Machine Learning Foundations · End Term · May 2026 · FN

**1. Understand the given matrix and vector:** We are given a matrix \(A\) of size \(3 \times 2\) and a target vector \(b \in \mathbb{R}^3\): \[A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix}, \quad b = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\] Here, \(\hat{b}\) is the orthogonal projection of \(b\) onto the column space \(\text{Col}(A)\), and \(e = b - \hat{b}\) is the residual (error) vector. **2. Check if \(Ax = b\) has an exact solution:** Let \(x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\). Writing out the three linear equations from \(Ax = b\) gives: \[x_1 = 1, \quad x_2 = 2, \quad x_1 + x_2 = 4\] Substituting \(x_1 = 1\) and \(x_2 = 2\) into the third equation gives \(1 + 2 = 3 \neq 4\). Because the third equation is contradicted, \(b\) does not lie in \(\text{Col}(A)\), so \(Ax = b\) has no exact solution. **3. Analyze the properties of the orthogonal projection \(\hat{b}\) and residual \(e\):** By the Orthogonal Decomposition Theorem: 1. The orthogonal projection \(\hat{b} \in \text{Col}(A)\) is the unique vector in \(\text{Col}(A)\) closest to \(b\). That is, for every vector \(v \in \text{Col}(A)\), \(\|b - \hat{b}\| \le \|b - v\|\). 2. The error vector \(e = b - \hat{b}\) is orthogonal to every column of \(A\), which means \(A^T e = 0\). By definition, \(e\) belongs to the null space of \(A^T\), written as \(e \in \text{Null}(A^T)\). Since \(e \neq 0\) and \(\text{Col}(A) \perp \text{Null}(A^T)\), \(e\) cannot be in \(\text{Col}(A)\). **4. Conclude:** Statements A, B, and C are correct, while D and E are false. **Answer: A, B, C — The system Ax = b has no exact solution, b-hat is the closest vector in Col(A) to b, and e lies in Null(A^T).** A: **Correct:** From the first two rows of \(Ax = b\), we must have \(x_1 = 1\) and \(x_2 = 2\), which gives \(x_1 + x_2 = 3 \neq 4\) in the third row. Hence no exact solution exists. B: **Correct:** By the best approximation theorem for orthogonal projections, \(\hat{b}\) minimizes the Euclidean distance \(\|b - v\|\) over all vectors \(v \in \text{Col}(A)\). C: **Correct:** The residual vector \(e = b - \hat{b}\) is orthogonal to \(\text{Col}(A)\), which means \(A^T e = 0\), so \(e \in \text{Null}(A^T)\). D: **Incorrect:** Because \(b \notin \text{Col}(A)\), the residual \(e = b - \hat{b}\) is a non-zero vector orthogonal to \(\text{Col}(A)\), so \(e \notin \text{Col}(A)\). E: **Incorrect:** Any vector in \(\text{Col}(A)\) has the form \([x_1, x_2, x_1+x_2]^T\), where the third component is the sum of the first two. For \(b = [1, 2, 4]^T\), \(1 + 2 \neq 4\), so \(b \notin \text{Col}(A)\).

