Machine Learning Foundations · Quiz 2 · Jan 2025

Let us find the **Taylor series expansion** of \( f(x) = \frac{1}{x} \) around \( x = 1 \) up to the fourth-degree term: **1. Compute the Derivatives of \( f(x) = x^{-1} \) at \( x = 1 \):** - \( f(x) = x^{-1} \implies f(1) = 1 \) - \( f'(x) = -x^{-2} \implies f'(1) = -1 \) - \( f''(x) = 2! x^{-3} \implies f''(1) = 2! \) - \( f^{(3)}(x) = -3! x^{-4} \implies f^{(3)}(1) = -3! \) - \( f^{(4)}(x) = 4! x^{-5} \implies f^{(4)}(1) = 4! \) **2. Substitute into the Taylor Series Formula \( \sum_{k=0}^4 \frac{f^{(k)}(1)}{k!}(x - 1)^k \):** \[ f(x) \approx 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4 \] **Answer: 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4.** A: **Incorrect:** Has the wrong signs on the odd and even power terms (`+ (x - 1) - (x - 1)^2 ...`). B: **Incorrect:** Uses `(x + 1)` instead of `(x - 1)` for an expansion centered at `x = 1`. C: **Correct:** The alternating geometric/Taylor series of `1/x = 1 / (1 + (x - 1))` around `x = 1` is `1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4`. D: **Incorrect:** Missing the alternating negative signs on the odd powers.

Let us find the eigenvalues of the Hermitian matrix \( A = \begin{bmatrix} 3 & i \\ -i & 5 \end{bmatrix} \): **1. Set Up the Characteristic Equation \( \det(A - \lambda I) = 0 \):** \[ \det\begin{bmatrix} 3 - \lambda & i \\ -i & 5 - \lambda \end{bmatrix} = (3 - \lambda)(5 - \lambda) - (i)(-i) = 0 \] Since \( i(-i) = -i^2 = 1 \): \[ (\lambda - 3)(\lambda - 5) - 1 = 0 \] \[ \lambda^2 - 8\lambda + 15 - 1 = 0 \implies \lambda^2 - 8\lambda + 14 = 0 \] **2. Solve the Quadratic Equation:** \[ \lambda = \frac{8 \pm \sqrt{64 - 4(14)}}{2} = \frac{8 \pm \sqrt{8}}{2} = \frac{8 \pm 2\sqrt{2}}{2} = 4 \pm \sqrt{2} \] **Answer: 4 + \sqrt{2} and 4 - \sqrt{2}.** A: **Incorrect:** `3` and `5` are the diagonal entries, but `A` is not a diagonal or triangular matrix. B: **Correct:** Solving `lambda^2 - 8*lambda + 14 = 0` yields `4 + sqrt(2)` and `4 - sqrt(2)`. C: **Incorrect:** The eigenvalues of a Hermitian matrix are always real numbers, never complex (`4 +- i`). D: **Incorrect:** For `2` and `6`, the determinant would be `2 * 6 = 12`, whereas `det(A) = 15 - 1 = 14`.

Let us apply **Schur's theorem** to the already upper-triangular complex matrix \( A = \begin{bmatrix} 1 & 2+i \\ 0 & 3 \end{bmatrix} \): **1. Eigenvalues on the Diagonal:** Since \( A \) is upper triangular, its eigenvalues are its diagonal entries: \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \). **2. General Upper Triangular Form Under Similarity Transformation:** By Schur's theorem (and general similarity transformations yielding an upper triangular matrix with eigenvalues ordered as \( 1, 3 \)), the diagonal entries must be the eigenvalues \( 1 \) and \( 3 \), while the superdiagonal entry can be some complex number \( x \) (for a unitary similarity \( U^* A U \), \( |x| = |2+i| = \sqrt{5} \), and for a general similarity transformation \( P^{-1} A P \), \( x \) can be any complex number including \( 0 \) or \( 2+i \)). Thus, the general upper triangular form is: \[ \begin{bmatrix} 1 & x \\ 0 & 3 \end{bmatrix}, \text{ for some complex } x \] **Answer: [[1, x], [0, 3]], for some complex x (Option D).** A: **Incorrect:** While `A` itself is one specific instance, a similarity transformation can change the upper-right entry to other complex values `x`. B: **Incorrect:** A diagonal matrix with `3` and `1` is only one special case with reordered eigenvalues. C: **Incorrect:** This is a lower triangular matrix, whereas Schur's theorem specifies an upper triangular form. D: **Correct:** Any upper triangular matrix similar to `A` (with eigenvalues `1, 3` on the diagonal) has the form `[[1, x], [0, 3]]` for some complex `x`.

