Mathematics for Data Science II · End Term · May 2026 · FN

**1. Set up:** In the displayed augmented matrix, row 3 equals twice row 1 plus row 2, including the right-hand side: 0=2(1)+(-2). **2. Work through the reasoning:** Thus there are two independent equations and three unknowns, with no inconsistency. Set x3=t; row 1 gives x1=1-2t and row 2 gives x2=-2+4t. There is one free parameter and infinitely many solutions. The coefficient matrix has rank 2 and is not invertible. Equivalence preserves rank, so every equivalent 3×3 matrix is also singular. **3. Check the result:** \[(x_1,x_2,x_3)=(1-2t,-2+4t,t),\quad t\in\mathbb R\] The question asks for the false statement. **Answer: B.** A: **Incorrect:** The free parameter gives infinitely many solutions, so this is true. B: **Correct:** False: rank 2 is below the matrix order 3. C: **Incorrect:** One arbitrary free variable determines the two pivot variables, so this is true. D: **Incorrect:** Rank preservation makes every equivalent matrix singular, so this is true.

**1. Set up:** The sequential criterion says that if f has limit L at a point, every sequence of distinct nearby domain points approaching that point has function values tending to L. **2. Work through the reasoning:** Therefore one such sequence whose function values fail to converge disproves existence of the limit. As usual, the approaching sequence must avoid the target point itself. **3. Check the result:** \[p_n\to p,\quad p_n\ne p,\quad f(p_n)\not\to L\quad\Longrightarrow\quad\lim_{x\to p}f(x)\ne L\] Equal limits along straight lines alone do not settle curved approaches, and failure of differentiability alone does not rule out continuity or a limit. **Answer: D.** A: **Incorrect:** Continuity guarantees a limit, the opposite of the question's requirement. B: **Incorrect:** Straight-line tests are not sufficient to establish or disprove a full multivariable limit by themselves. C: **Incorrect:** A function can be continuous and nondifferentiable, for example a norm at the origin. D: **Correct:** A nonconvergent image sequence contradicts the necessary sequential property of a limit.

**1. Set up:** A zero determinant is equivalent to row dependence, so some row can be solved as a linear combination of the others. **2. Work through the reasoning:** If adding another row leaves a row unchanged, the added row must be zero, forcing determinant zero. For a triangular matrix, determinant is the product of diagonal entries, so a zero product has a zero factor. In contrast, det(A)=det(A transpose) holds for all square matrices and implies no symmetry. **3. Check the result:** \[\det A=0\iff\text{rows of }A\text{ are dependent}\] Nonzero minors need not prevent cancellation in cofactor expansion. **Answer: A, C, D.** A: **Correct:** A nontrivial row dependence can be rearranged to express a row in terms of the other two. B: **Incorrect:** For example an invertible nonsymmetric upper-triangular matrix has the same determinant as its transpose. C: **Correct:** The operation leaves entries unchanged only if the added row is zero. D: **Correct:** A triangular determinant is its diagonal product. E: **Incorrect:** Rows (1,2,3),(1,2,3),(4,5,6) give determinant zero despite all three first-row minors being nonzero.

**1. Set up:** All five vectors have last coordinate zero, so their span lies in a three-dimensional subspace of R4. **2. Work through the reasoning:** The first three vectors (1,-1,1,0),(1,0,0,0),(1,1,1,0) are independent: their first three coordinates form a nonzero determinant. Hence dim(span S)=3. Every set of four vectors from this span is dependent. **3. Check the result:** \[\dim\operatorname{span}S=3<4\] A basis can always be selected from a finite spanning set, and that basis is both maximally independent and minimally spanning. **Answer: A, E.** A: **Correct:** Three independent vectors exist and the zero fourth coordinate gives the matching upper bound. B: **Incorrect:** The displayed independent triple rules out dimension 2. C: **Incorrect:** A basis subset does have both properties. D: **Incorrect:** No four vectors can be independent inside a three-dimensional span. E: **Correct:** Any four selected vectors are dependent.