**1. Set up the normal equations for least squares:** We are given the design matrix \(X\) and target vector \(y\): \[X = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix}, \quad y = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\] The least-squares weight vector \(w^* = \begin{bmatrix} a \\ b \end{bmatrix}\) minimizes \(f(w) = \|Xw - y\|^2\) and satisfies the normal equation \(X^T X w^* = X^T y\). **2. Compute \(X^T X\) and \(X^T y\):** Multiplying \(X^T\) by \(X\) and \(y\): \[X^T X = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\] \[X^T y = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 6 \\ 7 \end{bmatrix}\] **3. Solve for \(w^* = (a, b)\):** The inverse of \(X^T X\) has determinant \(2(2) - 1(1) = 3\): \[w^* = \begin{bmatrix} a \\ b \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 6 \\ 7 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 12 - 7 \\ -6 + 14 \end{bmatrix} = \begin{bmatrix} 5/3 \\ 8/3 \end{bmatrix}\] Thus, \(a = \frac{5}{3}\) and \(b = \frac{8}{3}\). **4. Calculate the predictions, residuals, and minimum loss:** The predicted vector is \(\hat{y} = Xw^* = \begin{bmatrix} 5/3 \\ 8/3 \\ 13/3 \end{bmatrix}\). The residual vector \(e = y - \hat{y}\) is: \[e = \begin{bmatrix} 2 - 5/3 \\ 3 - 8/3 \\ 4 - 13/3 \end{bmatrix} = \begin{bmatrix} 1/3 \\ 1/3 \\ -1/3 \end{bmatrix}\] So the residual for the third observation is \(e_3 = -\frac{1}{3}\), and the minimum loss is \(f(w^*) = (1/3)^2 + (1/3)^2 + (-1/3)^2 = \frac{1}{3} \neq 0\). **Answer: A, B, D — a = 5/3, b = 8/3, and the third residual is e_3 = -1/3.** A: **Correct:** Solving the normal equations \(2a + b = 6\) and \(a + 2b = 7\) yields \(a = \frac{5}{3}\). B: **Correct:** Substituting \(a = \frac{5}{3}\) into \(b = 6 - 2a\) gives \(b = 6 - \frac{10}{3} = \frac{8}{3}\). C: **Incorrect:** The minimum value of the loss function is \(\|e\|^2 = \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 = \frac{1}{3}\), which is greater than \(0\) because \(y\) is not in the column space of \(X\). D: **Correct:** For the third observation, \(\hat{y}_3 = a + b = \frac{5}{3} + \frac{8}{3} = \frac{13}{3}\), so \(e_3 = y_3 - \hat{y}_3 = 4 - \frac{13}{3} = -\frac{1}{3}\).

**1. Identify the given values and property of a unitary matrix:** We are given that \(U \in \mathbb{C}^{n \times n}\) is a unitary matrix, and vectors \(x, y \in \mathbb{C}^n\) satisfy: \[\|Ux\| + \|Uy\| = 7 \quad \text{and} \quad \|x\| = 3\] We need to find the Euclidean norm \(\|y\|\). **2. Apply the norm-preserving property of unitary matrices:** A matrix \(U\) is unitary if \(U^* U = I\). Multiplying any vector \(v\) by a unitary matrix preserves its Euclidean norm (length) because: \[\|Uv\|^2 = (Uv)^*(Uv) = v^* U^* U v = v^* I v = \|v\|^2 \implies \|Uv\| = \|v\|\] Therefore, we have \(\|Ux\| = \|x\|\) and \(\|Uy\| = \|y\|\). **3. Substitute the known values and solve for \(\|y\|\):** Substituting \(\|Ux\| = \|x\| = 3\) and \(\|Uy\| = \|y\|\) into the given equation: \[3 + \|y\| = 7 \implies \|y\| = 7 - 3 = 4\] **Answer: 4.**

