Machine Learning Foundations · Quiz 1 · Sep 2024

**Answer: A, D** **Topic:** Linear Algebra — *Rank-2 Matrices with Equal Column Space and Row Space* ### Step-by-Step Solution **Step 1:** Examine each $3 \times 3$ matrix to check if (1) $\text{rank}(M) = 2$ and (2) $\mathcal{C}(M) = \mathcal{R}(M)$: - **Option A ($\text{diag}(1, 1, 0)$)**: Has 2 non-zero pivots ($\text{rank} = 2$), and since it is symmetric, its column space and row space are both $\text{span}(e_1, e_2)$ — **Correct**. - **Option B ($\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{bmatrix}$)**: Compute $\det(B) = 1(24 - 25) - 2(12 - 15) + 3(10 - 12) = -1 + 6 - 6 = -1 \ne 0$, so $\text{rank}(B) = 3 \ne 2$ — Incorrect. - **Option C ($0_{3 \times 3}$)**: Has $\text{rank} = 0 \ne 2$ — Incorrect. - **Option D ($\text{diag}(2, 0, 2)$)**: Has 2 non-zero diagonal entries ($\text{rank} = 2$), and since it is symmetric, its column space and row space are both $\text{span}(e_1, e_3)$ — **Correct**. **Step 2:** Therefore, **A and D** are correct. **Key Takeaway / Tip:** Any symmetric matrix ($M = M^T$) automatically has its column space equal to its row space ($\mathcal{C}(M) = \mathcal{C}(M^T) = \mathcal{R}(M)$), so we only need to check that the rank is $2$. A: **Correct:** Option A (

) matches our derived result under Rank-2 Matrices with Equal Column Space and Row Space. Any symmetric matrix ($M = M^T$) automatically has its column space equal to its row space ($\mathcal{C}(M) = \mathcal{C}(M^T) = \mathcal{R}(M)$), so we only need to check that the rank is $2$. B: **Incorrect:** Option B (

) does not satisfy the conditions derived in the step-by-step analysis for Rank-2 Matrices with Equal Column Space and Row Space (correct choice: A, D). C: **Incorrect:** Option C (

) does not satisfy the conditions derived in the step-by-step analysis for Rank-2 Matrices with Equal Column Space and Row Space (correct choice: A, D). D: **Correct:** Option D (

) matches our derived result under Rank-2 Matrices with Equal Column Space and Row Space. Any symmetric matrix ($M = M^T$) automatically has its column space equal to its row space ($\mathcal{C}(M) = \mathcal{C}(M^T) = \mathcal{R}(M)$), so we only need to check that the rank is $2$.

**Answer: A, C** **Topic:** Linear Algebra — *Eigenvalues of a Matrix Polynomial and Triangular Matrices* ### Step-by-Step Solution **Step 1:** Given the upper triangular matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{bmatrix}$, its eigenvalues are simply its diagonal entries: $$\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 3$$ (Thus, **B is false**). **Step 2:** For any polynomial $p(x) = x^4 - 3x - 2$, the matrix $B = p(A) = A^4 - 3A - 2I$ has eigenvalues $p(\lambda_i) = \lambda_i^4 - 3\lambda_i - 2$: - For $\lambda_1 = 1$: $p(1) = 1^4 - 3(1) - 2 = 1 - 3 - 2 = -4$ - For $\lambda_2 = 2$: $p(2) = 2^4 - 3(2) - 2 = 16 - 6 - 2 = 8$ - For $\lambda_3 = 3$: $p(3) = 3^4 - 3(3) - 2 = 81 - 9 - 2 = 70$ Thus, **$B$ has eigenvalues $-4, 8, 70$** (**A is true**). **Step 3:** Furthermore, products and linear combinations of upper triangular matrices are always **upper triangular** (**C is true**, **D is false**). **Step 4:** Therefore, **A and C** are true. **Key Takeaway / Tip:** If $\lambda$ is an eigenvalue of $A$, then $p(\lambda)$ is an eigenvalue of the matrix polynomial $p(A)$—there is no need to multiply $A^4$ explicitly! A: **Correct:** Option A (B has eigenvalues -4, 8, 70.) matches our derived result under Eigenvalues of a Matrix Polynomial and Triangular Matrices. If $\lambda$ is an eigenvalue of $A$, then $p(\lambda)$ is an eigenvalue of the matrix polynomial $p(A)$—there is no need to multiply $A^4$ explicitly! B: **Incorrect:** Option B (A has eigenvalues 1, 16, 81.) does not satisfy the conditions derived in the step-by-step analysis for Eigenvalues of a Matrix Polynomial and Triangular Matrices (correct choice: A, C). C: **Correct:** Option C (B is an upper triangular matrix.) matches our derived result under Eigenvalues of a Matrix Polynomial and Triangular Matrices. If $\lambda$ is an eigenvalue of $A$, then $p(\lambda)$ is an eigenvalue of the matrix polynomial $p(A)$—there is no need to multiply $A^4$ explicitly! D: **Incorrect:** Option D (B is a lower triangular matrix.) does not satisfy the conditions derived in the step-by-step analysis for Eigenvalues of a Matrix Polynomial and Triangular Matrices (correct choice: A, C).