**1. Set up:** With alpha=(3,1), the first condition is 3gamma1+gamma2=4. **2. Work through the reasoning:** Since beta=(6,2)=2alpha, its condition is exactly twice the first, imposing no further restriction. All solutions are gamma=(t,4-3t), t real. **3. Check the result:** \[\gamma=(t,4-3t),\quad t\in\mathbb R\] The free real parameter gives infinitely many vectors; (1,1) is just one member of the line. **Answer: B, D.** A: **Incorrect:** The explicit parameterised line gives valid solutions. B: **Correct:** One independent equation in two variables leaves infinitely many solutions. C: **Incorrect:** The second condition is redundant rather than making the solution unique. D: **Correct:** Substitution gives 3t+(4-3t)=4 and twice that equals 8. E: **Incorrect:** For example (0,4) also satisfies both conditions.

**1. Set up:** A is smooth and its gradient vanishes only at the origin. **2. Work through the reasoning:** D is polynomial; its gradient equations y(2x+y)=0 and x(x+2y)=0 also give only the origin. B=sin x+cos y is smooth but has critical points (pi/2+kpi,m pi) for all integers k,m, so infinitely many. C tends to 1 at the origin because sin(r²)/r²→1 but is assigned 0 there, so it is discontinuous and hence not differentiable. **3. Check the result:** \[\nabla(\sin x+\cos y)=(\cos x,-\sin y)=0\iff (x,y)=\left(\frac\pi2+k\pi,m\pi\right)\] Therefore B and C each violate at least one condition. **Answer: B, C.** A: **Incorrect:** Smooth everywhere with a single critical point, so it satisfies both conditions. B: **Correct:** Has infinitely many critical points, violating finiteness. C: **Correct:** Fails continuity and therefore differentiability at the origin. D: **Incorrect:** Smooth polynomial with only one simultaneous gradient zero.

**1. Set up:** Write v1=(1,0,2,1) and v2=(0,1,1,1). **2. Work through the reasoning:** The remaining vectors satisfy v3=2v1+v2 and v4=3v1+v2. v1 and v2 are independent because their first two coordinates are the two standard coordinate directions. Hence the span has dimension 2. **3. Check the result:** \[v_3=2v_1+v_2,\quad v_4=3v_1+v_2,\quad\dim\operatorname{span}S=2\] Any basis for it has exactly two vectors, so remove 4-2=2, retaining v1 and v2 for example. **Answer: 2.**

**1. Set up:** The four standard-basis images are c1=(1,2,0),c2=(0,1,1),c3=(1,3,1)=c1+c2,c4=(0,0,1). c1,c2,c4 are independent: their 3×3 column determinant is 1. **2. Work through the reasoning:** Thus rank(T)=3. **3. Check the result:** \[\dim\ker T=\dim\mathbb R^4-\operatorname{rank}T=4-3=1\] The domain has dimension 4, so rank-nullity gives nullity=4-3=1. **Answer: 1.**

**1. Set up:** Perpendicular vectors have zero dot product. **2. Work through the reasoning:** For u=(1,3,-5) and v=(5,c,7), u·v=5+3c-35=3c-30. Setting this to zero gives c=10. **3. Check the result:** \[(1,3,-5)\cdot(5,c,7)=3c-30=0\] Substituting checks 5+30-35=0. **Answer: 10.**

**1. Set up:** For f=x^4+kxy, the Hessian is [[12x²,k],[k,0]]. **2. Work through the reasoning:** At (1,2) its determinant is -k², which is zero exactly when k=0; this is the formal zero-discriminant answer expected by the question. There is a qualification: the usual extremum Hessian test requires a critical point. Here the gradient at (1,2) is (4+2k,k), which is nonzero for every k, so this point is never a critical point and that extremum-classification test does not actually apply there. **3. Check the result:** \[H_f(1,2)=\begin{bmatrix}12&k\\k&0\end{bmatrix},\qquad\det H_f=-k^2\] The answer 0 refers specifically to singularity of the displayed Hessian. **Answer: 0.**

**1. Set up:** Express u=(3,2,1) in B=((1,0,1),(1,1,0),(0,1,1)). **2. Work through the reasoning:** The coefficients are 1,2,0, since (3,2,1)=(1,0,1)+2(1,1,0). Apply linearity: T(u)=(1,2,0)+2(0,1,1)=(1,4,2). **3. Check the result:** \[T(3,2,1)=T(1,0,1)+2T(1,1,0)=(1,4,2)\] Thus a+2b-c=1+8-2=7. **Answer: 7.**