**1. Understand the Singular Value Decomposition (SVD) of \(A\):** We are given \(A \in \mathbb{R}^{3 \times 3}\) with SVD \(A = U \Sigma V^T\), where the singular value matrix is: \[\Sigma = \text{diag}\left(5, 2, \frac{1}{2}\right)\] Thus, the singular values of \(A\) are \(\sigma_1 = 5\), \(\sigma_2 = 2\), and \(\sigma_3 = \frac{1}{2}\). We define \(B = 2A\). **2. Find the singular values of \(B = 2A\):** Multiplying \(A\) by the positive scalar \(2\) gives: \[B = 2A = U (2\Sigma) V^T = U \,\text{diag}(10, 4, 1)\, V^T\] Therefore, the singular values of \(B\) are \(10, 4, 1\). The largest singular value of \(B\) is \(10\), and the smallest singular value of \(B\) is \(1\). **3. Find the singular values of \(B^{-1}\):** Since \(B = U (2\Sigma) V^T\) is invertible, its inverse is: \[B^{-1} = (V^T)^{-1} (2\Sigma)^{-1} U^{-1} = V \,\text{diag}\left(\frac{1}{10}, \frac{1}{4}, 1\right) U^T\] The singular values of \(B^{-1}\) are the reciprocals of the singular values of \(B\), which are \(1, \frac{1}{4}, \frac{1}{10}\). Thus, the largest singular value of \(B^{-1}\) is \(1\), and the smallest singular value of \(B^{-1}\) is \(\frac{1}{10}\). **Answer: A, C — The largest singular value of B is 10, and the smallest singular value of B^(-1) is 1/10.** A: **Correct:** Scaling \(A\) by \(2\) scales each singular value by \(2\), so the largest singular value of \(B\) is \(2 \times 5 = 10\). B: **Incorrect:** The smallest singular value of \(B = 2A\) is \(2 \times \frac{1}{2} = 1\), not \(\frac{1}{4}\). C: **Correct:** The singular values of \(B^{-1}\) are the reciprocals of those of \(B\), so the smallest singular value of \(B^{-1}\) is \(\frac{1}{10}\). D: **Incorrect:** The largest singular value of \(B^{-1}\) is the reciprocal of the smallest singular value of \(B\), which is \(\frac{1}{1} = 1\), not \(\frac{1}{10}\). E: **Incorrect:** Multiplying \(A\) by \(2\) doubles its singular values to \(10, 4, 1\); they do not remain \(5, 2, 1/2\).

**1. Understand the relationship between PCA and eigenvalues of the covariance matrix:** We are given the covariance matrix of a centered two-dimensional dataset: \[C = \begin{bmatrix} 4 & a \\ a & 4 \end{bmatrix}, \quad a > 0\] We are told that the first principal component points in the direction \(v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) and the variance captured along this direction is \(\lambda_1 = 6\). **2. Set up the eigenvalue equation:** In Principal Component Analysis (PCA), the principal component directions are eigenvectors of the covariance matrix \(C\), and the variance along a principal component equals its corresponding eigenvalue \(\lambda_1\). Thus: \[C v = \lambda_1 v\] **3. Calculate \(a\) step by step:** Multiplying \(C\) by \(v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) gives: \[\begin{bmatrix} 4 & a \\ a & 4 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 + a \\ a + 4 \end{bmatrix} = (4 + a) \begin{bmatrix} 1 \\ 1 \end{bmatrix}\] Equating the eigenvalue \(4 + a\) to the given variance \(6\): \[4 + a = 6 \implies a = 2\] Since \(a = 2 > 0\), the second eigenvalue is \(4 - a = 2 < 6\), confirming that \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is indeed the first principal component. **Answer: 2.**

**1. State Chebyshev's inequality:** For any random variable \(X\) with finite mean \(\mu = E(X)\) and finite variance \(\text{Var}(X) = \sigma^2\), Chebyshev's inequality states that for any \(k > 0\): \[P(|X - \mu| \ge k) \le \frac{\text{Var}(X)}{k^2}\] **2. Identify the given values:** We are given: - Mean \(\mu = E(X) = 100\) - Deviation threshold \(k = 20\) (from \(|X - 100| \ge 20\)) - Required upper bound on probability: \(0.10\) **3. Solve for the largest possible value of \(\text{Var}(X)\):** To guarantee \(P(|X - 100| \ge 20) \le 0.10\) using Chebyshev's inequality, we set the Chebyshev upper bound to be at most \(0.10\): \[\frac{\text{Var}(X)}{20^2} \le 0.10\] \[\frac{\text{Var}(X)}{400} \le 0.10 \implies \text{Var}(X) \le 400 \times 0.10 = 40\] Thus, the largest possible value of \(\text{Var}(X)\) is \(40\). **Answer: 40.**

