Interleaved — Phase 1

AI-ML interleaved practice

printable — key stays hidden on paper

Instructions. Solve each problem on its own. These problems deliberately jump between subtopics — before computing, identify which tool each problem needs. Show all work. Use $\|\cdot\|$ notation for norms. Total: 50 marks.

1. (5 marks) Let $\mathbf{u} = (3, -4)$ and $\mathbf{v} = (1, 2)$ . (a) Compute $\mathbf{u}\cdot\mathbf{v}$ . (b) Find the angle $\theta$ between them (in terms of $\arccos$ ). (c) Find the scalar projection of $\mathbf{u}$ onto $\mathbf{v}$ .

2. (6 marks) For the matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix},$ determine whether $A$ is symmetric, and whether $A$ is positive definite. Justify using the appropriate criterion.

3. (4 marks) Let $\mathbf{x} = (1, -3, 4, -2)$ . Compute $\|\mathbf{x}\|_1$ , $\|\mathbf{x}\|_2$ , and $\|\mathbf{x}\|_\infty$ .

4. (6 marks) Let $f(x, y) = x^2 y + \sin(xy)$ . Compute the gradient $\nabla f$ , and evaluate it at $(1, 0)$ .

5. (5 marks) Given $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix},$ compute $\det(A)$ , state whether $A$ is invertible, and if so compute $A^{-1}$ .

6. (5 marks) Find all eigenvalues of $B = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix},$ and give an eigenvector for the largest eigenvalue.

7. (5 marks) Determine the rank of $C = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{pmatrix}$ using Gaussian elimination, and state the dimension of its null space.

8. (4 marks) Let $g(t) = (t^2 + 1)\,e^{3t}$ . Compute $g'(t)$ using the appropriate differentiation rule.

9. (5 marks) Given $A$ is $3\times 4$ and $B$ is $4 \times 2$ . State the dimensions of $AB$ and of $(AB)^\top$ . If $\mathbf{v}\in\mathbb{R}^2$ , is $AB\mathbf{v}$ defined, and what is its dimension? Also state $\operatorname{tr}$ requirements: for which of $AB$ or $BA$ is the trace defined?

10. (5 marks) Let $f(x,y) = 3x^2 + 2xy + y^2$ . Compute the Hessian matrix $H$ and state whether $f$ is convex.

Answer keyMark scheme & solutions

1. (Subtopic 1.1.3 — Dot product / projection / angle) (a) $\mathbf{u}\cdot\mathbf{v} = 3(1) + (-4)(2) = 3 - 8 = -5$ . (b) $\|\mathbf{u}\| = 5$ , $\|\mathbf{v}\| = \sqrt{5}$ . $\cos\theta = \dfrac{-5}{5\sqrt5} = \dfrac{-1}{\sqrt5}$ , so $\theta = \arccos(-1/\sqrt5) \approx 116.6^\circ$ . (c) Scalar projection of $\mathbf u$ onto $\mathbf v$ $= \dfrac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|} = \dfrac{-5}{\sqrt5} = -\sqrt5$ . Why: the phrase "angle / projection" signals the dot-product toolkit, not norms alone.

2. (Subtopics 1.1.6 symmetric + 1.1.17 positive definite) $A = A^\top$ (mirror across diagonal) → symmetric. Positive definite test via leading principal minors: $\Delta_1 = 2 > 0$ ; $\Delta_2 = \det\begin{pmatrix}2&1\\1&3\end{pmatrix} = 5 > 0$ ; $\Delta_3 = \det A$ . $\det A = 2(3\cdot2 - 1) - 1(1\cdot2 - 0) + 0 = 2(5) - 2 = 8 > 0$ . All minors positive → positive definite. Why: symmetric structure is required before applying the Sylvester (minors) criterion — recognizing that ordering is the point.

3. (Subtopic 1.1.4 — norms) $\|\mathbf x\|_1 = 1+3+4+2 = 10$ . $\|\mathbf x\|_2 = \sqrt{1+9+16+4} = \sqrt{30}\approx 5.48$ . $\|\mathbf x\|_\infty = \max(1,3,4,2) = 4$ . Why: three different norm definitions on the same vector — forces recall of each formula.