Let \( A \) be a complex square matrix and \( A^* = \overline{A}^T \) be its conjugate transpose. **Relationship Between the Eigenvalues of \( A \) and \( A^* \):** - For any complex number \( \lambda \), taking the conjugate transpose of \( A - \lambda I \) gives \( (A - \lambda I)^* = A^* - \overline{\lambda} I \). - Since \( \det(M^*) = \overline{\det(M)} \), we have: \[ \det(A^* - \overline{\lambda} I) = \overline{\det(A - \lambda I)} \] - Therefore, \( \det(A - \lambda I) = 0 \iff \det(A^* - \overline{\lambda} I) = 0 \). - This proves that **the eigenvalues of \( A^* \) are the complex conjugates (\( \overline{\lambda} \)) of the eigenvalues of \( A \)**. **Answer: The eigenvalues of A* are the complex conjugates of the eigenvalues of A.** A: **Correct:** Since `det(A* - conj(lambda) I) = conj(det(A - lambda I))`, the eigenvalues of `A*` are the complex conjugates of the eigenvalues of `A`. B: **Incorrect:** The eigenvalues of `A*` are only equal to the eigenvalues of `A` if all eigenvalues of `A` happen to be real. C: **Incorrect:** Reciprocals of eigenvalues correspond to the inverse matrix `A^{-1}`, not `A*`. D: **Incorrect:** Taking the conjugate transpose does not take square roots of eigenvalues.

Let us analyze the matrix \( A = \begin{bmatrix} 1 & 1 & i \\ 1 & -1 & -i \\ -i & -i & 1 \end{bmatrix} \): **1. Check if \( A = A^* \) (Hermitian):** The conjugate transpose \( A^* = \overline{A}^T \) is: \[ A^* = \begin{bmatrix} 1 & 1 & i \\ 1 & -1 & i \\ -i & i & 1 \end{bmatrix} \] Notice that \( A_{23} = -i \) while \( (A^*)_{23} = \overline{A_{32}} = \overline{-i} = i \neq -i \). Thus \( A \neq A^* \) (**Option B is false**). **2. Check the Determinant of \( A \) (Option D):** Compare Row 1 \( R_1 = [1, 1, i] \) and Row 3 \( R_3 = [-i, -i, 1] \): Multiply Row 1 by \( -i \): \[ -i \cdot R_1 = [-i(1), \; -i(1), \; -i(i)] = [-i, \; -i, \; 1] = R_3 \] Since Row 3 is a scalar multiple of Row 1 (\( R_3 = -i R_1 \)), the rows of \( A \) are linearly dependent! Therefore, **\( \det(A) = 0 \)** (**Option D is true**, and since \( \det(A) = 0 \), \( A \) is singular and **not unitary**, so **Option A is false**). **3. Find All Eigenvalues of \( A \) (Option C):** - Trace of \( A \): \( \text{tr}(A) = 1 + (-1) + 1 = 1 \). - Characteristic polynomial \( \det(A - \lambda I) = 0 \): Since \( \det(A) = 0 \), \( \lambda = 0 \) is an eigenvalue. Let us compute the sum of principal \( 2 \times 2 \) minors \( M_{11} + M_{22} + M_{33} \): - \( M_{11} = \det\begin{bmatrix} -1 & -i \\ -i & 1 \end{bmatrix} = -1 - (-i)^2 = -1 - (-1) = 0 \) - \( M_{22} = \det\begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix} = 1 - (-i^2) = 1 - 1 = 0 \) - \( M_{33} = \det\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = -1 - 1 = -2 \) Thus, the characteristic equation is: \[ -\lambda^3 + \lambda^2 - (-2)\lambda + 0 = 0 \implies \lambda(\lambda^2 - \lambda - 2) = 0 \implies \lambda(\lambda - 2)(\lambda + 1) = 0 \] - The eigenvalues of \( A \) are \( \lambda = 0, 2, -1 \), all of which are **real**! (**Option C is true**). **Answer: Every eigenvalue of A is real, and The determinant of A is zero (Options C and D).** A: **Incorrect:** Since `det(A) = 0`, `A` is not invertible and therefore cannot be unitary. B: **Incorrect:** `A_{23} = -i` whereas `(A*)_{23} = conj(A_{32}) = conj(-i) = +i`, so `A != A*`. C: **Correct:** The characteristic polynomial of `A` is `-lambda(lambda - 2)(lambda + 1) = 0`, so its eigenvalues are `0, 2, -1`, which are all real. D: **Correct:** Row 3 is `-i` times Row 1, so `A` is rank-deficient and `det(A) = 0`.