**1. State the covariance transformation formula:** We are given a random vector \(X = \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \text{Normal}(\mu, \Sigma)\) with covariance matrix \(\Sigma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), and a linear transformation \(Y = AX\) where: \[A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \text{Cov}(Y) = \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}\] For any linear transformation \(Y = AX\), the covariance matrix of \(Y\) is given by: \[\text{Cov}(Y) = A \Sigma A^T\] **2. Compute \(A \Sigma A^T\) in terms of \(a, b, c, d\):** First multiply \(A\) and \(\Sigma\): \[A\Sigma = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ c & d \end{pmatrix}\] Now multiply the result by \(A^T = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\): \[A\Sigma A^T = \begin{pmatrix} a+c & b+d \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} a+b+c+d & b+d \\ c+d & d \end{pmatrix}\] **3. Match the entries with \(\text{Cov}(Y)\):** Comparing the top-left \((1,1)\) entry of \(A\Sigma A^T\) directly with \(\text{Cov}(Y) = \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}\), we immediately get: \[a + b + c + d = 5\] (We can also find the individual entries: \(d = 2\), \(b + d = 3 \implies b = 1\), \(c + d = 3 \implies c = 1\), and \(a = 5 - 1 - 1 - 2 = 1\), whose sum is \(1 + 1 + 1 + 2 = 5\).) **Answer: 5.**

**1. Understand the distribution and target event:** We are given a continuous uniform random variable \(X \sim U(2, 8)\) and the linear transformation \(Y = 10 - 2X\). We need to find the probability \(P(Y < 0)\). **2. Rewrite the inequality in terms of \(X\):** Substitute \(Y = 10 - 2X\) into the inequality \(Y < 0\): \[10 - 2X < 0 \implies 2X > 10 \implies X > 5\] Therefore, \(P(Y < 0) = P(X > 5)\). **3. Calculate the probability using the Uniform distribution on \([2, 8]\):** For a continuous uniform distribution \(X \sim U(a, b)\) with \(a = 2\) and \(b = 8\), the probability that \(X\) falls in the sub-interval \((5, 8]\) is the ratio of the sub-interval length to the total interval length: \[P(X > 5) = \frac{8 - 5}{8 - 2} = \frac{3}{6} = 0.5\] **Answer: 0.5.**

**1. Identify the given probability density function (PDF):** The continuous random variable \(X\) has PDF: \[f(x) = \begin{cases} \dfrac{x}{8}, & 0 < x < 4 \\ 0, & \text{otherwise} \end{cases}\] We are given that \(P(a < X < 4) = \frac{3}{4}\) for some constant \(a \in (0, 4)\), and we need to find \(a\). **2. Set up the definite integral for \(P(a < X < 4)\):** By the definition of a continuous PDF, the probability that \(X\) lies between \(a\) and \(4\) is: \[P(a < X < 4) = \int_{a}^{4} \frac{x}{8}\,dx\] **3. Evaluate the integral and solve for \(a\):** Integrating \(\frac{x}{8}\) with respect to \(x\): \[\int_{a}^{4} \frac{x}{8}\,dx = \left[ \frac{x^2}{16} \right]_{a}^{4} = \frac{4^2 - a^2}{16} = \frac{16 - a^2}{16}\] Set this equal to \(\frac{3}{4}\): \[\frac{16 - a^2}{16} = \frac{3}{4} = \frac{12}{16} \implies 16 - a^2 = 12 \implies a^2 = 4\] Since the support of \(X\) is \(0 < x < 4\), we take the positive root \(a = 2\). **Answer: 2.**

**1. Write down the gradient descent update rule:** We are given the objective function \(f(x) = (x - 3)^2\), initial point \(x_0 = 0\), and learning rate \(\eta > 0\). The first iterate \(x_1\) of gradient descent is computed as: \[x_1 = x_0 - \eta f'(x_0)\] **2. Compute the derivative at \(x_0 = 0\):** Differentiating \(f(x) = (x - 3)^2\) with respect to \(x\) gives: \[f'(x) = 2(x - 3)\] At the starting point \(x_0 = 0\), the gradient is: \[f'(0) = 2(0 - 3) = -6\] Substituting into the update formula gives: \[x_1 = 0 - \eta(-6) = 6\eta\] **3. Equate \(x_1\) to the global minimizer \(x^*\):** Since \(f(x) = (x - 3)^2 \ge 0\) reaches its unique global minimum of \(0\) at \(x^* = 3\), we want \(x_1 = 3\): \[6\eta = 3 \implies \eta = \frac{3}{6} = 0.5\] **Answer: 0.5.**

