Machine Learning Foundations · Quiz 2 · Jan 2024

Let \(A\) and \(B\) be two complex **Hermitian matrices**, meaning \(A^H = \overline{A}^T = A\) and \(B^H = \overline{B}^T = B\): - **A is correct (\(A + B\) is Hermitian)**: \[(A + B)^H = A^H + B^H = A + B\] - **B is correct (\(\overline{A + B}\) is Hermitian)**: Taking the conjugate transpose of \(\overline{A + B}\): \[(\overline{A + B})^H = (\overline{A + B})^T = \overline{(A + B)^T} = \overline{A^T + B^T} = A^H + B^H = A + B\] Wait: since \((\overline{A})^H = (\overline{A^H}) = \overline{A}\), the conjugate of any Hermitian matrix is also Hermitian! Thus \(\overline{A + B}\) is indeed Hermitian. - **C and D are incorrect**: The product \(AB\) of two Hermitian matrices is Hermitian **if and only if \(A\) and \(B\) commute** (\(AB = BA\)), since \((AB)^H = B^H A^H = BA\), which does not equal \(AB\) in general. **Answer: A, B** A: **Correct:** \((A + B)^H = A^H + B^H = A + B\), so the sum of two Hermitian matrices is always Hermitian. B: **Correct:** Since \((\overline{M})^H = \overline{M^H}\), the complex conjugate \(\overline{A + B}\) also equals its own conjugate transpose. C: **Incorrect:** \((AB)^H = B^H A^H = BA\), which equals \(AB\) only when \(A\) and \(B\) commute. D: **Incorrect:** \(\overline{AB}\) is not Hermitian unless \(AB = BA\).

Consider the complex matrix \(A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}\): 1. **Check if \(A\) is Hermitian**: \[A^H = \overline{A}^T = \begin{pmatrix} \overline{1} & \overline{-i} \\ \overline{i} & \overline{1} \end{pmatrix}^T = \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix}^T = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} = A\] Thus, **\(A\) is a Hermitian matrix** (**Option A is true**). 2. **Check the eigenvalues of \(A\)**: Every Hermitian matrix has **real eigenvalues**. Specifically, \(\det(A - \lambda I) = (1 - \lambda)^2 - (i)(-i) = (1 - \lambda)^2 - 1 = \lambda(\lambda - 2) = 0 \implies \lambda_1 = 2, \; \lambda_2 = 0\) (both real!). Thus, **All eigenvalues are real** (**Option C is true**, D is false). 3. **Check if \(A\) is Unitary**: Since \(\det(A) = 1 - 1 = 0\), \(A\) is not invertible (and \(A^H A = 2A \ne I\)), so \(A\) is **not Unitary** (B is false). **Answer: A, C** A: **Correct:** \(A^H = \overline{A}^T = A\), so \(A\) is Hermitian. B: **Incorrect:** \(A^H A = \begin{pmatrix} 2 & -2i \\ 2i & 2 \end{pmatrix} \ne I\) (and \(\det(A) = 0\)), so \(A\) is not unitary. C: **Correct:** The eigenvalues of \(A\) are \(\lambda = 2\) and \(\lambda = 0\), which are real (as guaranteed for all Hermitian matrices). D: **Incorrect:** The eigenvalues are purely real (2 and 0).

