................ 164
E11.10 - first-order linear price adjustment .................................................................................... 166
Chapter 12 - Cheat Sheet and Canonical Proofs .................................................................................... 169
E12.1 - method selection............................................................................................................... 169
E12.2 - ODE classification ............................................................................................................. 170
E12.3 - shadow price / Lagrange multiplier ..................................................................................... 172
E12.4 - eigenvalues vs invertibility .................................................................................................. 173
Reflex 1 - Read the question, label the type ........................................................................................ 175
Reflex 2 - Always state $\nabla f=\vec 0$ before classifying ................................................................ 175
Reflex 3 - For changes of variable, write down the Jacobian first .......................................................... 175
Reflex 4 - For ODEs, classify before computing .................................................................................. 175
Reflex 5 - Show your method, not just your answer ............................................................................. 175
4
,Chapter 1 - Linear Geometry
Solutions to exercises E1.1 onwards. Refer back to Chapter 1 of the Summary for the corresponding theory,
definitions, and worked examples.
E1.1 - vector arithmetic, length
Problem statement.
"⃗ = (1, −2,2) and 𝑣⃗ = (3,0, −4) in ℝ! , compute 𝑢
Given 𝑢 "⃗ + 𝑣⃗, 2𝑢
"⃗ − 𝑣⃗, and |𝑢
"⃗|.
(Type: warm-up)
Solution.
Step 1. What is being asked. Two vectors of ℝ! are given, 𝑢 = (1, −2,2) and 𝑣 = (3,0, −4), and three
separate objects are requested: the sum 𝑢 + 𝑣, the combination 2𝑢 − 𝑣, and the length |𝑢|. Before touching
any arithmetic, note the type of each answer: 𝑢 + 𝑣 and 2𝑢 − 𝑣 are again vectors of ℝ! , so each must come
out as a triple, while |𝑢| is a single non-negative real number.
Deciding in advance whether the answer is a vector or a scalar is the cheapest possible error check,
because a triple where a number was expected signals a wrong formula immediately.
Step 2. Strategy. Addition and scalar multiplication in ℝ" are defined slot by slot, so both vector answers
are obtained by doing ordinary arithmetic in each coordinate independently. The length uses a different
definition, the square root of the dot product of 𝑢 with itself, so it is computed separately at the end.
Vector arithmetic in Ch. 1.1 of the Summary is defined component-wise, and length is defined together with
the dot product in Ch. 1.3, so the two parts of this exercise rest on two different definitions.
Step 3. Compute the sum coordinate by coordinate:
𝑢 + 𝑣 = (1 + 3, −2 + 0, 2 + (−4)) = (4, −2, −2)
This is exactly the definition 𝑢 + 𝑣 = (𝑢# + 𝑣# , … , 𝑢" + 𝑣" ) of Ch. 1.1, applied with no rearrangement.
Step 4. Compute the scalar multiple first, then subtract. Scaling gives 2𝑢 = (2 ⋅ 1, 2 ⋅ (−2), 2 ⋅ 2) =
(2, −4,4).
Doing the scaling as a separate written line prevents the common slip of multiplying only the first coordinate
by 2, which is easy to make when the whole expression is evaluated in one go.
Step 5. Now subtract 𝑣 slot by slot:
2𝑢 − 𝑣 = (2, −4,4) − (3,0, −4) = (2 − 3, −4 − 0, 4 − (−4)) = (−1, −4,8)
Subtraction is addition of (−1)𝑣, so it too acts component-wise. The trap sits in the third slot, where 4 −
(−4) = 8 and not 0.
Step 6. Set up the length. By definition |𝑢| = √𝑢 ⋅ 𝑢 = 7𝑢#$ + 𝑢$$ + 𝑢!$ , so here
|𝑢| = 71$ + (−2)$ + 2$ = √1 + 4 + 4 = √9
Every squared term is non-negative, which is why the sum under the root can never be negative and the
length is always defined.
5
,Step 7. Evaluate the root: √9 = 3, so |𝑢| = 3.
The length is by convention the positive square root, so −3 is not an admissible answer even though it also
squares to 9.