Let us review **Schur's Unitary Triangularization Theorem**: - **Option A**: Schur's Theorem states that **every** complex square matrix \( A \in \mathbb{C}^{n \times n} \) is unitarily similar to an upper triangular matrix \( T = U^* A U \) (where \( U \) is unitary). (**True**) - **Option B**: Because \( T = U^* A U = U^{-1} A U \) is similar to \( A \) and upper triangular, the diagonal entries of the Schur form \( T \) are the eigenvalues of \( A \). (**True**) - **Option C**: Schur's Theorem holds for **all** square matrices, even those with repeated eigenvalues and defective eigenspaces (unlike diagonalization!). (**False**) - **Option D**: Unitary similarity \( U^* A U \) and eigenvalues are only defined for **square matrices**. (**True**) **Answer: Options A, B, and D.** A: **Correct:** Schur's Theorem guarantees that every square matrix is unitarily similar to an upper triangular matrix. B: **Correct:** Similar matrices share the same eigenvalues, and the eigenvalues of an upper triangular matrix lie on its main diagonal. C: **Incorrect:** Even matrices with repeated eigenvalues (including non-diagonalizable matrices) are unitarily similar to an upper triangular matrix by Schur's Theorem. D: **Correct:** Schur's Theorem applies only to square matrices because similarity transformations and eigenvalues are defined only for square matrices.

Let us check the properties of singular values \( \sigma_1, \sigma_2, \dots, \sigma_n \) (especially for a positive definite matrix \( A \)): - **Option A (\( \sigma_i \ge 0 \) for all \( i \))**: By definition, singular values are non-negative (\( \sigma_i \ge 0 \), and for a positive definite matrix \( \sigma_i > 0 \), which satisfies \( \sigma_i \ge 0 \)). (**True**) - **Option B (\( A \) has at least one zero singular value)**: A positive definite matrix is strictly invertible with all eigenvalues \( \lambda_i > 0 \), so all its singular values \( \sigma_i = \lambda_i > 0 \) are strictly positive (none are zero). (**False**) - **Option C (The eigenvalues of \( A \) must be negative)**: By definition, all eigenvalues of a positive definite matrix are strictly **positive**. (**False**) - **Option D (If \( \sigma_i = 0 \) for all \( i \), then \( A \) is the zero matrix)**: As a conditional statement ("If all singular values of a matrix are 0, then it is the zero matrix"), this is a fundamental property of SVD (\( A = U \Sigma V^T = 0 \)), and is vacuously true for positive definite matrices as well. (**True**) **Answer: Options A and D.** A: **Correct:** Singular values are always non-negative (`sigma_i >= 0` for all `i`). B: **Incorrect:** A positive definite matrix is full rank, so all its singular values are strictly positive (`sigma_i > 0`). C: **Incorrect:** The eigenvalues of a positive definite matrix are strictly positive, not negative. D: **Correct:** A matrix has all zero singular values if and only if it is the zero matrix.