**Answer: B, C** **Topic:** Linear Algebra — *Eigenvectors and Matrix Power via Diagonalization* ### Step-by-Step Solution **Step 1:** Test whether $v_1 = (1, -1)^T$ and $v_2 = (1, 1)^T$ are eigenvectors of $A = \begin{bmatrix} 6 & 3 \\ -2 & 1 \end{bmatrix}$: - $A v_1 = \begin{bmatrix} 6(1) + 3(-1) \\ -2(1) + 1(-1) \end{bmatrix} = \begin{bmatrix} 3 \\ -3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = 3 v_1 \implies v_1$ **is an eigenvector with $\lambda_1 = 3$**. - $A v_2 = \begin{bmatrix} 6(1) + 3(1) \\ -2(1) + 1(1) \end{bmatrix} = \begin{bmatrix} 9 \\ -1 \end{bmatrix} \ne \lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix} \implies v_2$ is **not** an eigenvector. Thus, **Only $v_1$ is an eigenvector of $A$** (**B is true**, **A is false**). **Step 2:** Find the second eigenvalue and eigenvector of $A$: since $\text{tr}(A) = 6 + 1 = 7$ and $\lambda_1 = 3$, the second eigenvalue is $\lambda_2 = 7 - 3 = 4$. Solving $(A - 4I)u_2 = 0 \implies 2x + 3y = 0 \implies u_2 = (3, -2)^T$. **Step 3:** Write the diagonalization $A = P D P^{-1}$ with $P = \begin{pmatrix} 1 & 3 \\ -1 & -2 \end{pmatrix}$ ($\det(P) = -2 - (-3) = 1$, so $P^{-1} = \begin{pmatrix} -2 & -3 \\ 1 & 1 \end{pmatrix}$) and $D = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix}$: $$A^{50} = P D^{50} P^{-1} = \begin{pmatrix} 1 & 3 \\ -1 & -2 \end{pmatrix} \begin{pmatrix} 3^{50} & 0 \\ 0 & 4^{50} \end{pmatrix} \begin{pmatrix} -2 & -3 \\ 1 & 1 \end{pmatrix}$$ (**C is true**, **D is false**). **Step 4:** Therefore, **B and C** are correct. **Key Takeaway / Tip:** When $P = \begin{pmatrix} 1 & 3 \\ -1 & -2 \end{pmatrix}$ has determinant $-2 + 3 = 1$, its inverse is simply $\begin{pmatrix} -2 & -3 \\ 1 & 1 \end{pmatrix}$. A: **Incorrect:** Option A (

) does not satisfy the conditions derived in the step-by-step analysis for Eigenvectors and Matrix Power via Diagonalization (correct choice: B, C). B: **Correct:** Option B (

) matches our derived result under Eigenvectors and Matrix Power via Diagonalization. When $P = \begin{pmatrix} 1 & 3 \\ -1 & -2 \end{pmatrix}$ has determinant $-2 + 3 = 1$, its inverse is simply $\begin{pmatrix} -2 & -3 \\ 1 & 1 \end{pmatrix}$. C: **Correct:** Option C (

) matches our derived result under Eigenvectors and Matrix Power via Diagonalization. When $P = \begin{pmatrix} 1 & 3 \\ -1 & -2 \end{pmatrix}$ has determinant $-2 + 3 = 1$, its inverse is simply $\begin{pmatrix} -2 & -3 \\ 1 & 1 \end{pmatrix}$. D: **Incorrect:** Option D (

) does not satisfy the conditions derived in the step-by-step analysis for Eigenvectors and Matrix Power via Diagonalization (correct choice: B, C).