Step 8. Sanity check. Each coordinate of 𝑢 has absolute value at most 2, so the length must lie between the
largest single coordinate, 2, and √3 ⋅ 2 ≈ 3.46. The value 3 sits inside that range. As a second check, adding
𝑣 to the answer (4, −2, −2) returns (4 − 3, −2 − 0, −2 + 4) = (1, −2,2) = 𝑢, so the sum was computed
correctly.
Reversing an operation and recovering the original input is the standard self-check for component-wise
arithmetic and costs only one extra line.
Result
𝑢 + 𝑣 = (4, −2, −2), 2𝑢 − 𝑣 = (−1, −4,8), |𝑢| = 3.
E1.2 - dot product, angle
Problem statement.
"⃗ = (1,1,0) and 𝑣⃗ = (1,0,1) in ℝ! .
Compute the angle between 𝑢
(Type: warm-up)
Solution.
Step 1. What is being asked. The exercise asks for the angle between 𝑢 = (1,1,0) and 𝑣 = (1,0,1) in ℝ! . The
answer is a single angle, conventionally taken in [0, 𝜋], and it is reported here both in degrees and in radians.
The angle formula returns a cosine, and restricting to [0, 𝜋] is what makes the inverse cosine single valued,
so the answer is unique.
%⋅'
Step 2. Strategy. The angle formula of Ch. 1.3 states that cos𝜃 = |%| |'| whenever both vectors are non-zero.
This needs three separate ingredients: the dot product 𝑢 ⋅ 𝑣, the length |𝑢|, and the length |𝑣|. Compute all
three first, then assemble.
Assembling the formula only after every ingredient is known avoids mixing a dot product with a length in the
same line, which is where sign and root errors normally appear.
Step 3. Dot product. Multiply matching coordinates and add:
𝑢⋅𝑣 =1⋅1+1⋅0+0⋅1=1+0+0=1
This is the definition 𝑢 ⋅ 𝑣 = ∑* 𝑢* 𝑣* from Ch. 1.3. Note that the result is a number, not a vector.
Step 4. Length of 𝑢: |𝑢| = √1$ + 1$ + 0$ = √2. Length of 𝑣: |𝑣| = √1$ + 0$ + 1$ = √2. Both vectors have
exactly two coordinates equal to 1 and one equal to 0, which is why the two lengths agree.
Length is √𝑣 ⋅ 𝑣, so a coordinate equal to 0 contributes nothing and only the two unit slots matter.
6
,Step 5. Assemble the formula:
𝑢⋅𝑣 1 1
cos𝜃 = = =
|𝑢| |𝑣| √2 ⋅ √2 2
The product √2 ⋅ √2 = 2 because squaring undoes the square root of a non-negative number, so no radical
survives in the final cosine.
Step 6. Invert the cosine. On [0, 𝜋] the only angle with cosine 1⁄2 is
𝜋
𝜃 = arccos(1⁄2) = 60∘ =
3
This is one of the standard values of the cosine and should be recognised rather than approximated, since
exam answers are expected in exact form.
Step 7. Name the trap. Do not divide by |𝑢|$ or forget one of the two lengths. Leaving out a factor of √2
would give cos𝜃 = 1⁄√2 and the wrong answer 45∘ . Both lengths belong in the denominator because the
formula normalises each vector separately.
The cosine must lie in [−1,1] for every pair of vectors, and this is guaranteed only when both lengths divide
out.
Step 8. Sanity check. The dot product is positive, so the angle must be strictly acute, and 60∘ < 90∘
#
confirms this. Numerically cos60∘ = 0.5 and = 0.5, so the two sides agree.
√$⋅√$
The sign of the dot product alone decides acute against obtuse, as recorded in the concept check of Ch. 1.3,
so it gives a free consistency test on the final angle.
Result
𝜃 = 60∘ = 𝜋/3.
E1.3 - perpendicularity / parameter
Problem statement.
For which value of 𝑘 ∈ ℝ are (𝑘, 1) and (4,3) perpendicular in ℝ$ ?
(Type: warm-up)
Solution.
Step 1. What is being asked. The two vectors (𝑘, 1) and (4,3) in ℝ$ contain one unknown, 𝑘, and the
exercise asks for the value or values of 𝑘 that make them perpendicular. The answer is therefore a number,
or possibly a set of numbers, and not a vector.
Recognising that the unknown sits inside a vector but that the answer is a scalar keeps the bookkeeping
straight when the condition is written down.