**1. Set up:** Track determinant multipliers for the operations producing I. **2. Work through the reasoning:** Swapping two rows multiplies by -1; adding a multiple of another row has factor 1; dividing a row by 6 has factor 1/6; multiplying one row by 3 has factor 3. Thus det(I)=(-1)(1)(1/6)(3)det(A)=-(1/2)det(A). **3. Check the result:** \[1=-\frac12\det A\quad\Longrightarrow\quad\det A=-2\] Since det(I)=1, det(A)=-2. **Answer: -2.**

**1. Set up:** For nonzero x,y, f=x³/(3x²y)=x/(3y). **2. Work through the reasoning:** Along x=y≠0 it is 1/3, whereas along the axes it is defined as zero. These distinct approach values show f is not continuous at the origin. g is the polynomial x^4+x³y+xy³+y^4 away from the origin and is assigned its polynomial value 0 at the origin, so it is continuous. **3. Check the result:** \[f(t,t)=\frac13\ne0=f(t,0)\quad(t\ne0)\] Exactly Statement 2 is true. **Answer: 1.**

**1. Set up:** A zero-change direction is orthogonal to the gradient and hence also to the given steepest-descent direction (12,3,-4)/13. **2. Work through the reasoning:** For (a,-4,3), the dot product condition is (12a-12-12)/13=0. Therefore 12a=24 and a=2. **3. Check the result:** \[(a,-4,3)\cdot\frac{(12,3,-4)}{13}=0\quad\Longrightarrow\quad12a=24\] Normalising this direction would not change whether its directional derivative is zero. **Answer: 2.**

**1. Set up:** The maximum directional derivative over unit directions equals the gradient norm, by Cauchy–Schwarz, attained in the steepest-ascent direction. **2. Work through the reasoning:** That derivative is given as 13, so ||gradient f||=13. The requested 2||gradient f|| is therefore 26. **3. Check the result:** \[\max_{\|u\|=1}D_u f=\|\nabla f\|=13\quad\Longrightarrow\quad2\|\nabla f\|=26\] The supplied descent direction fixes orientation, while this magnitude is supplied by the maximum rate. **Answer: 26.**

**1. Set up:** Here f=exp(x³+y²). **2. Work through the reasoning:** At (1,1), f=e² and its partial derivatives are 3e² and 2e². The tangent affine plane has equation z=e²+3e²(x-1)+2e²(y-1). Its associated direction subspace is obtained by removing the constant offset: z=3e²x+2e²y. **3. Check the result:** \[z=e^2+3e^2(x-1)+2e^2(y-1),\qquad V:\ z=3e^2x+2e^2y\] The question asks for this parallel subspace, not the translated tangent plane itself. **Answer: B.** A: **Incorrect:** Omits the e² factor in both slopes. B: **Correct:** Uses the two correct tangent slopes and passes through the origin. C: **Incorrect:** Interchanges and changes signs of the slopes. D: **Incorrect:** Uses incorrect negative slopes and omits e².

**1. Set up:** Linearise f=exp(x³+y²) at (0,0). **2. Work through the reasoning:** Its value is 1 and both first partial derivatives are zero there. Thus its first-order approximation is the constant L(x,y)=1. In particular f(0.1,-0.2)≈1.00 by this requested linear approximation. **3. Check the result:** \[L(x,y)=f(0,0)+\nabla f(0,0)\cdot(x,y)=1\] The actual value includes higher-order effects, but those are not part of the first-order estimate. **Answer: 1.00.**

**1. Set up:** The function x²+2xy+y² equals (x+y)². **2. Work through the reasoning:** A square is always nonnegative, and every nonnegative value t is achieved by choosing x=sqrt(t),y=0. **3. Check the result:** \[f(x,y)=(x+y)^2,\qquad\operatorname{range}f=[0,\infty)\] The range is therefore exactly [0,infinity), including zero when x=-y. **Answer: B.** A: **Incorrect:** Includes negative values and excludes the attainable value zero. B: **Correct:** Exactly the nonnegative real numbers. C: **Incorrect:** Excludes positive squares and includes negative numbers, which are impossible. D: **Incorrect:** Incorrectly excludes zero.

**1. Set up:** On D, x ranges from 0 to 2 and y from 2 to 4, so x+y ranges from 2 to 6 and is positive. **2. Work through the reasoning:** The square (x+y)² increases with this sum, so the maximum is 6²=36, attained at (2,4). **3. Check the result:** \[2\le x+y\le6\quad\Longrightarrow\quad f(x,y)\le36=f(2,4)\] The bound x+y<=6 together with a feasible equality point proves global maximality. **Answer: 36.**