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

**1. Set up:** In the displayed augmented matrix, row 3 equals twice row 1 plus row 2, including the right-hand entry: 6=2(1)+4. **2. Work through the reasoning:** Thus there are two independent consistent equations. Put x3=t; then x1=1-2t and x2=4+3t. There is one free variable and infinitely many solutions. The coefficient matrix has rank 2, so A is not invertible. Any equivalent 3×3 matrix has the same rank and is also singular. **3. Check the result:** \[(x_1,x_2,x_3)=(1-2t,4+3t,t),\quad t\in\mathbb R\] B is the false statement requested. **Answer: B.** A: **Incorrect:** The free parameter proves this statement true. B: **Correct:** False: the matrix has rank 2 rather than 3. C: **Incorrect:** One free parameter determines every solution, so this is true. D: **Incorrect:** Rank preservation keeps every equivalent matrix singular, so this is true.

**1. Set up:** A genuine first-order linear approximation means differentiability: f(p+h)=f(p)+L(h)+o(||h||). **2. Work through the reasoning:** As h→0, the linear term and remainder tend to zero, so f(p+h)→f(p), proving continuity and a limit. Equal limits on all straight lines do not control all curved approaches. Failure of differentiability does not itself decide limit existence. **3. Check the result:** \[f(p+h)=f(p)+L(h)+o(\|h\|)\quad\Longrightarrow\quad f(p+h)\to f(p)\] A divergent image along an approaching punctured sequence would instead disprove a limit. **Answer: A.** A: **Correct:** Differentiability implies continuity and therefore a limit. B: **Incorrect:** Matching line limits alone is insufficient for a full multivariable limit. C: **Incorrect:** Nondifferentiability can coexist with continuity or discontinuity. D: **Incorrect:** Under the standard punctured-sequence convention this contradicts limit existence.

**1. Set up:** Zero determinant means dependent columns, allowing one to be expressed using the others. **2. Work through the reasoning:** If replacing a column ci by 3(ci+cj) leaves A unchanged, then ci=3ci+3cj, so 2ci+3cj=0: a nontrivial column dependence and determinant zero. A triangular determinant is its diagonal product. Equality of determinant with its transpose holds universally and does not imply symmetry. **3. Check the result:** \[c_i=3(c_i+c_j)\quad\Longrightarrow\quad2c_i+3c_j=0\quad\Longrightarrow\quad\det A=0\] Nonzero first-column minors alone do not prevent cancellation in the determinant expansion. **Answer: A, C, D.** A: **Correct:** Singularity implies a nontrivial dependence that can be solved for a column. B: **Incorrect:** An invertible nonsymmetric upper-triangular matrix is a counterexample. C: **Correct:** The unchanged-column condition gives 2ci+3cj=0, forcing singularity. D: **Correct:** A zero product of diagonal entries has a zero diagonal factor. E: **Incorrect:** Transpose the matrix with rows (1,2,3),(1,2,3),(4,5,6): its determinant is zero while all first-column minors are nonzero.

**1. Set up:** Every vector has fourth coordinate zero, so the span has dimension at most 3. **2. Work through the reasoning:** The first three vectors' first-three-coordinate matrix has determinant 7, so they are independent and the dimension is exactly 3. Consequently no four-vector subset is independent, but every four-vector subset is dependent. **3. Check the result:** \[\det[v_1\ v_2\ v_3]_{\text{first 3 coordinates}}=7\ne0\] A basis subset exists and is both maximally independent and minimally spanning. **Answer: A, E.** A: **Correct:** The independent triple and zero fourth coordinate establish dimension 3. B: **Incorrect:** The independent triple rules out dimension 2. C: **Incorrect:** A basis selected from S has both stated properties. D: **Incorrect:** Four vectors cannot be independent in this three-dimensional span. E: **Correct:** A four-vector dependent subset exists; in fact all such subsets 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:** B is a function of x alone: it is -1-2x for x<-1, x² on [-1,1], and 2x-1 for x>1. **2. Work through the reasoning:** Values and first derivatives match at x=±1, so it is differentiable everywhere; at x=0 both partials vanish for every real y, giving infinitely many critical points. C=sin x+cos y is smooth and has an infinite grid of critical points. A is discontinuous at the origin because it grows without bound along x=y>0. **3. Check the result:** \[B_x(0,y)=B_y(0,y)=0\quad\text{for every }y\in\mathbb R\] D=xy+x+y is smooth but has only one critical point, (-1,-1). **Answer: B, C.** A: **Incorrect:** Fails differentiability at the origin because it is not even continuous there. B: **Correct:** Differentiable at its joins and has the entire line x=0 as critical points. C: **Correct:** Its gradient (cos x,-sin y) vanishes on an infinite grid. D: **Incorrect:** The gradient (y+1,x+1) vanishes only at (-1,-1).