Step 2. Strategy. Perpendicularity is not tested with angles here. The rule in Ch. 1.3 says that for non-zero 𝑢
%⋅'
and 𝑣 the angle satisfies cos𝜃 = , and since cos90∘ = 0 and the lengths are non-zero, 𝑢 ⊥ 𝑣 holds
|%||'|
exactly when 𝑢 ⋅ 𝑣 = 0. So translate the geometric condition into the single algebraic equation 𝑢 ⋅ 𝑣 = 0.
7
,Setting the dot product to zero replaces a trigonometric condition by a linear equation, which is far easier to
solve and involves no square roots at all.
Step 3. Write out the dot product with the unknown carried along:
(𝑘, 1) ⋅ (4,3) = 𝑘 ⋅ 4 + 1 ⋅ 3 = 4𝑘 + 3
The dot product is defined component-wise in Ch. 1.3, and it stays a legitimate expression when one
coordinate is a symbol rather than a number.
Step 4. Impose the perpendicularity condition, which gives the equation 4𝑘 + 3 = 0.
This is the translation step: the geometry has now become a single linear equation in one unknown, so a
unique solution is expected.
Step 5. Solve it. Subtract 3 from both sides to get 4𝑘 = −3, then divide both sides by 4:
3
𝑘=−
4
Both operations are reversible, since 4 ≠ 0, so no solution is created or lost and the value found is the only
one.
Step 6. Name the trap. Perpendicular is not the same as parallel. Requiring (𝑘, 1) to be a multiple of (4,3)
would give 𝑘 = 4⁄3, which is the answer to a different question. Also note that no square roots and no
lengths are needed here, because the lengths cancel out of the condition cos𝜃 = 0.
The angle formula has the lengths only in the denominator, so a zero numerator already forces the angle to
be 90∘ regardless of how long the vectors are.
Step 7. Check by substitution. With 𝑘 = − 3⁄4 the first vector is (− 3⁄4 , 1) and
(− 3⁄4 , 1) ⋅ (4,3) = −3 + 3 = 0
so the two vectors are indeed perpendicular.
Substituting back into the original condition is the definitive test, and it takes a single line.
Result
𝑘 = −3/4.
8
, E1.4 - line through points, membership
Problem statement.
Let 𝐴 = (1,0,2,1) and 𝐵 = (3,2,0,5) in ℝ- . Does 𝑃 = (2,1,1,3) lie on the line through 𝐴 and 𝐵?
(Type: mixed)
Solution.
Step 1. What is being asked. The line through 𝐴 = (1,0,2,1) and 𝐵 = (3,2,0,5) lives in ℝ- , and the question is
whether the point 𝑃 = (2,1,1,3) lies on it. The answer is yes or no, and if yes it should come with the
parameter value 𝑡 that locates 𝑃 on the line.
A membership question always has a binary answer, but stating the witnessing parameter turns the answer
into a verifiable proof rather than a claim.
Step 2. Strategy. Follow the membership test of Procedure 1-C in Ch. 1.2. Write the line in parametric form
{𝐴 + 𝑡𝑑}, translate the problem so that the base point sits at the origin, and then test whether the translated
point is a scalar multiple of the direction.
Subtracting the base point removes the offset, so the remaining question is purely about direction and can
be answered coordinate by coordinate.
Step 3. Compute the direction from the two given points:
𝑑 = 𝐵 − 𝐴 = (3 − 1, 2 − 0, 0 − 2, 5 − 1) = (2,2, −2,4)
so the line is {𝐴 + 𝑡𝑑: 𝑡 ∈ ℝ}.
Any non-zero multiple of 𝐵 − 𝐴 is a valid direction for the line through 𝐴 and 𝐵, and this particular choice
makes 𝑡 = 0 land on 𝐴 and 𝑡 = 1 land on 𝐵.
Step 4. Translate the test point:
𝑃 − 𝐴 = (2 − 1, 1 − 0, 1 − 2, 3 − 1) = (1,1, −1,2)
The question is now whether there is a single scalar 𝑡 with (1,1, −1,2) = 𝑡(2,2, −2,4).
This is precisely the translated form of the membership condition in Procedure 1-C: 𝑃 lies on the line exactly
when 𝑃 − 𝐴 is built from the direction vector.
