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

**1. Set up:** Express (x,y,z) in the supplied input basis: \((x,y,z)=a(1,0,1)+b(1,1,0)+c(0,1,1)\). **2. Work through the reasoning:** Solving x=a+b,y=b+c,z=a+c gives a=(x-y+z)/2,b=(x+y-z)/2,c=(-x+y+z)/2. By linearity, the images of these basis vectors are e1,e2,e3, so T(x,y,z)=(a,b,c). **3. Check the result:** \[T(x,y,z)=\left(\frac{x-y+z}{2},\frac{x+y-z}{2},\frac{-x+y+z}{2}\right)\] Thus the factor 1/2 and the stated coordinate order are essential. **Answer: B.** A: **Incorrect:** Omits the required factor 1/2. B: **Correct:** Gives the three solved coefficients in the correct order. C: **Incorrect:** Interchanges the first two coordinates. D: **Incorrect:** Multiplies by 2 instead of dividing by 2.

**1. Set up:** For nonzero (x,y), put r=sqrt(x²+y²). **2. Work through the reasoning:** As (x,y)→(0,0), r→0 and sin(r)/r→1, regardless of path. Thus the limit is 1 although f(0,0)=0; the function is not continuous at the origin. **3. Check the result:** \[\lim_{(x,y)\to(0,0)}\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}=1\ne f(0,0)\] At (0,pi), the function is continuous with value sin(pi)/pi=0, while at (pi/2,0) its value is 2/pi, not 0. **Answer: D.** A: **Incorrect:** At (0,pi) the continuous expression tends to 0, not 1. B: **Incorrect:** The origin's assigned value is 0 but the limit is 1. C: **Incorrect:** At radius pi/2 the expression is 2/pi, not 0. D: **Correct:** The radial limit sin(r)/r at zero is 1.

**1. Set up:** Columns are the B2-coordinates of T applied to successive B1 vectors. **2. Work through the reasoning:** T(1,0,1)=(1,-3), T(1,1,0)=(-1,1), and T(0,1,1)=(-2,-2). Write any (u,v)=a(1,1)+b(1,-1), so a=(u+v)/2,b=(u-v)/2. The three coordinate columns are (-1,2),(0,-1),(-2,0). **3. Check the result:** \[[T]_{B_2\leftarrow B_1}=\begin{bmatrix}-1&0&-2\\2&-1&0\end{bmatrix}\] Therefore the representation is \(\begin{bmatrix}-1&0&-2\\2&-1&0\end{bmatrix}\). **Answer: C.** A: **Incorrect:** Its first column does not represent (1,-3) in B2. B: **Incorrect:** The third column has the wrong first-coordinate sign. C: **Correct:** Matches all three coordinate columns. D: **Incorrect:** The second column should be (0,-1), not (0,1).

**1. Set up:** The bound |f(x,y)|<=|x-y|²<=2(x²+y²) proves continuity at the origin. **2. Work through the reasoning:** For the x-partial, f(h,0)/h=h sin(1/h)→0. For the y-partial, f(0,h)/h=-h sin(1/h)→0. Thus both partials are zero. To prove differentiability, divide the error from the zero linear map by r=sqrt(x²+y²): |f|/r<=2r→0. **3. Check the result:** \[\frac{|f(x,y)-f(0,0)-0|}{\sqrt{x^2+y^2}}\le2\sqrt{x^2+y^2}\longrightarrow0\] This establishes more than the existence of partial derivatives. **Answer: A, B, D.** A: **Correct:** The quadratic bound tends to zero. B: **Correct:** Both partial derivatives equal 0. C: **Incorrect:** The two partial derivatives are equal, not unequal. D: **Correct:** The first-order remainder divided by distance tends to zero.