**1. Set up:** Let v1=(1,2,0,1),v2=(0,1,2,1). **2. Work through the reasoning:** The third vector equals 2v1+v2. The fourth, (3,8,4,4), is not in their span: its first two coordinates force coefficients 3 and 2, which predict fourth coordinate 5 rather than 4. Thus v1,v2,v4 are independent and the span has dimension 3. **3. Check the result:** \[v_3=2v_1+v_2,\qquad v_4\notin\operatorname{span}\{v_1,v_2\},\qquad\dim\operatorname{span}S=3\] From four supplied vectors, remove the dependent third vector, so the minimum removal count is 1. **Answer: 1.**

**1. Set up:** Use the four displayed standard-basis images as matrix columns. **2. Work through the reasoning:** The first three columns are (2,1,1),(1,1,0),(1,2,1), whose determinant is 2≠0. Therefore rank(T)=3, the full codomain dimension. Its domain has dimension 4, so nullity=4-3=1 by rank-nullity. **3. Check the result:** \[\det\begin{bmatrix}2&1&1\\1&1&2\\1&0&1\end{bmatrix}=2\ne0,\qquad\dim\ker T=4-3=1\] The fourth column cannot raise rank above 3. **Answer: 1.**

**1. Set up:** The dot product of (2,-1,4) and (3,c,-2) is 6-c-8=-c-2. **2. Work through the reasoning:** Perpendicularity requires -c-2=0, giving c=-2. **3. Check the result:** \[(2,-1,4)\cdot(3,c,-2)=-c-2=0\quad\Longrightarrow\quad c=-2\] Substitute to check 6+2-8=0. **Answer: -2.**

**1. Set up:** The Hessian of f=kx²+y² is [[2k,0],[0,2]], with determinant 4k. **2. Work through the reasoning:** It is singular exactly at k=0, the formal inconclusive-discriminant value requested by the source. However, the ordinary second-derivative test for a local extremum presumes a critical point: at (2,1), the gradient is (4k,2), never zero. Thus no extremum classification by that test is applicable at this point for any k. **3. Check the result:** \[H_f=\begin{bmatrix}2k&0\\0&2\end{bmatrix},\qquad\det H_f=4k\] The numeric answer 0 refers to the zero Hessian determinant, not to establishing an extremum. **Answer: 0.**

**1. Set up:** The input u=(2,3,1) equals 2(1,1,0)+(0,1,1)+0(1,0,1). **2. Work through the reasoning:** Apply the given images linearly: T(u)=2(1,0,2)+(2,1,0)=(4,1,4). Hence 2a-b+c=2(4)-1+4=11. **3. Check the result:** \[T(2,3,1)=2(1,0,2)+(2,1,0)=(4,1,4),\qquad2a-b+c=11\] The order of the supplied basis and its corresponding images must be preserved. **Answer: 11.**