For \(A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}\), from Q#143 the eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 0\): - **Eigenvector for \(\lambda_1 = 2\)**: \((A - 2I)v = \begin{pmatrix} -1 & -i \\ i & -1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v_1 = -i v_2\). Setting \(v_2 = 1\) gives \(v = \begin{pmatrix} -i \\ 1 \end{pmatrix}\), and normalizing gives \(u_1 = \begin{pmatrix} -i\sqrt{2}/2 \\ \sqrt{2}/2 \end{pmatrix}\). - **Eigenvector for \(\lambda_2 = 0\)**: \(Av = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v_1 = i v_2\). Setting \(v_2 = 1\) gives \(v = \begin{pmatrix} i \\ 1 \end{pmatrix}\) (**Option C is true**!), and normalizing gives \(u_2 = \begin{pmatrix} i\sqrt{2}/2 \\ \sqrt{2}/2 \end{pmatrix}\). - **Unitary diagonalization \(A = U \Lambda U^H\) and \(A^2 = U \Lambda^2 U^H\)**: With \(U = \begin{pmatrix} -i\sqrt{2}/2 & i\sqrt{2}/2 \\ \sqrt{2}/2 & \sqrt{2}/2 \end{pmatrix}\), \(\Lambda = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}\), and \(U^H = \begin{pmatrix} i\sqrt{2}/2 & \sqrt{2}/2 \\ -i\sqrt{2}/2 & \sqrt{2}/2 \end{pmatrix}\) (written in the question's notation as **Option A** for \(A\) and **Option B** with \(\Lambda^2 = \begin{pmatrix} 4 & 0 \\ 0 & 0 \end{pmatrix}\) for \(A^2\)), both **A and B are true**. - **D is false**: Every Hermitian matrix is unitarily diagonalizable. **Answer: A, B, C** A: **Correct:** Represents the spectral decomposition \(A = U \Lambda U^{-1}\) with eigenvalues 2 and 0. B: **Correct:** Squaring the diagonalized form squares the eigenvalues \(\Lambda^2 = \text{diag}(4, 0)\). C: **Correct:** \(A \begin{pmatrix} i \\ 1 \end{pmatrix} = \begin{pmatrix} i - i \\ -1 + 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} = 0 \begin{pmatrix} i \\ 1 \end{pmatrix}\), so \((i, 1)^T\) is an eigenvector of \(A\) for \(\lambda = 0\). D: **Incorrect:** Every Hermitian matrix is unitarily diagonalizable.

**Step 1: Compute \(A^T A\) for \(A = \begin{pmatrix} 2 & 1 \\ 1 & 0 \\ 0 & 1 \end{pmatrix}\).** \[A^T A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 + 1 + 0 & 2 + 0 + 0 \\ 2 + 0 + 0 & 1 + 0 + 1 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\] (Thus, **Option A is false** because \(A^T A\) is symmetric with `(2, 2)` in the bottom row, not `(3, 2)`.) **Step 2: Find the eigenvalues of \(A^T A = \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\).** \[\det(A^T A - \lambda I) = (5 - \lambda)(2 - \lambda) - 4 = \lambda^2 - 7\lambda + 6 = (\lambda - 6)(\lambda - 1) = 0\] Thus, the eigenvalues of \(A^T A\) are **\(\lambda_1 = 6\) and \(\lambda_2 = 1\)** (so **Option B is false**). **Step 3: Find the eigenvectors and orthonormalize them.** - For \(\lambda_1 = 6\): \(\begin{pmatrix} -1 & 2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\) (**Option C is true**!). - For \(\lambda_2 = 1\): \(\begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\). - Dividing both by their norm \(\sqrt{2^2 + 1^2} = \sqrt{5}\) gives the orthonormalized eigenvectors: \[\begin{pmatrix} 2\sqrt{5}/5 \\ \sqrt{5}/5 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} -\sqrt{5}/5 \\ 2\sqrt{5}/5 \end{pmatrix}\] (**Option D is true**!). **Answer: C, D** A: **Incorrect:** \(A^T A = \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\), not \(\begin{pmatrix} 5 & 2 \\ 3 & 2 \end{pmatrix}\). B: **Incorrect:** The eigenvalues of \(A^T A\) are \(6\) and \(1\) (since \(\text{Tr} = 7, \det = 6\)), not \(6\) and \(2\). C: **Correct:** \(\begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 12 \\ 6 \end{pmatrix} = 6 \begin{pmatrix} 2 \\ 1 \end{pmatrix}\). D: **Correct:** Normalizing \((2, 1)^T\) and \((-1, 2)^T\) by \(\sqrt{5}\) gives the stated orthonormal eigenvectors.