**1. Set up:** Finite-dimensional real vector spaces are isomorphic exactly when their dimensions agree. **2. Work through the reasoning:** W1 is a 4-entry matrix space with one independent homogeneous equation, so dimension 3. W2 has two free parameters (x,y), since z=x+y and w=-x, so dimension 2. W3 has nine entries and three independent row-sum conditions, so dimension 6. W4 has determinant -2 and rank 3, so its column space has dimension 3. **3. Check the result:** \[(\dim W_1,\dim W_2,\dim W_3,\dim W_4,\dim W_5)=(3,2,6,3,3)\] Every vector is orthogonal to the zero vector, so W5=R³. **Answer: A, D, E.** A: **Correct:** Four coordinates minus one independent constraint gives dimension 3. B: **Incorrect:** It has dimension 2, not 3. C: **Incorrect:** Three independent row constraints leave dimension 6. D: **Correct:** The displayed three columns are independent, giving dimension 3. E: **Correct:** Orthogonality to zero imposes no restriction.

**1. Set up:** A rotation is an orthogonal transformation and preserves Euclidean norm. **2. Work through the reasoning:** If u=Rv and R transpose R=I, then ||u||²=v transpose R transpose Rv=v transpose v=||v||². Therefore the 70-degree rotation about the z-axis leaves the length 5 unchanged. **3. Check the result:** \[\|Rv\|^2=v^{\mathsf T}R^{\mathsf T}Rv=\|v\|^2=25\] It changes direction, not magnitude. **Answer: 5.**

**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:** \[u\cdot v=5+3c-35=0\quad\Longrightarrow\quad c=10\] Substituting checks 5+30-35=0. **Answer: 10.**

**1. Set up:** For f=3xyz+x²+y²+z², the gradient is (3yz+2x,3xz+2y,3xy+2z). **2. Work through the reasoning:** At (1,-1,0) it equals (2,-2,-3), whose norm is sqrt(17). The unit steepest-descent direction is minus the normalised gradient: (a,b,c)=(-2,2,3)/sqrt(17). **3. Check the result:** \[-\frac{\nabla f}{\|\nabla f\|}=\frac{(-2,2,3)}{\sqrt{17}},\qquad\sqrt{17}(a+b+c)=3\] Thus sqrt(17)(a+b+c)=-2+2+3=3. **Answer: 3.**

**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:** \[1=\det I=-\frac12\det A\quad\Longrightarrow\quad\det A=-2\] Since det(I)=1, det(A)=-2. **Answer: -2.**

**1. Set up:** For f(x,y)=x³+y², f(1,1)=2, fx(1,1)=3 and fy(1,1)=2. **2. Work through the reasoning:** The tangent plane is z=2+3(x-1)+2(y-1)=3x+2y-3. Test every candidate in this plane equation, rather than the original curved surface. **3. Check the result:** \[z=f(1,1)+f_x(1,1)(x-1)+f_y(1,1)(y-1)=3x+2y-3\] At (3,-3) it gives z=0 and at (0,0) it gives z=-3. **Answer: C, D.** A: **Incorrect:** For (1,0), the plane requires z=0, not 1. B: **Incorrect:** For (1,1), the plane requires z=2, not 0. C: **Correct:** 3(3)+2(-3)-3=0. D: **Correct:** At x=y=0 the plane height is -3.

**1. Set up:** At (0,0), f=0 and both partial derivatives of x³+y² are zero. **2. Work through the reasoning:** The tangent plane is therefore z=0. Its direction subspace is span{(1,0,0),(0,1,0)}, containing two independent directions, so its dimension is 2. **3. Check the result:** \[V=\operatorname{span}\{(1,0,0),(0,1,0)\},\qquad\dim V=2\] Being embedded in three-dimensional space does not make a plane three-dimensional. **Answer: 2.**

**1. Set up:** The inequalities -1<=x<=1 and 2<=y<=6 describe the Cartesian product of two intervals. **2. Work through the reasoning:** The boundary consists of horizontal and vertical line segments forming a rectangle with side lengths 2 and 4. **3. Check the result:** \[D=[-1,1]\times[2,6]\] Because the lengths differ, it is not a square. **Answer: B — Selected: A rectangle.** A: **Incorrect:** The two side lengths are unequal. B: **Correct:** The four boundary segments form a 2-by-4 rectangle. C: **Incorrect:** The boundary has four sides rather than three. D: **Incorrect:** The coordinate bounds give straight edges rather than a circular equation.