**1. Set up:** The operations taking A to I multiply its determinant by -1 for the row swap, 1 for adding another row's multiple, 1/10 for dividing row 3, and 3 for multiplying row 1. **2. Work through the reasoning:** Therefore 1=det(I)=-(3/10)det(A), so det(A)=-10/3≈-3.333333. Round to two decimal places: -3.33. **3. Check the result:** \[1=-\frac3{10}\det A\quad\Longrightarrow\quad\det A=-\frac{10}{3}\approx-3.33\] The added multiple has no effect on determinant regardless of whether that multiple is 2 or 3. **Answer: -3.33.**

**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 direction of no first-order change is orthogonal to the gradient and hence to its steepest-ascent direction. **2. Work through the reasoning:** Dot (3,b,6) with (-1,2,2)/9: (-3+2b+12)/9=(2b+9)/9. Set it equal to zero to obtain 2b=-9. **3. Check the result:** \[(3,b,6)\cdot\frac{(-1,2,2)}9=0\quad\Longrightarrow\quad2b=-9\] Only the orientation of the supplied ascent vector matters here; scaling it to unit length would not change orthogonality. **Answer: -9.**

**1. Set up:** At the specified point the gradient is nonzero because a steepest-ascent direction is supplied. **2. Work through the reasoning:** The vectors with zero directional derivative form the orthogonal complement of that gradient, a plane through the origin in R³. Its unit vectors form a circle, but the span of that circle is the whole plane. Hence dim(span S)=2. **3. Check the result:** \[\operatorname{span}S=(\nabla f)^\perp,\qquad\dim(\nabla f)^\perp=3-1=2\] The set of unit directions itself is not a subspace, which is why the question asks for its span. **Answer: 2.**

**1. Set up:** The graph of f(x,y,z)=x³+y²+z lies in R4 with output coordinate w. **2. Work through the reasoning:** At (1,0,-1), f=0 and the gradient is (3,0,1). **3. Check the result:** \[w=f(p)+\nabla f(p)\cdot((x,y,z)-p)=3x+z-2\] Its tangent affine hyperplane is \[w=0+3(x-1)+0(y-0)+(z+1)=3x+z-2.\] Because the input has three coordinates, this tangent hyperplane has three independent input directions even though the source calls it a plane. **Answer: A.** A: **Correct:** Uses the correct value and gradient at the specified point. B: **Incorrect:** Adds a y term and gives the z slope the wrong sign. C: **Incorrect:** Uses incorrect x and y slopes. D: **Incorrect:** Has incorrect signs and offset.

**1. Set up:** Linearise f=x³+y²+z at p=(1,-1,0). **2. Work through the reasoning:** Its value there is 2 and its gradient is (3,-2,1). The displacement to (0.5,-0.7,0.1) is (-0.5,0.3,0.1). Therefore the first-order estimate is 2+3(-0.5)-2(0.3)+0.1=0.00. **3. Check the result:** \[L(0.5,-0.7,0.1)=2+3(-0.5)-2(0.3)+0.1=0\] This is the requested linear approximation, not the exact function value at the new point. **Answer: 0.00.**

**1. Set up:** For f=x^4-y^4, every nonnegative t can be obtained with x=t^(1/4),y=0. **2. Work through the reasoning:** Every negative t can be obtained with x=0,y=(-t)^(1/4). Zero is obtained at x=y. **3. Check the result:** \[f(x,0)=x^4\ge0,\quad f(0,y)=-y^4\le0,\quad\operatorname{range}f=\mathbb R\] Therefore its range is all real numbers, R. **Answer: B.** A: **Incorrect:** Positive values and zero are also attainable. B: **Correct:** Both signs and zero are attainable, covering all real values. C: **Incorrect:** Zero is attained, so it must not be excluded. D: **Incorrect:** Negative values and zero are also attainable.

**1. Set up:** On the square -1<=x,y<=1, x^4>=0 and y^4<=1, so f=x^4-y^4>=-1. **2. Work through the reasoning:** Equality occurs at x=0,y=1 or y=-1, all feasible points. Thus the absolute minimum is -1. **3. Check the result:** \[x^4-y^4\ge0-1=-1,\qquad f(0,\pm1)=-1\] A global lower bound plus a feasible equality point proves the result. **Answer: -1.**