**Answer: A, C, D** **Topic:** Linear Algebra — *Orthogonal Projection Matrix onto a Line in R^3* ### Step-by-Step Solution **Step 1:** Given $a = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ and $b = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$, the matrix that projects onto the line spanned by $a$ is: $$P = \frac{a a^T}{a^T a} = \frac{1}{1^2 + (-1)^2 + 0^2} \begin{bmatrix} 1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ **Step 2:** Evaluate each option: - **A is true**: $P^T = P$, so $P$ is a symmetric matrix. - **B is false**: The projection of $b$ onto $a$ is $\frac{a^T b}{a^T a} a = \frac{1(1) + (-1)(2) + 0(-1)}{2} a = -\frac{1}{2} a = \frac{1}{2}(-a) \ne \frac{1}{4}(-a)$. - **C is true**: Because $P$ is the outer product of a single non-zero vector $a$ with itself, $\text{rank}(P) = 1$. - **D is true**: A vector $(x, y, z)^T$ is orthogonal to $a = (1, -1, 0)^T$ iff $1(x) + (-1)(y) + 0(z) = 0 \iff x = y$, which is the plane $x = y$. **Step 3:** Therefore, **A, C, and D** are true. **Key Takeaway / Tip:** The orthogonal complement of a non-zero vector $a = (a_1, a_2, a_3)^T$ in $\mathbb{R}^3$ is the plane $a_1 x + a_2 y + a_3 z = 0$, which for $a = (1, -1, 0)^T$ is $x - y = 0 \iff x = y$. A: **Correct:** Option A (

) matches our derived result under Orthogonal Projection Matrix onto a Line in R^3. The orthogonal complement of a non-zero vector $a = (a_1, a_2, a_3)^T$ in $\mathbb{R}^3$ is the plane $a_1 x + a_2 y + a_3 z = 0$, which for $a = (1, -1, 0)^T$ is $x - y = 0 \iff x = y$. B: **Incorrect:** Option B (

) does not satisfy the conditions derived in the step-by-step analysis for Orthogonal Projection Matrix onto a Line in R^3 (correct choice: A, C, D). C: **Correct:** Option C (

) matches our derived result under Orthogonal Projection Matrix onto a Line in R^3. The orthogonal complement of a non-zero vector $a = (a_1, a_2, a_3)^T$ in $\mathbb{R}^3$ is the plane $a_1 x + a_2 y + a_3 z = 0$, which for $a = (1, -1, 0)^T$ is $x - y = 0 \iff x = y$. D: **Correct:** Option D (

) matches our derived result under Orthogonal Projection Matrix onto a Line in R^3. The orthogonal complement of a non-zero vector $a = (a_1, a_2, a_3)^T$ in $\mathbb{R}^3$ is the plane $a_1 x + a_2 y + a_3 z = 0$, which for $a = (1, -1, 0)^T$ is $x - y = 0 \iff x = y$.

**Answer: A, F** **Topic:** Linear Regression — *Least Squares Fit on a Constant-Target Dataset* ### Step-by-Step Solution **Step 1:** Inspect the given dataset: $(x_1, y_1) = (1, 1)$, $(x_2, y_2) = (1, 1)$, and $(x_3, y_3) = (-1, 1)$. **Step 2:** Every single data point has $y_i = 1$! Therefore, the horizontal line $\hat{y} = 1$ (slope $0$, intercept $1$) passes **exactly** through all three points (**A is true**, and B, C, D are false). **Step 3:** Since $\hat{y}_i = 1 = y_i$ for all $i = 1, 2, 3$, the error (residual) vector is: $$e = \hat{y} - y = \begin{bmatrix} 1 - 1 \\ 1 - 1 \\ 1 - 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ (**F is true**, **E is false**). **Step 4:** Therefore, **A and F** are correct. **Key Takeaway / Tip:** Whenever all target values $y_i$ are equal to a constant $c$, the least-squares regression line is simply $\hat{y} = c$ with zero error vector. A: **Correct:** Option A (