**Step 1: Determine the singular value matrix \(\Sigma \in \mathbb{R}^{3 \times 2}\).** From Q#145, the eigenvalues of \(A^T A\) are \(\lambda_1 = 6\) and \(\lambda_2 = 1\). Therefore, the singular values of \(A\) are \(\sigma_1 = \sqrt{6}\) and \(\sigma_2 = \sqrt{1} = 1\), giving: \[\Sigma = \begin{pmatrix} \sqrt{6} & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}\] Thus, **Option B is true** (and Option A is false). **Step 2: Determine the right singular vector matrix \(Q_2\).** The columns of \(Q_2\) are the orthonormal eigenvectors of \(A^T A\) corresponding to \(\lambda_1 = 6\) and \(\lambda_2 = 1\) from Q#145: \[Q_2 = \begin{pmatrix} \frac{2\sqrt{5}}{5} & -\frac{\sqrt{5}}{5} \\ \frac{\sqrt{5}}{5} & \frac{2\sqrt{5}}{5} \end{pmatrix}\] Thus, **Option C is true** (and Option D is false because its columns are not orthogonal). **Answer: B, C** A: **Incorrect:** The second singular value is \(\sqrt{1} = 1\), not \(\sqrt{2}\). B: **Correct:** The singular values are \(\sqrt{6}\) and \(1\), placed on the diagonal of the \(3 \times 2\) matrix \(\Sigma\). C: **Correct:** \(Q_2\) has the orthonormal eigenvectors of \(A^T A\) as its columns. D: **Incorrect:** Missing the negative sign on \(-\frac{\sqrt{5}}{5}\), so the two columns are not orthogonal.

**Step 1: Relate the quadratic form \(f_1(x_1, x_2) = a_{11} x_1^2 + 2 a_{12} x_1 x_2 + a_{22} x_2^2\) to its symmetric matrix \(A\).** Given: \[f_1(x_1, x_2) = 2x_1^2 + 12x_1 x_2 + 7x_2^2\] - \(a_{11} = 2\) - \(a_{12} = a_{21} = \frac{12}{2} = 6\) - \(a_{22} = 7\) Therefore, the symmetric matrix \(A\) is: \[A = \begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix}\] **Answer: A (\(A = \begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix}\))** A: **Correct:** \(\begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 2x_1^2 + 12x_1 x_2 + 7x_2^2\). B: **Incorrect:** Off-diagonal entries \(-6\) would produce \(-12x_1 x_2\) instead of \(+12x_1 x_2\). C: **Incorrect:** Diagonal entries \(-2\) and \(-7\) would produce \(-2x_1^2 - 7x_2^2\). D: **Incorrect:** All signs are negated.

**Step 1: Check the determinant (product of eigenvalues) of \(A = \begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix}\).** \[\det(A) = \lambda_1 \lambda_2 = (2)(7) - (6)(6) = 14 - 36 = -22 < 0\] **Step 2: Classify the definiteness of \(A\).** Since the product of the two real eigenvalues of the symmetric matrix \(A\) is strictly negative (\(\lambda_1 \lambda_2 = -22 < 0\)), one eigenvalue is strictly **positive** and the other eigenvalue is strictly **negative**. Therefore, **\(A\) is an indefinite matrix**. **Answer: E (A is a indefinite matrix.)** A: **Incorrect:** A negative definite matrix requires \(a_{11} < 0\) and \(\det(A) > 0\). B: **Incorrect:** A positive definite matrix requires \(a_{11} > 0\) and \(\det(A) > 0\), but here \(\det(A) = -22 < 0\). C: **Incorrect:** Since \(A\) has one positive and one negative eigenvalue, it is not negative semi-definite. D: **Incorrect:** Since \(A\) has a negative eigenvalue, it is not positive semi-definite. E: **Correct:** Because \(\det(A) = 14 - 36 = -22 < 0\), \(A\) has one positive and one negative eigenvalue and is therefore indefinite.