Step 5. Read the vector equation coordinate by coordinate. It splits into four scalar equations, 1 = 2𝑡, 1 =
2𝑡, −1 = −2𝑡 and 2 = 4𝑡, that is four equations in the single unknown 𝑡.
The system is heavily overdetermined, four conditions on one unknown, and that is exactly why membership
can fail: any one disagreement is fatal.
Step 6. Solve from the easiest coordinate and then verify the rest. The first coordinate gives 𝑡 = 1⁄2.
Substituting this single value into the remaining three: 2 ⋅ 1⁄2 = 1 matches the required 1; −2 ⋅ 1⁄2 = −1
matches the required −1; 4 ⋅ 1⁄2 = 2 matches the required 2.
Solving once and then checking, rather than solving four times, is what Procedure 1-C prescribes, and it
makes any inconsistency visible immediately.
9
MATHEMATICS II
SUMMARY SOLUTIONS MANUAL
KOEN HANEGREEFS
VUB
,Table of contents
Table of contents ..................................................................................................................................... 1
Chapter 1 - Linear Geometry ..................................................................................................................... 5
E1.1 - vector arithmetic, length .......................................................................................................... 5
E1.2 - dot product, angle ................................................................................................................... 6
E1.3 - perpendicularity / parameter .................................................................................................... 7
E1.4 - line through points, membership .............................................................................................. 9
E1.5 - plane through three points ..................................................................................................... 10
E1.6 - hyperplane equation to parametrise ....................................................................................... 12
E1.7 - degenerate parametrisation ................................................................................................... 13
E1.8 - hyperplane through points + line ............................................................................................ 15
Chapter 2 - Vector Spaces ...................................................................................................................... 17
E2.1 - subspace test........................................................................................................................ 17
E2.2 - subspace test, fails................................................................................................................ 18
E2.3 - linear independence in R^3 .................................................................................................... 20
E2.4 - basis for solution space ......................................................................................................... 21
E2.5 - basis of P_2 ........................................................................................................................... 23
E2.6 - subspace check in P_3........................................................................................................... 24
E2.7 - spanning set vs basis ............................................................................................................. 26
E2.8 - coordinate representation ..................................................................................................... 27
Chapter 3 - Maps Between Spaces.......................................................................................................... 29
E3.1 - linearity check ....................................................................................................................... 29
E3.2 - linearity fails .......................................................................................................................... 30
E3.3 - rank, nullity ........................................................................................................................... 32
E3.4 - kernel & range ....................................................................................................................... 33
E3.5 - matrix representation ............................................................................................................ 34
E3.6 - change of basis ..................................................................................................................... 36
E3.7 - homomorphism check ........................................................................................................... 38
E3.8 - construct homomorphism ..................................................................................................... 39
Chapter 4 - Matrix Operations ................................................................................................................. 42
E4.1 - matrix arithmetic ................................................................................................................... 42
1
, E4.2 - matrix product ....................................................................................................................... 43
E4.3 - non-commutativity ................................................................................................................ 45
E4.4 - 2x2 inverse ............................................................................................................................ 46
E4.5 - 3x3 inverse via Gauss-Jordan ................................................................................................. 48
E4.6 - elementary matrices .............................................................................................................. 50
E4.7 - matrix-product dimensions .................................................................................................... 52
E4.8 - composition = product ........................................................................................................... 53
Chapter 5 - Linear Systems ..................................................................................................................... 56
E5.1 - 2x2 Gauss ............................................................................................................................. 56
E5.2 - Gauss with free variable ......................................................................................................... 58
E5.3 - inconsistent system............................................................................................................... 60
E5.4 - particular + homogeneous ..................................................................................................... 61
E5.5 - Gauss-Jordan ........................................................................................................................ 63
E5.6 - kernel as solution set ............................................................................................................. 65
E5.7 - back-substitution mistake ..................................................................................................... 66
E5.8 - full Gauss workflow ............................................................................................................... 68
Chapter 6 - Determinants ....................................................................................................................... 70
E6.1 - 2x2 determinant .................................................................................................................... 70
E6.2 - 3x3 Sarrus ............................................................................................................................. 71
E6.3 - Laplace expansion ................................................................................................................. 72
E6.4 - triangular determinant ........................................................................................................... 74
E6.5 - row reduction for determinant ................................................................................................ 75
E6.6 - inverse via adjugate ............................................................................................................... 77
E6.7 - cofactor sign ......................................................................................................................... 79
E6.8 - determinant + Cramer + interpretation.................................................................................... 80
Chapter 7 - Eigen-Theory ........................................................................................................................ 83