) matches our derived result under Least Squares Fit on a Constant-Target Dataset. Whenever all target values $y_i$ are equal to a constant $c$, the least-squares regression line is simply $\hat{y} = c$ with zero error vector. B: **Incorrect:** Option B (

) does not satisfy the conditions derived in the step-by-step analysis for Least Squares Fit on a Constant-Target Dataset (correct choice: A, F). C: **Incorrect:** Option C (

) does not satisfy the conditions derived in the step-by-step analysis for Least Squares Fit on a Constant-Target Dataset (correct choice: A, F). D: **Incorrect:** Option D (

) does not satisfy the conditions derived in the step-by-step analysis for Least Squares Fit on a Constant-Target Dataset (correct choice: A, F). E: **Incorrect:** Option E (

) does not satisfy the conditions derived in the step-by-step analysis for Least Squares Fit on a Constant-Target Dataset (correct choice: A, F). F: **Correct:** Option F (

) matches our derived result under Least Squares Fit on a Constant-Target Dataset. Whenever all target values $y_i$ are equal to a constant $c$, the least-squares regression line is simply $\hat{y} = c$ with zero error vector.

**Answer: A, B, D** **Topic:** Calculus — *Evaluating Limits for Finite Values* ### Step-by-Step Solution **Step 1:** Evaluate each limit: - **Option A ($\lim_{x \to 1} \frac{x^{1/3} - 1}{x^{1/2} - 1}$)**: Using L'Hôpital's rule (or $\lim_{x \to 1} \frac{x^p - 1}{x^q - 1} = \frac{p}{q}$): $$\lim_{x \to 1} \frac{\frac{1}{3} x^{-2/3}}{\frac{1}{2} x^{-1/2}} = \frac{1/3}{1/2} = \frac{2}{3}\text{ (Finite — Correct)}$$ - **Option B ($\lim_{x \to 1} \frac{1 - x^{1/2}}{1 - x}$)**: Factor $1 - x = (1 - x^{1/2})(1 + x^{1/2})$: $$\lim_{x \to 1} \frac{1}{1 + x^{1/2}} = \frac{1}{2}\text{ (Finite — Correct)}$$ - **Option C ($\lim_{x \to \infty} \frac{x^2 - 7x}{x + 1}$)**: Degree of numerator ($2$) $>$ degree of denominator ($1$), so the limit diverges to $+\infty$ (Infinite). - **Option D ($\lim_{x \to a} \frac{x^2 - a^2}{x^4 - a^4}$ for $a > 0$)**: Factor $x^4 - a^4 = (x^2 - a^2)(x^2 + a^2)$: $$\lim_{x \to a} \frac{1}{x^2 + a^2} = \frac{1}{2a^2}\text{ (Finite since }a > 0\text{ — Correct)}$$ **Step 2:** Therefore, **A, B, and D** yield finite values. **Key Takeaway / Tip:** Use algebraic factoring or L'Hôpital's Rule on $0/0$ indeterminate forms at $x = 1$ and $x = a$. A: **Correct:** Option A (

) matches our derived result under Evaluating Limits for Finite Values. Use algebraic factoring or L'Hôpital's Rule on $0/0$ indeterminate forms at $x = 1$ and $x = a$. B: **Correct:** Option B (

) matches our derived result under Evaluating Limits for Finite Values. Use algebraic factoring or L'Hôpital's Rule on $0/0$ indeterminate forms at $x = 1$ and $x = a$. C: **Incorrect:** Option C (

) does not satisfy the conditions derived in the step-by-step analysis for Evaluating Limits for Finite Values (correct choice: A, B, D). D: **Correct:** Option D (

) matches our derived result under Evaluating Limits for Finite Values. Use algebraic factoring or L'Hôpital's Rule on $0/0$ indeterminate forms at $x = 1$ and $x = a$. E: **Incorrect:** Option E (

) does not satisfy the conditions derived in the step-by-step analysis for Evaluating Limits for Finite Values (correct choice: A, B, D).