**1. Set up:** For f=x²+xy+y² on the rectangle, fy=x+2y>=-1+4=3>0, so the maximum lies on y=6. **2. Work through the reasoning:** There f(x,6)=x²+6x+36 has derivative 2x+6>0 throughout [-1,1], so its maximum is at x=1. Hence M=f(1,6)=1+6+36=43. **3. Check the result:** \[f(x,y)\le f(x,6)\le f(1,6)=43\] Monotonicity checks both boundary reductions and ensures the value is global. **Answer: 43.**

**1. Set up:** Since fy=x+2y>0 everywhere in the rectangle, the minimum lies on y=2. **2. Work through the reasoning:** On that edge f=x²+2x+4=(x+1)²+3. Its minimum over -1<=x<=1 occurs at x=-1, giving m=3. **3. Check the result:** \[f(x,y)\ge f(x,2)=(x+1)^2+3\ge3\] The unconstrained critical point (0,0) lies outside this rectangle and is not a feasible minimum candidate. **Answer: 3.**

**1. Set up:** Here f=exp(x²+y²+z²). **2. Work through the reasoning:** S1 is false because the exponent is nonnegative, so f>=1 and zero is not in its range. S2 is true because the exponential of this polynomial is differentiable everywhere in R³. S3 is false: the gradient at the origin is (0,0,0), so every directional derivative there is 0 rather than 2. **3. Check the result:** \[\operatorname{range}(e^{x^2+y^2+z^2})=[1,\infty),\qquad\nabla f(0,0,0)=0\] Exactly one statement is correct. **Answer: 1.**

**1. Set up:** The gradient is 2exp(x²+y²+z²)(x,y,z). **2. Work through the reasoning:** At (-1,-1,1), it is 2e³(-1,-1,1). Dot with the supplied unit vector (1,1,1)/sqrt(3) to get the rate \(-2e^3/\sqrt3\). It is written as k e³, so k=-2/sqrt(3)≈-1.154701. **3. Check the result:** \[D_u f(-1,-1,1)=2e^3(-1,-1,1)\cdot\frac{(1,1,1)}{\sqrt3}=-\frac{2e^3}{\sqrt3}\] Rounded to two decimals, k=-1.15. **Answer: -1.15.**

**1. Set up:** For A=[[3,0,-1],[1,2,0],[1,1,1]], expand along the first row. det(A)=3(2×1-0×1)+(-1)(1×1-2×1)=6+1=7. **2. Work through the reasoning:** The third-column cofactor has positive sign in this row, and its negative entry must still be included. **3. Check the result:** \[\det A=3(2)+(-1)(-1)=7\] A nonzero determinant also proves A has rank 3. **Answer: 7.**

**1. Set up:** The columns of B are u=(1,2,5), v=(1,4,-3), and w=(1,3,1). **2. Work through the reasoning:** They satisfy w=(u+v)/2, so rank is at most 2. The first two columns are independent: their first two entries have determinant 1×4-1×2=2≠0. **3. Check the result:** \[w=\frac12(u+v),\quad\det\begin{bmatrix}1&1\\2&4\end{bmatrix}=2\ne0\] Therefore rank(B)=2, not 1 or 3. **Answer: 2.**

**1. Set up:** A has nonzero determinant 7 and rank 3, whereas B has rank 2. **2. Work through the reasoning:** Matrix equivalence preserves rank, and similarity does as well, so these two are not equivalent or similar. A's homogeneous system has only the zero solution; B's nullity is 3-2=1, giving infinitely many homogeneous solutions. **3. Check the result:** \[\operatorname{rank}A=3,\quad\operatorname{rank}B=2,\quad\dim\ker A=0,\quad\dim\ker B=1\] B's column space is a two-dimensional plane through the origin, so all three of its columns lie in that plane. **Answer: B, D.** A: **Incorrect:** Different ranks rule out equivalence as well as similarity. B: **Correct:** Invertibility of A and positive nullity of B give the stated homogeneous-solution comparison. C: **Incorrect:** Reverses the independence properties of the two matrices' columns. D: **Correct:** All B columns lie in its two-dimensional column space.