We are given the function: \[ f(x, y) = x^2 + 3y^2 - 4xy + 5 \] with initial point \( (x_0, y_0) = (2, 3) \) and learning rate \( \eta = 0.1 \). **1. Compute the Gradient \( \nabla f(x, y) \):** - \( \frac{\partial f}{\partial x} = 2x - 4y \) - \( \frac{\partial f}{\partial y} = 6y - 4x \) **2. Evaluate the Gradient at \( (x_0, y_0) = (2, 3) \):** - \( \frac{\partial f}{\partial x}(2, 3) = 2(2) - 4(3) = 4 - 12 = -8 \) - \( \frac{\partial f}{\partial y}(2, 3) = 6(3) - 4(2) = 18 - 8 = 10 \) **3. Perform One Iteration of Gradient Descent:** - \( x_1 = x_0 - \eta \frac{\partial f}{\partial x}(2, 3) = 2 - 0.1(-8) = 2 + 0.8 = 2.8 \) - \( y_1 = y_0 - \eta \frac{\partial f}{\partial y}(2, 3) = 3 - 0.1(10) = 3 - 1.0 = 2.0 \) **4. Compute \( f(x_1, y_1) = f(2.8, 2.0) \):** \[ f(2.8, 2.0) = (2.8)^2 + 3(2.0)^2 - 4(2.8)(2.0) + 5 \] \[ f(2.8, 2.0) = 7.84 + 12.0 - 22.4 + 5 = 2.44 \] **Answer: 2.44.**

We want to find the minimum value of the function: \[ f(x) = e^{x^2 - 8x + 15} \] **1. Minimize the Exponent \( g(x) = x^2 - 8x + 15 \):** Since the exponential function \( e^u \) is strictly increasing for all real \( u \), \( f(x) = e^{g(x)} \) attains its minimum at the exact same point where the quadratic exponent \( g(x) = x^2 - 8x + 15 \) attains its minimum. **2. Complete the Square (or Set Derivative to Zero):** - \( g'(x) = 2x - 8 = 0 \implies x = 4 \). - Evaluating \( g(4) \): \[ g_{\min} = g(4) = 4^2 - 8(4) + 15 = 16 - 32 + 15 = -1 \] **3. Find \( \alpha \):** The minimum value of \( f(x) \) is \( e^{-1} = e^\alpha \), so **\( \alpha = -1 \)**. **Answer: -1.**

We are given the dataset of \( n = 3 \) points in \( \mathbb{R}^2 \): \( x_1 = \begin{pmatrix} 2 \\ 4 \end{pmatrix}, \; x_2 = \begin{pmatrix} 3 \\ 6 \end{pmatrix}, \; x_3 = \begin{pmatrix} 5 \\ 10 \end{pmatrix} \). **1. Compute the Sample Mean \( \mu \):** \[ \mu = \frac{1}{3} \left[ \begin{pmatrix} 2 \\ 4 \end{pmatrix} + \begin{pmatrix} 3 \\ 6 \end{pmatrix} + \begin{pmatrix} 5 \\ 10 \end{pmatrix} \right] = \frac{1}{3} \begin{pmatrix} 10 \\ 20 \end{pmatrix} = \begin{pmatrix} 10/3 \\ 20/3 \end{pmatrix} \] **2. Center the Data Points \( \bar{x}_i = x_i - \mu \):** - \( \bar{x}_1 = \begin{pmatrix} 2 - 10/3 \\ 4 - 20/3 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -4 \\ -8 \end{pmatrix} \) - \( \bar{x}_2 = \begin{pmatrix} 3 - 10/3 \\ 6 - 20/3 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -1 \\ -2 \end{pmatrix} \) - \( \bar{x}_3 = \begin{pmatrix} 5 - 10/3 \\ 10 - 20/3 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 5 \\ 10 \end{pmatrix} \) **3. Compute the Covariance Matrix \( C = \frac{1}{n} \sum_{i=1}^3 \bar{x}_i \bar{x}_i^T \) (or \( \frac{1}{n-1} \); let us check the denominator):** - First, sum the outer products of the numerators \( (-4, -8)^T, (-1, -2)^T, (5, 10)^T \): - Top-left entry: \( (-4)^2 + (-1)^2 + 5^2 = 16 + 1 + 25 = 42 \) - Off-diagonal entry: \( (-4)(-8) + (-1)(-2) + (5)(10) = 32 + 2 + 50 = 84 \) - Bottom-right entry: \( (-8)^2 + (-2)^2 + 10^2 = 64 + 4 + 100 = 168 \) - Dividing by \( n \cdot 3^2 = 3 \times 9 = 27 \): \[ C = \frac{1}{27} \begin{pmatrix} 42 & 84 \\ 84 & 168 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} 14 & 28 \\ 28 & 56 \end{pmatrix} \] **Answer: C = (1/9) [[14, 28], [28, 56]] (Option C).** A: **Incorrect:** Does not properly center the data around the mean `(10/3, 20/3)`. B: **Incorrect:** The points are not identical, so the covariance matrix is non-zero. C: **Correct:** Centering the three points around `(10/3, 20/3)` and computing `(1/3) sum (x_i - mu)(x_i - mu)^T` gives `(1/9) [[14, 28], [28, 56]]`. D: **Incorrect:** Has incorrect numerator entries `10, 20, 40` instead of `14, 28, 56`.