E7.1 - 2x2 eigenvalues ..................................................................................................................... 83
E7.2 - eigenvectors.......................................................................................................................... 85
E7.3 - characteristic polynomial ...................................................................................................... 87
E7.4 - diagonalisation ...................................................................................................................... 89
E7.5 - 3x3 eigenvalues ..................................................................................................................... 91
E7.6 - non-diagonalisable ................................................................................................................ 93
2
, E7.7 - characteristic polynomial mistake .......................................................................................... 95
E7.8 - diagonalise + compute power ................................................................................................ 97
E7.9 - transition matrix & coordinate change .................................................................................... 99
E7.10 - diagonalisation via eigenbasis ............................................................................................ 101
Chapter 8 - Multivariable Calculus ........................................................................................................ 104
E8.1 - partial derivatives ................................................................................................................ 104
E8.2 - gradient .............................................................................................................................. 106
E8.3 - homogeneity ....................................................................................................................... 108
E8.4 - critical points + Hessian ....................................................................................................... 109
E8.5 - directional derivative ........................................................................................................... 111
E8.6 - chain rule ............................................................................................................................ 113
E8.7 - Hessian sign rule ................................................................................................................. 115
E8.8 - contour-plot signs ............................................................................................................... 117
E8.9 - multivariable limit non-existence ......................................................................................... 119
E8.10 - chain rule in polar coordinates ........................................................................................... 120
Chapter 9 - Optimisation ...................................................................................................................... 123
E9.1 - unconstrained min .............................................................................................................. 123
E9.2 - saddle point ........................................................................................................................ 124
E9.3 - Lagrange - single constraint ................................................................................................. 126
E9.4 - constrained - substitution & Lagrange .................................................................................. 127
E9.5 - open container cost min ...................................................................................................... 129
E9.6 - Hessian classification with leading minors ........................................................................... 131
E9.7 - Lagrange - missing condition ................................................................................................ 132
E9.8 - marketing-budget Lagrangian ............................................................................................... 134
Chapter 10 - Multiple Integrals .............................................................................................................. 136
E10.1 - rectangular double integral ................................................................................................ 136
E10.2 - iterated integral ................................................................................................................. 137
E10.3 - polar conversion ................................................................................................................ 138
E10.4 - polar - circular region ......................................................................................................... 140
E10.5 - Jacobian - linear substitution.............................................................................................. 141
E10.6 - polar - variable bounds ...................................................................................................... 143
E10.7 - missing Jacobian ............................................................................................................... 145
3
, E10.8 - iterated then sub ............................................................................................................... 146
E10.9 - reverse the order of integration ........................................................................................... 148
Chapter 11 - Ordinary Differential Equations ......................................................................................... 151
E11.1 - separable ODE .................................................................................................................. 151
E11.2 - 1st-order linear .................................................................................................................. 153
E11.3 - 2nd-order homogeneous, distinct roots .............................................................................. 155
E11.4 - Bernoulli ODE .................................................................................................................... 156
E11.5 - homogeneous-substitution ODE ........................................................................................ 158
E11.6 - 2nd-order - complex roots .................................................................................................. 160
E11.7 - ODE classification ............................................................................................................. 161
E11.8 - 1st-order linear with finance setup ..................................................................................... 163
E11.9 - debt with constant repayment ............................................................................................ 164
E11.10 - first-order linear price adjustment .................................................................................... 166
Chapter 12 - Cheat Sheet and Canonical Proofs .................................................................................... 169
E12.1 - method selection............................................................................................................... 169
E12.2 - ODE classification ............................................................................................................. 170
E12.3 - shadow price / Lagrange multiplier ..................................................................................... 172
E12.4 - eigenvalues vs invertibility .................................................................................................. 173
Reflex 1 - Read the question, label the type ........................................................................................ 175
Reflex 2 - Always state $\nabla f=\vec 0$ before classifying ................................................................ 175
Reflex 3 - For changes of variable, write down the Jacobian first .......................................................... 175
Reflex 4 - For ODEs, classify before computing .................................................................................. 175
Reflex 5 - Show your method, not just your answer ............................................................................. 175
4
,Chapter 1 - Linear Geometry
Solutions to exercises E1.1 onwards. Refer back to Chapter 1 of the Summary for the corresponding theory,
definitions, and worked examples.