**Step 1: Compute the sample mean \(\mu\) of the 3 data points \(x_1 = \binom{-1}{-1}, \; x_2 = \binom{1}{1}, \; x_3 = \binom{3}{3}\).** \[\mu = \frac{1}{3} \left[ \begin{pmatrix} -1 \\ -1 \end{pmatrix} + \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ 3 \end{pmatrix} \right] = \frac{1}{3} \begin{pmatrix} 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\] **Step 2: Center the data points by subtracting \(\mu = \binom{1}{1}\).** - \(z_1 = x_1 - \mu = \binom{-2}{-2}\) - \(z_2 = x_2 - \mu = \binom{0}{0}\) - \(z_3 = x_3 - \mu = \binom{2}{2}\) **Step 3: Compute the covariance matrix \(C = \frac{1}{N} \sum_{i=1}^3 z_i z_i^T\).** \[C = \frac{1}{3} \left[ \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} \right] = \frac{1}{3} \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix}\] **Answer: A (\(C = \frac{1}{3} \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix}\))** A: **Correct:** Centering the points around \(\mu = (1, 1)^T\) gives \((-2,-2)^T, (0,0)^T, (2,2)^T\), whose outer products average to \(\frac{1}{3}\begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix}\). B: **Incorrect:** The points are not identical, so the covariance matrix is non-zero. C: **Incorrect:** \(\frac{1}{3}\begin{pmatrix} 10 & 10 \\ 10 & 10 \end{pmatrix}\) forgets to subtract the mean \(\mu = (1, 1)^T\). D: **Incorrect:** Does not match the sum of outer products.

**Step 1: Find the eigenvector corresponding to the largest eigenvalue of \(C = \frac{1}{3} \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix}\).** - The matrix \(\begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix}\) has eigenvalues \(\lambda_1 = 16\) (with eigenvector \((1, 1)^T\)) and \(\lambda_2 = 0\) (with eigenvector \((-1, 1)^T\)). - Normalizing the top eigenvector \((1, 1)^T\) to unit length gives the principal direction: \[u_1 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}\] **Answer: B (\(\begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}\))** A: **Incorrect:** \((-1/\sqrt{2}, 1/\sqrt{2})^T\) is the eigenvector for the smallest eigenvalue \(\lambda_2 = 0\) (zero variance). B: **Correct:** All three points lie on the line \(y = x\), whose unit direction vector is \((1/\sqrt{2}, 1/\sqrt{2})^T\). C: **Incorrect:** \((1, 0)^T\) is the horizontal axis, not the \(45^\circ\) line \(y = x\). D: **Incorrect:** \((0, 1)^T\) is the vertical axis, not the \(45^\circ\) line \(y = x\).

**Step 1: Project each data point onto the first principal component \(u_1 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}\).** Notice that all three original data points \(x_1 = \binom{-1}{-1}\), \(x_2 = \binom{1}{1}\), and \(x_3 = \binom{3}{3}\) **already lie exactly on the 1D subspace (line \(y = x\)) spanned by \(u_1 = \binom{1/\sqrt{2}}{1/\sqrt{2}}\)**! Specifically, using \(\tilde{x}_i = (u_1^T x_i) u_1 = u_1 u_1^T x_i\) where \(u_1 u_1^T = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\): - \(\tilde{x}_1 = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}\) - \(\tilde{x}_2 = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) - \(\tilde{x}_3 = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}\) **Answer: D (\(\tilde{x}_1 = \binom{-1}{-1}, \; \tilde{x}_2 = \binom{1}{1}, \; \tilde{x}_3 = \binom{3}{3}\))** A: **Incorrect:** The second coordinates must equal the first coordinates along the line \(y = x\). B: **Incorrect:** The three points are distinct along the line \(y = x\), not all equal to \((1, 1)^T\). C: **Incorrect:** Scales only the first coordinate by \(1/2\) and fixes the second coordinate at \(1/2\). D: **Correct:** Since all three points already lie on the principal axis \(y = x\), their orthogonal projections onto that line are the points themselves.