The **first principal direction** in PCA is the unit eigenvector corresponding to the largest eigenvalue of the covariance matrix \( C = \frac{1}{9}\begin{pmatrix} 14 & 28 \\ 28 & 56 \end{pmatrix} = \frac{14}{9}\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \). **1. Observe the Linear Relationship of the Data Points:** Notice that all three data points \( \begin{pmatrix} 2 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 6 \end{pmatrix}, \begin{pmatrix} 5 \\ 10 \end{pmatrix} \) lie **exactly on the line \( x_2 = 2x_1 \)** (i.e., in the direction of the vector \( v = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \)). **2. Normalize \( v = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \) to Unit Length:** \[ u_1 = \frac{v}{\|v\|} = \frac{1}{\sqrt{1^2 + 2^2}} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \frac{1}{\sqrt{5}} \begin{pmatrix} 1 \\ 2 \end{pmatrix} \] **Answer: (1 / \sqrt{5}) [1, 2]^T (Option B).** A: **Incorrect:** `[2, 1]^T / sqrt(5)` is along `x_1 = 2x_2`, whereas the points satisfy `x_2 = 2x_1`. B: **Correct:** All points lie along `[1, 2]^T`, which when normalized gives `(1/sqrt(5)) [1, 2]^T`. C: **Incorrect:** `[1, 0]^T` is the horizontal axis, not the direction of maximum variance. D: **Incorrect:** `[0, 1]^T` is the vertical axis, not the direction of maximum variance.

Let us find the orthogonal projections \( \tilde{x}_1, \tilde{x}_2, \tilde{x}_3 \) of the data points onto the 1D subspace spanned by the first principal component \( u_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ 2 \end{pmatrix} \): **1. Projection Formula:** For any point \( x_i \), its projection onto the subspace spanned by unit vector \( u_1 \) is: \[ \tilde{x}_i = (u_1^T x_i) u_1 \] **2. Notice That Each \( x_i \) Already Lies in \( \text{span}(u_1) \):** - \( x_1 = \begin{pmatrix} 2 \\ 4 \end{pmatrix} = 2\sqrt{5} \, u_1 \implies \tilde{x}_1 = x_1 = \begin{pmatrix} 2 \\ 4 \end{pmatrix} \) - \( x_2 = \begin{pmatrix} 3 \\ 6 \end{pmatrix} = 3\sqrt{5} \, u_1 \implies \tilde{x}_2 = x_2 = \begin{pmatrix} 3 \\ 6 \end{pmatrix} \) - \( x_3 = \begin{pmatrix} 5 \\ 10 \end{pmatrix} = 5\sqrt{5} \, u_1 \implies \tilde{x}_3 = x_3 = \begin{pmatrix} 5 \\ 10 \end{pmatrix} \) Because all three original data points already lie directly on the line spanned by \( u_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ 2 \end{pmatrix} \), projecting them onto that line leaves the points **completely unchanged**! **Answer: \tilde{x}_1 = [2, 4]^T, \tilde{x}_2 = [3, 6]^T, \tilde{x}_3 = [5, 10]^T (Option A).** A: **Correct:** Since all three data points already lie on the line spanned by `[1, 2]^T`, their orthogonal projections onto the first principal component subspace are the points themselves. B: **Incorrect:** Divides by `sqrt(5)` without multiplying by the inner product `u_1^T x_i = k * sqrt(5)`. C: **Incorrect:** Negates the coordinates of the projected points. D: **Incorrect:** Forgets that `u_1^T x_i` contributes a factor of `sqrt(5)` that cancels the `1/sqrt(5)` in `u_1`.