E1.1 - vector arithmetic, length
Problem statement.
"⃗ = (1, −2,2) and 𝑣⃗ = (3,0, −4) in ℝ! , compute 𝑢
Given 𝑢 "⃗ + 𝑣⃗, 2𝑢
"⃗ − 𝑣⃗, and |𝑢
"⃗|.
(Type: warm-up)
Solution.
Step 1. What is being asked. Two vectors of ℝ! are given, 𝑢 = (1, −2,2) and 𝑣 = (3,0, −4), and three
separate objects are requested: the sum 𝑢 + 𝑣, the combination 2𝑢 − 𝑣, and the length |𝑢|. Before touching
any arithmetic, note the type of each answer: 𝑢 + 𝑣 and 2𝑢 − 𝑣 are again vectors of ℝ! , so each must come
out as a triple, while |𝑢| is a single non-negative real number.
Deciding in advance whether the answer is a vector or a scalar is the cheapest possible error check,
because a triple where a number was expected signals a wrong formula immediately.
Step 2. Strategy. Addition and scalar multiplication in ℝ" are defined slot by slot, so both vector answers
are obtained by doing ordinary arithmetic in each coordinate independently. The length uses a different
definition, the square root of the dot product of 𝑢 with itself, so it is computed separately at the end.
Vector arithmetic in Ch. 1.1 of the Summary is defined component-wise, and length is defined together with
the dot product in Ch. 1.3, so the two parts of this exercise rest on two different definitions.
Step 3. Compute the sum coordinate by coordinate:
𝑢 + 𝑣 = (1 + 3, −2 + 0, 2 + (−4)) = (4, −2, −2)
This is exactly the definition 𝑢 + 𝑣 = (𝑢# + 𝑣# , … , 𝑢" + 𝑣" ) of Ch. 1.1, applied with no rearrangement.
Step 4. Compute the scalar multiple first, then subtract. Scaling gives 2𝑢 = (2 ⋅ 1, 2 ⋅ (−2), 2 ⋅ 2) =
(2, −4,4).
Doing the scaling as a separate written line prevents the common slip of multiplying only the first coordinate
by 2, which is easy to make when the whole expression is evaluated in one go.
Step 5. Now subtract 𝑣 slot by slot:
2𝑢 − 𝑣 = (2, −4,4) − (3,0, −4) = (2 − 3, −4 − 0, 4 − (−4)) = (−1, −4,8)
Subtraction is addition of (−1)𝑣, so it too acts component-wise. The trap sits in the third slot, where 4 −
(−4) = 8 and not 0.
Step 6. Set up the length. By definition |𝑢| = √𝑢 ⋅ 𝑢 = 7𝑢#$ + 𝑢$$ + 𝑢!$ , so here
|𝑢| = 71$ + (−2)$ + 2$ = √1 + 4 + 4 = √9
Every squared term is non-negative, which is why the sum under the root can never be negative and the
length is always defined.
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,Step 7. Evaluate the root: √9 = 3, so |𝑢| = 3.
The length is by convention the positive square root, so −3 is not an admissible answer even though it also
squares to 9.
Step 8. Sanity check. Each coordinate of 𝑢 has absolute value at most 2, so the length must lie between the
largest single coordinate, 2, and √3 ⋅ 2 ≈ 3.46. The value 3 sits inside that range. As a second check, adding
𝑣 to the answer (4, −2, −2) returns (4 − 3, −2 − 0, −2 + 4) = (1, −2,2) = 𝑢, so the sum was computed
correctly.
Reversing an operation and recovering the original input is the standard self-check for component-wise
arithmetic and costs only one extra line.
Result
𝑢 + 𝑣 = (4, −2, −2), 2𝑢 − 𝑣 = (−1, −4,8), |𝑢| = 3.
E1.2 - dot product, angle
Problem statement.
"⃗ = (1,1,0) and 𝑣⃗ = (1,0,1) in ℝ! .
Compute the angle between 𝑢
(Type: warm-up)
Solution.