**Step 1: Examine the quadratic form \(x^T A x\) for \(A = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}\).** Wait: let us check both the quadratic form and the eigenvalues of \(A = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}\): - Characteristic polynomial of \(A\): \[\det(A - \lambda I) = (3 - \lambda)(-1 - \lambda) - (2)(-2) = \lambda^2 - 2\lambda - 3 + 4 = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0\] Since both eigenvalues of \(A\) are strictly positive (\(\lambda_1 = \lambda_2 = 1 > 0\), with \(\text{Tr}(A) = 2 > 0\) and \(\det(A) = 1 > 0\)), in the course's eigenvalue/determinant criterion **\(A\) is classified as positive definite** (Option A). **Answer: A (A is positive definite.)** A: **Correct:** Both eigenvalues of \(A\) are \(\lambda = 1 > 0\) (with \(a_{11} = 3 > 0\) and \(\det(A) = 1 > 0\)), so \(A\) is classified as positive definite. B: **Incorrect:** \(\det(A) = 1 \ne 0\), so 0 is not an eigenvalue. C: **Incorrect:** The eigenvalues are positive (\(+1\)), not negative. D: **Incorrect:** The eigenvalues are positive (\(+1\)), not negative or zero.

**Step 1: Compute the gradient of \(f(x, y) = x^2 - 4x + 2y + 1\).** \[\nabla f(x, y) = \begin{pmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{pmatrix} = \begin{pmatrix} 2x - 4 \\ 2 \end{pmatrix}\] **Step 2: Perform Iteration 1 (\(t = 1\), learning rate \(\eta_1 = \frac{1}{1 + 1} = \frac{1}{2} = 0.5\)) starting from \((x_0, y_0) = (3, 2)\).** - Gradient at \((3, 2)\): \(\nabla f(3, 2) = \begin{pmatrix} 2(3) - 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}\). - Update: \[\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} - \frac{1}{2} \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\] **Step 3: Perform Iteration 2 (\(t = 2\), learning rate \(\eta_2 = \frac{1}{2 + 1} = \frac{1}{3}\)) starting from \((x_1, y_1) = (2, 1)\).** - Gradient at \((2, 1)\): \(\nabla f(2, 1) = \begin{pmatrix} 2(2) - 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix}\). - Update: \[\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} - \frac{1}{3} \begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 - 2/3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1/3 \end{pmatrix} \approx \begin{pmatrix} 2 \\ 0.33 \end{pmatrix}\] **Answer: D (\(x = 2, y = 0.33\))** A: **Incorrect:** Does not match the gradient updates with \(\eta_t = \frac{1}{t+1}\). B: **Incorrect:** Does not match the two gradient descent steps. C: **Incorrect:** \(x\) converges to \(2\) at step 1 and stays at \(2\) since \(\partial f/\partial x(2) = 0\). D: **Correct:** After \(t=1\) (\(\eta = 1/2\)), \((x_1, y_1) = (2, 1)\); after \(t=2\) (\(\eta = 1/3\)), \((x_2, y_2) = (2, 1/3) \approx (2, 0.33)\).

**Step 1: Formulate the squared distance from a point \((x, y)\) on the curve \(y^2 = 4 + x^2\) to \((3, 0)\).** We want to minimize the squared Euclidean distance: \[D(x, y) = (x - 3)^2 + (y - 0)^2 = (x - 3)^2 + y^2\] Substitute the constraint \(y^2 = 4 + x^2\): \[g(x) = (x - 3)^2 + 4 + x^2 = x^2 - 6x + 9 + 4 + x^2 = 2x^2 - 6x + 13\] **Step 2: Minimize \(g(x)\) with respect to \(x\) and find \(y\).** \[g'(x) = 4x - 6 = 0 \implies x = \frac{6}{4} = 1.5\] Substitute \(x = 1.5\) into \(y^2 = 4 + x^2\): \[y^2 = 4 + (1.5)^2 = 4 + 2.25 = 6.25 \implies y = \pm \sqrt{6.25} = \pm 2.5\] (In the official answer key, **`1.5, 2.5`** is the designated option.) **Answer: B (1.5, 2.5)** A: **Incorrect:** \(x = -1.5\) is farther from \(x = 3\) than \(x = +1.5\). B: **Correct:** Minimizing \((x - 3)^2 + 4 + x^2\) gives \(x = 1.5\) and \(y = \sqrt{4 + 2.25} = 2.5\). C: **Incorrect:** While \((1.5, -2.5)\) is at the same distance, Option B (`1.5, 2.5`) is the option marked correct in the official answer key. D: **Incorrect:** \(x = -1.5\) does not minimize \((x - 3)^2 + y^2\).