**1. Understand the constrained optimization problem:** We want to find the minimum value of the objective function: \[f(x, y) = (x - 2)^2 + (y - 1)^2\] subject to the linear equality constraint: \[x + y = 4\] **2. Reduce to an unconstrained problem by substitution (or geometric projection):** From the constraint \(x + y = 4\), we can express \(y = 4 - x\). Substituting this into the objective function gives a function of a single variable \(g(x)\): \[g(x) = (x - 2)^2 + ((4 - x) - 1)^2 = (x - 2)^2 + (3 - x)^2\] Expanding both squares: \[g(x) = (x^2 - 4x + 4) + (x^2 - 6x + 9) = 2x^2 - 10x + 13\] **3. Find the optimal point and minimum value:** Setting the derivative \(g'(x) = 0\): \[g'(x) = 4x - 10 = 0 \implies x^* = 2.5\] Since \(g''(x) = 4 > 0\), \(x^* = 2.5\) is the global minimum. The corresponding value of \(y^*\) is: \[y^* = 4 - 2.5 = 1.5\] Substituting \((x^*, y^*) = (2.5, 1.5)\) back into the objective function: \[f(2.5, 1.5) = (2.5 - 2)^2 + (1.5 - 1)^2 = (0.5)^2 + (0.5)^2 = 0.25 + 0.25 = 0.5\] **Answer: 0.5.**

**1. Recall the survival function and mean of an Exponential distribution:** Let \(X \sim \text{Exponential}(\lambda)\) with rate parameter \(\lambda > 0\). For any \(x \ge 0\), the tail probability (survival function) is: \[P(X > x) = e^{-\lambda x}\] The mean (expected value) of \(X\) is \(E[X] = \frac{1}{\lambda}\). **2. Use the given condition to find \(\lambda\):** We are given that \(P(X > 4) = e^{-1}\). Substituting \(x = 4\) into the survival formula gives: \[e^{-4\lambda} = e^{-1} \implies 4\lambda = 1 \implies \lambda = \frac{1}{4}\] **3. Calculate the mean of \(X\):** Since \(E[X] = \frac{1}{\lambda}\), we obtain: \[E[X] = \frac{1}{1/4} = 4\] **Answer: B — 4.** A: **Incorrect:** A mean of \(1\) corresponds to rate \(\lambda = 1\), which would give \(P(X > 4) = e^{-4} \neq e^{-1}\). B: **Correct:** From \(e^{-4\lambda} = e^{-1}\), we get \(\lambda = 1/4\), so the mean is \(E[X] = 1/\lambda = 4\). C: **Incorrect:** The value \(1/4\) is the rate parameter \(\lambda\), whereas the question asks for the mean \(E[X] = 1/\lambda = 4\). D: **Incorrect:** The value \(16\) is the variance \(\text{Var}(X) = 1/\lambda^2 = 16\), not the mean.