Step 1. What is being asked. The exercise asks for the angle between 𝑢 = (1,1,0) and 𝑣 = (1,0,1) in ℝ! . The
answer is a single angle, conventionally taken in [0, 𝜋], and it is reported here both in degrees and in radians.
The angle formula returns a cosine, and restricting to [0, 𝜋] is what makes the inverse cosine single valued,
so the answer is unique.
%⋅'
Step 2. Strategy. The angle formula of Ch. 1.3 states that cos𝜃 = |%| |'| whenever both vectors are non-zero.
This needs three separate ingredients: the dot product 𝑢 ⋅ 𝑣, the length |𝑢|, and the length |𝑣|. Compute all
three first, then assemble.
Assembling the formula only after every ingredient is known avoids mixing a dot product with a length in the
same line, which is where sign and root errors normally appear.
Step 3. Dot product. Multiply matching coordinates and add:
𝑢⋅𝑣 =1⋅1+1⋅0+0⋅1=1+0+0=1
This is the definition 𝑢 ⋅ 𝑣 = ∑* 𝑢* 𝑣* from Ch. 1.3. Note that the result is a number, not a vector.
Step 4. Length of 𝑢: |𝑢| = √1$ + 1$ + 0$ = √2. Length of 𝑣: |𝑣| = √1$ + 0$ + 1$ = √2. Both vectors have
exactly two coordinates equal to 1 and one equal to 0, which is why the two lengths agree.
Length is √𝑣 ⋅ 𝑣, so a coordinate equal to 0 contributes nothing and only the two unit slots matter.
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,Step 5. Assemble the formula:
𝑢⋅𝑣 1 1
cos𝜃 = = =
|𝑢| |𝑣| √2 ⋅ √2 2
The product √2 ⋅ √2 = 2 because squaring undoes the square root of a non-negative number, so no radical
survives in the final cosine.
Step 6. Invert the cosine. On [0, 𝜋] the only angle with cosine 1⁄2 is
𝜋
𝜃 = arccos(1⁄2) = 60∘ =
3
This is one of the standard values of the cosine and should be recognised rather than approximated, since
exam answers are expected in exact form.
Step 7. Name the trap. Do not divide by |𝑢|$ or forget one of the two lengths. Leaving out a factor of √2
would give cos𝜃 = 1⁄√2 and the wrong answer 45∘ . Both lengths belong in the denominator because the
formula normalises each vector separately.
The cosine must lie in [−1,1] for every pair of vectors, and this is guaranteed only when both lengths divide
out.
Step 8. Sanity check. The dot product is positive, so the angle must be strictly acute, and 60∘ < 90∘
#
confirms this. Numerically cos60∘ = 0.5 and = 0.5, so the two sides agree.
√$⋅√$
The sign of the dot product alone decides acute against obtuse, as recorded in the concept check of Ch. 1.3,
so it gives a free consistency test on the final angle.
Result
𝜃 = 60∘ = 𝜋/3.
E1.3 - perpendicularity / parameter
Problem statement.
For which value of 𝑘 ∈ ℝ are (𝑘, 1) and (4,3) perpendicular in ℝ$ ?
(Type: warm-up)
Solution.
Step 1. What is being asked. The two vectors (𝑘, 1) and (4,3) in ℝ$ contain one unknown, 𝑘, and the
exercise asks for the value or values of 𝑘 that make them perpendicular. The answer is therefore a number,
or possibly a set of numbers, and not a vector.
Recognising that the unknown sits inside a vector but that the answer is a scalar keeps the bookkeeping
straight when the condition is written down.
Step 2. Strategy. Perpendicularity is not tested with angles here. The rule in Ch. 1.3 says that for non-zero 𝑢
%⋅'
and 𝑣 the angle satisfies cos𝜃 = , and since cos90∘ = 0 and the lengths are non-zero, 𝑢 ⊥ 𝑣 holds
|%||'|
exactly when 𝑢 ⋅ 𝑣 = 0. So translate the geometric condition into the single algebraic equation 𝑢 ⋅ 𝑣 = 0.
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,Setting the dot product to zero replaces a trigonometric condition by a linear equation, which is far easier to
solve and involves no square roots at all.
Step 3. Write out the dot product with the unknown carried along:
(𝑘, 1) ⋅ (4,3) = 𝑘 ⋅ 4 + 1 ⋅ 3 = 4𝑘 + 3
The dot product is defined component-wise in Ch. 1.3, and it stays a legitimate expression when one
coordinate is a symbol rather than a number.