4. (Subtopic 1.2.4 gradient, using 1.2.3 partials) $f_x = 2xy + y\cos(xy)$ , $f_y = x^2 + x\cos(xy)$ . $\nabla f = (2xy + y\cos(xy),\; x^2 + x\cos(xy))$ . At $(1,0)$ : $f_x = 0 + 0 = 0$ ; $f_y = 1 + 1\cdot\cos(0) = 1 + 1 = 2$ . So $\nabla f(1,0) = (0, 2)$ . Why: gradient = vector of partials; chain rule inside $\sin(xy)$ .

5. (Subtopics 1.1.9 determinant + 1.1.8 inverse) $\det A = 1\cdot4 - 2\cdot3 = 4 - 6 = -2 \neq 0$ → invertible. $A^{-1} = \dfrac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$ . Why: nonzero determinant is the invertibility condition; then the 2×2 adjugate formula.

6. (Subtopic 1.1.13 — eigenvalues/eigenvectors) $\det(B - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 -7\lambda + 10 = (\lambda-5)(\lambda-2)$ . Eigenvalues $\lambda = 5, 2$ . Largest is $5$ : $(B-5I)\mathbf x = \begin{pmatrix}-1&1\\2&-2\end{pmatrix}\mathbf x = 0 \Rightarrow x_1 = x_2$ . Eigenvector $(1,1)$ . Why: characteristic polynomial, not decomposition — problem only asks eigenvalues/vectors.

7. (Subtopics 1.1.10 rank/null space + 1.1.12 Gaussian elimination) Row-reduce: $R_2 \to R_2 - 2R_1 = (0,0,0)$ ; $R_3 \to R_3 - R_1 = (0,-1,-2)$ . Rows: $(1,2,3),(0,-1,-2),(0,0,0)$ → 2 pivots → rank 2. Nullity $= n - \text{rank} = 3 - 2 = 1$ . Why: elimination reveals rank; rank–nullity gives null-space dimension.

8. (Subtopic 1.2.2 — product + chain rule) $g'(t) = (2t)e^{3t} + (t^2+1)(3e^{3t}) = e^{3t}(2t + 3t^2 + 3) = e^{3t}(3t^2 + 2t + 3)$ . Why: product rule with a chain-rule factor $\frac{d}{dt}e^{3t}=3e^{3t}$ .

9. (Subtopics 1.1.5 matrix mult dims + 1.1.7 transpose + 1.1.16 trace) $A_{3\times4}B_{4\times2} = (AB)_{3\times2}$ . $(AB)^\top$ is $2\times3$ . $AB\mathbf v$ : $(3\times2)(2\times1) = 3\times1$ vector — defined. Trace requires a square matrix. $AB$ is $3\times2$ (not square) → trace undefined. $BA$ is $4\times4$ (square) → $\operatorname{tr}(BA)$ defined. Why: dimensional-compatibility bookkeeping plus square-only trace rule.

10. (Subtopics 1.2.6 Hessian + 1.1.17 relation to PD/convexity) $f_x = 6x + 2y$ , $f_y = 2x + 2y$ . $H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 2 & 2 \end{pmatrix}$ . Minors: $6>0$ , $\det = 12 - 4 = 8 > 0$ → PD → $f$ is convex (strictly). Why: connects second-derivative Hessian to positive-definiteness convexity test.

[
  {"claim":"Problem 2: det A = 8 and A positive definite (all leading minors positive)","code":"import sympy as sp\nA=sp.Matrix([[2,1,0],[1,3,1],[0,1,2]])\nm1=A[0,0]; m2=A[:2,:2].det(); m3=A.det()\nresult=(m3==8) and all(x>0 for x in [m1,m2,m3])"},
  {"claim":"Problem 6: eigenvalues of B are {2,5}","code":"import sympy as sp\nB=sp.Matrix([[4,1],[2,3]])\nresult=set(B.eigenvals().keys())=={sp.Integer(2),sp.Integer(5)}"},
  {"claim":"Problem 10: Hessian is [[6,2],[2,2]] and positive definite","code":"import sympy as sp\nx,y=sp.symbols('x y')\nf=3*x**2+2*x*y+y**2\nH=sp.hessian(f,(x,y))\nmz=H[0,0]>0 and H.det()>0\nresult=(H==sp.Matrix([[6,2],[2,2]])) and mz"}
]