**Answer: A** **Topic:** Linear Algebra — *Diagonalizability of a 3x3 Upper Triangular Matrix* ### Step-by-Step Solution **Step 1:** Inspect $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 6 \end{bmatrix}$: its eigenvalues are the diagonal entries $\lambda_1 = 2$ (algebraic multiplicity $2$) and $\lambda_2 = 6$ (algebraic multiplicity $1$). **Step 2:** $A$ is diagonalizable if and only if the geometric multiplicity of $\lambda_1 = 2$ equals its algebraic multiplicity ($2$), i.e., $\dim(\mathcal{N}(A - 2I)) = 3 - \text{rank}(A - 2I) = 2 \iff \text{rank}(A - 2I) = 1$. **Step 3:** Compute $A - 2I$: $$A - 2I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 4 \end{bmatrix}$$ Since Columns 1 and 2 are entirely zero, $\text{rank}(A - 2I) = 1$, so $\dim(\mathcal{N}(A - 2I)) = 3 - 1 = 2$ (spanned by $e_1 = [1, 0, 0]^T$ and $e_2 = [0, 1, 0]^T$). **Step 4:** Because the geometric multiplicity equals the algebraic multiplicity for every eigenvalue, **$A$ is diagonalizable** (**A: Yes**). **Key Takeaway / Tip:** Do not assume the presence of an off-diagonal `1` makes $A$ a defective Jordan block—here the `1` is in Position `(2, 3)` linking eigenvalue `2` and eigenvalue `6`, NOT linking the two `2`s! A: **Correct:** Option A (Yes) matches our derived result under Diagonalizability of a 3x3 Upper Triangular Matrix. Do not assume the presence of an off-diagonal `1` makes $A$ a defective Jordan block—here the `1` is in Position `(2, 3)` linking eigenvalue `2` and eigenvalue `6`, NOT linking the two `2`s! B: **Incorrect:** Option B (No) does not satisfy the conditions derived in the step-by-step analysis for Diagonalizability of a 3x3 Upper Triangular Matrix (correct choice: A). C: **Incorrect:** Option C (Insufficient information) does not satisfy the conditions derived in the step-by-step analysis for Diagonalizability of a 3x3 Upper Triangular Matrix (correct choice: A).

**Answer: A** **Topic:** Linear Classifiers — *Evaluating Misclassification Error of Linear Sign Classifiers* ### Step-by-Step Solution **Step 1:** Test Option **A**: $\hat{y} = \text{Sign}(x_1 + x_2 + x_3 - 5)$ (where $\text{Sign}(0) = +1$) on all 6 data points: 1. $[0, 1, 1]$ ($Y = -1$): $0 + 1 + 1 - 5 = -3 < 0 \implies \hat{y} = -1$ (Correct) 2. $[0, 1, 0]$ ($Y = -1$): $0 + 1 + 0 - 5 = -4 < 0 \implies \hat{y} = -1$ (Correct) 3. $[1, 1, 3]$ ($Y = +1$): $1 + 1 + 3 - 5 = 0 \implies \text{Sign}(0) = +1$ (Correct) 4. $[1, 0, 5]$ ($Y = +1$): $1 + 0 + 5 - 5 = +1 > 0 \implies \hat{y} = +1$ (Correct) 5. $[0, 0, 6]$ ($Y = +1$): $0 + 0 + 6 - 5 = +1 > 0 \implies \hat{y} = +1$ (Correct) 6. $[1, 0, 4]$ ($Y = +1$): $1 + 0 + 4 - 5 = 0 \implies \text{Sign}(0) = +1$ (Correct) **Step 2:** Option **A** classifies all 6 points with **0 errors** (100% accuracy), whereas Option B misclassifies the positive points ($x_1 + x_2 - 2 \le 0$), Option C misclassifies the negative points ($x_1 + x_2 + x_3 > 0$), and Option D misclassifies $[0, 1, 1]$ ($1 - 1 - 0 = 0 \implies +1 \ne -1$). **Step 3:** Therefore, **A** has the least misclassification error. **Key Takeaway / Tip:** Notice that the sum of coordinates $x_1 + x_2 + x_3$ is $\le 2$ for the negative points and $\ge 5$ for all four positive points, so $x_1 + x_2 + x_3 - 5$ separates them cleanly. A: **Correct:** Option A (

) matches our derived result under Evaluating Misclassification Error of Linear Sign Classifiers. Notice that the sum of coordinates $x_1 + x_2 + x_3$ is $\le 2$ for the negative points and $\ge 5$ for all four positive points, so $x_1 + x_2 + x_3 - 5$ separates them cleanly. B: **Incorrect:** Option B (

) does not satisfy the conditions derived in the step-by-step analysis for Evaluating Misclassification Error of Linear Sign Classifiers (correct choice: A). C: **Incorrect:** Option C (

) does not satisfy the conditions derived in the step-by-step analysis for Evaluating Misclassification Error of Linear Sign Classifiers (correct choice: A). D: **Incorrect:** Option D (

) does not satisfy the conditions derived in the step-by-step analysis for Evaluating Misclassification Error of Linear Sign Classifiers (correct choice: A).