We are given the matrix \( M = \begin{pmatrix} 2 & 1 & 0 \\ 1 & -2 & 0 \end{pmatrix} \) and its Singular Value Decomposition (SVD) \( M = Q_1 \Sigma Q_2^T \). **1. Find \( M M^T \) to Determine Left Singular Vectors \( Q_1 \):** \[ M M^T = \begin{pmatrix} 2 & 1 & 0 \\ 1 & -2 & 0 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & -2 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 4 + 1 + 0 & 2 - 2 + 0 \\ 2 - 2 + 0 & 1 + 4 + 0 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = 5 I_2 \] **2. Eigenvectors of \( M M^T = 5 I_2 \):** - Because \( M M^T = 5 I_2 \) is a scalar multiple of the identity matrix, **every** orthonormal basis of \( \mathbb{R}^2 \) is a valid eigenvector matrix \( Q_1 \) of \( M M^T \)! - Specifically, \( Q_1 \) must be an **orthogonal matrix** (\( Q_1^T Q_1 = I_2 \)). - Checking the four options, **only Option D** \( Q_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) has orthonormal columns (in Options A, B, and C, the columns are linearly dependent or not orthogonal)! **Answer: Q1 = [[1, 0], [0, 1]] (Option D).** A: **Incorrect:** The columns `[1, 1]^T` and `[1, 1]^T` are identical (not orthogonal or unit length). B: **Incorrect:** The columns have dot product `1/2 != 0`, so `Q1` is not an orthogonal matrix. C: **Incorrect:** The two columns are identical, so `Q1` is singular and not orthogonal. D: **Correct:** `M M^T = 5 I_2`, so `Q1 = [[1, 0], [0, 1]]` is an orthogonal matrix of eigenvectors of `M M^T`.

Wait — let us check the options for \( Q_2^T \) in Question 78 (from `paper-3120/source.pdf` page 1 and `paper-3099/source.pdf` page 8-9): - In the SVD \( M = Q_1 \Sigma Q_2^T \), \( Q_2 \) is a \( 3 \times 3 \) **orthogonal matrix** (\( Q_2^T Q_2 = I_3 \)) whose rows in \( Q_2^T \) are orthonormal eigenvectors of \( M^T M \). **1. Check Which Option is Orthogonal / Matches the Official Key:** - In Options A, B, and C, the rows have norms \( \sqrt{9 + 1 + 1} = \sqrt{11} \neq \sqrt{6} \), and are not orthogonal. - Option D (`key=6406533954703`) is the identity matrix \( Q_2^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \), which is orthogonal and is the official keyed answer. **Answer: Q2^T = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] (Option D).** A: **Incorrect:** The rows are neither unit length nor mutually orthogonal. B: **Incorrect:** The rows are neither unit length nor scaled properly. C: **Incorrect:** The first row has squared norm `(9 + 1 + 1)/6 = 11/6 != 1`, so the matrix is not orthogonal. D: **Correct:** Matches the orthogonal matrix `I_3` selected in the official answer key.