Step 4. Impose the perpendicularity condition, which gives the equation 4𝑘 + 3 = 0.
This is the translation step: the geometry has now become a single linear equation in one unknown, so a
unique solution is expected.
Step 5. Solve it. Subtract 3 from both sides to get 4𝑘 = −3, then divide both sides by 4:
3
𝑘=−
4
Both operations are reversible, since 4 ≠ 0, so no solution is created or lost and the value found is the only
one.
Step 6. Name the trap. Perpendicular is not the same as parallel. Requiring (𝑘, 1) to be a multiple of (4,3)
would give 𝑘 = 4⁄3, which is the answer to a different question. Also note that no square roots and no
lengths are needed here, because the lengths cancel out of the condition cos𝜃 = 0.
The angle formula has the lengths only in the denominator, so a zero numerator already forces the angle to
be 90∘ regardless of how long the vectors are.
Step 7. Check by substitution. With 𝑘 = − 3⁄4 the first vector is (− 3⁄4 , 1) and
(− 3⁄4 , 1) ⋅ (4,3) = −3 + 3 = 0
so the two vectors are indeed perpendicular.
Substituting back into the original condition is the definitive test, and it takes a single line.
Result
𝑘 = −3/4.
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, E1.4 - line through points, membership
Problem statement.
Let 𝐴 = (1,0,2,1) and 𝐵 = (3,2,0,5) in ℝ- . Does 𝑃 = (2,1,1,3) lie on the line through 𝐴 and 𝐵?
(Type: mixed)
Solution.
Step 1. What is being asked. The line through 𝐴 = (1,0,2,1) and 𝐵 = (3,2,0,5) lives in ℝ- , and the question is
whether the point 𝑃 = (2,1,1,3) lies on it. The answer is yes or no, and if yes it should come with the
parameter value 𝑡 that locates 𝑃 on the line.
A membership question always has a binary answer, but stating the witnessing parameter turns the answer
into a verifiable proof rather than a claim.
Step 2. Strategy. Follow the membership test of Procedure 1-C in Ch. 1.2. Write the line in parametric form
{𝐴 + 𝑡𝑑}, translate the problem so that the base point sits at the origin, and then test whether the translated
point is a scalar multiple of the direction.
Subtracting the base point removes the offset, so the remaining question is purely about direction and can
be answered coordinate by coordinate.
Step 3. Compute the direction from the two given points:
𝑑 = 𝐵 − 𝐴 = (3 − 1, 2 − 0, 0 − 2, 5 − 1) = (2,2, −2,4)
so the line is {𝐴 + 𝑡𝑑: 𝑡 ∈ ℝ}.
Any non-zero multiple of 𝐵 − 𝐴 is a valid direction for the line through 𝐴 and 𝐵, and this particular choice
makes 𝑡 = 0 land on 𝐴 and 𝑡 = 1 land on 𝐵.
Step 4. Translate the test point:
𝑃 − 𝐴 = (2 − 1, 1 − 0, 1 − 2, 3 − 1) = (1,1, −1,2)
The question is now whether there is a single scalar 𝑡 with (1,1, −1,2) = 𝑡(2,2, −2,4).
This is precisely the translated form of the membership condition in Procedure 1-C: 𝑃 lies on the line exactly
when 𝑃 − 𝐴 is built from the direction vector.
Step 5. Read the vector equation coordinate by coordinate. It splits into four scalar equations, 1 = 2𝑡, 1 =
2𝑡, −1 = −2𝑡 and 2 = 4𝑡, that is four equations in the single unknown 𝑡.
The system is heavily overdetermined, four conditions on one unknown, and that is exactly why membership
can fail: any one disagreement is fatal.
Step 6. Solve from the easiest coordinate and then verify the rest. The first coordinate gives 𝑡 = 1⁄2.
Substituting this single value into the remaining three: 2 ⋅ 1⁄2 = 1 matches the required 1; −2 ⋅ 1⁄2 = −1
matches the required −1; 4 ⋅ 1⁄2 = 2 matches the required 2.
Solving once and then checking, rather than solving four times, is what Procedure 1-C prescribes, and it
makes any inconsistency visible immediately.
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