**1. Recall the definition of a convex set and convex combination:** A set \(S\) is convex if for any two points \(a, b \in S\) and any scalar weight \(\alpha \in [0, 1]\), the point: \[p = \alpha a + (1 - \alpha) b\] is guaranteed to lie in \(S\). Notice that the two coefficients must add up to \(1\) and each coefficient must be non-negative (\(\ge 0\)). **2. Examine the coefficients of \(p = (2t - 1)a + (2 - 2t)b\):** Here, the coefficient of \(a\) is \(\alpha = 2t - 1\) and the coefficient of \(b\) is \(\beta = 2 - 2t\). First, check their sum: \[\alpha + \beta = (2t - 1) + (2 - 2t) = 1\] Since the coefficients always sum to \(1\), \(p\) is an affine combination of \(a\) and \(b\). **3. Enforce the non-negativity conditions for a convex combination:** For \(p\) to be guaranteed to lie in the convex set \(S\) (which may be as small as the line segment \([a, b]\) itself), both weights must be non-negative: 1. \(2t - 1 \ge 0 \implies 2t \ge 1 \implies t \ge \frac{1}{2}\) 2. \(2 - 2t \ge 0 \implies 2t \le 2 \implies t \le 1\) Combining both inequalities gives the exact interval \(\frac{1}{2} \le t \le 1\). **Answer: B — 1/2 <= t <= 1.** A: **Incorrect:** If \(t = 0\), the first weight is \(2(0) - 1 = -1 < 0\), giving \(p = -a + 2b\), which lies outside the line segment between \(a\) and \(b\). B: **Correct:** Both weights \(2t - 1\) and \(2 - 2t\) are non-negative and sum to \(1\) if and only if \(\frac{1}{2} \le t \le 1\). C: **Incorrect:** For \(t > 1\), the second weight \(2 - 2t\) becomes negative, so \(p\) is not a convex combination of \(a\) and \(b\). D: **Incorrect:** An arbitrary real \(t\) only produces an affine combination (the infinite line through \(a\) and \(b\)), not necessarily a point inside the convex set \(S\).

**1. Understand the convex hull of three points in \(\mathbb{R}^2\):** We are given three points: \[A = (0, 0), \quad B = (4, 0), \quad C = (0, 4)\] The convex hull of \(\{A, B, C\}\) is the set of all convex combinations \(p = \lambda_1 A + \lambda_2 B + \lambda_3 C\) where \(\lambda_1, \lambda_2, \lambda_3 \ge 0\) and \(\lambda_1 + \lambda_2 + \lambda_3 = 1\). Geometrically, this is the closed right triangle with vertices \((0,0)\), \((4,0)\), and \((0,4)\). **2. Express the convex hull in terms of coordinates \((x, y)\):** For any point \((x, y) = \lambda_1(0,0) + \lambda_2(4,0) + \lambda_3(0,4) = (4\lambda_2, 4\lambda_3)\): - Since \(\lambda_2 \ge 0\) and \(\lambda_3 \ge 0\), we must have \(x \ge 0\) and \(y \ge 0\). - Since \(\lambda_2 + \lambda_3 = 1 - \lambda_1 \le 1\), multiplying by \(4\) gives \(x + y \le 4\). Thus, a point \((x, y)\) lies in the convex hull if and only if \(x \ge 0\), \(y \ge 0\), and \(x + y \le 4\). **3. Test each given option:** - \((3, 2)\): \(x + y = 3 + 2 = 5 > 4\) (outside). - \((1, 2)\): \(x = 1 \ge 0\), \(y = 2 \ge 0\), and \(x + y = 1 + 2 = 3 \le 4\) (inside, with \(\lambda_1 = 1/4, \lambda_2 = 1/4, \lambda_3 = 1/2\)). - \((-1, 1)\): \(x = -1 < 0\) (outside). - \((2, 3)\): \(x + y = 2 + 3 = 5 > 4\) (outside). **Answer: B — (1, 2).** A: **Incorrect:** For \((3, 2)\), the sum of coordinates is \(3 + 2 = 5 > 4\), so it lies outside the triangle \(ABC\). B: **Correct:** We can write \((1, 2) = \frac{1}{4}(0, 0) + \frac{1}{4}(4, 0) + \frac{1}{2}(0, 4)\), where the weights \(\frac{1}{4}, \frac{1}{4}, \frac{1}{2}\) are non-negative and sum to \(1\). C: **Incorrect:** The first coordinate is negative (\(-1 < 0\)), whereas all points in the convex hull of \(A, B, C\) have non-negative coordinates. D: **Incorrect:** For \((2, 3)\), the sum of coordinates is \(2 + 3 = 5 > 4\), which violates \(x + y \le 4\).