**Answer: C** **Topic:** Calculus — *Continuity of a Piecewise Function* ### Step-by-Step Solution **Step 1:** For $f(x) = \begin{cases} \frac{x - b}{b + 1}, & x < 0 \\ x^2 + b, & x \ge 0 \end{cases}$ to be continuous at every $x$, we need $b \ne -1$ and the left-hand limit at $x = 0$ must equal the right-hand limit (value) at $x = 0$: **Step 2:** Equate the two branches at $x = 0$: $$\lim_{x \to 0^-} \frac{x - b}{b + 1} = \frac{-b}{b + 1}, \quad \lim_{x \to 0^+} (x^2 + b) = b$$ **Step 3:** Solve $\frac{-b}{b + 1} = b$ for $b$: $$-b = b(b + 1) \implies b^2 + 2b = 0 \implies b(b + 2) = 0 \implies b = 0 \text{ or } b = -2$$ **Step 4:** Among the given options (`3, 2, -2, 1`), **`-2`** (**C**) is the correct value. **Key Takeaway / Tip:** Substitute $b = -2$ to verify: at $x = 0$, the upper branch gives $\frac{0 - (-2)}{-2 + 1} = \frac{2}{-1} = -2$, and the lower branch gives $0^2 + (-2) = -2$. A: **Incorrect:** Option A (3) does not satisfy the conditions derived in the step-by-step analysis for Continuity of a Piecewise Function (correct choice: C). B: **Incorrect:** Option B (2) does not satisfy the conditions derived in the step-by-step analysis for Continuity of a Piecewise Function (correct choice: C). C: **Correct:** Option C (-2) matches our derived result under Continuity of a Piecewise Function. Substitute $b = -2$ to verify: at $x = 0$, the upper branch gives $\frac{0 - (-2)}{-2 + 1} = \frac{2}{-1} = -2$, and the lower branch gives $0^2 + (-2) = -2$. D: **Incorrect:** Option D (1) does not satisfy the conditions derived in the step-by-step analysis for Continuity of a Piecewise Function (correct choice: C).

**Answer: B** **Topic:** Multivariate Calculus — *Unit Direction of Maximum Directional Derivative* ### Step-by-Step Solution **Step 1:** Recall that the maximum directional derivative of a differentiable function $f(x, y)$ at a point $(x_0, y_0)$ occurs in the **unit direction of the gradient vector** $\nabla f(x_0, y_0)$. **Step 2:** Compute the gradient of $f(x, y) = x^2 + 3y^2$ and evaluate at $(1, 2)$: $$\nabla f(x, y) = (2x, \; 6y) \implies \nabla f(1, 2) = (2(1), \; 6(2)) = (2, 12)$$ **Step 3:** Normalize $(2, 12)$ to a unit vector by dividing by its norm $\|(2, 12)\| = \sqrt{2^2 + 12^2} = \sqrt{4 + 144} = \sqrt{148} = 2\sqrt{37}$: $$u = \frac{(2, 12)}{2\sqrt{37}} = \left(\frac{1}{\sqrt{37}}, \; \frac{6}{\sqrt{37}}\right)$$ **Step 4:** This matches Option **B**. **Key Takeaway / Tip:** Factor out $2$ from $\nabla f(1, 2) = 2(1, 6)$ first so you only have to normalize $(1, 6)$, which has length $\sqrt{1^2 + 6^2} = \sqrt{37}$! A: **Incorrect:** Option A (

) does not satisfy the conditions derived in the step-by-step analysis for Unit Direction of Maximum Directional Derivative (correct choice: B). B: **Correct:** Option B (

) matches our derived result under Unit Direction of Maximum Directional Derivative. Factor out $2$ from $\nabla f(1, 2) = 2(1, 6)$ first so you only have to normalize $(1, 6)$, which has length $\sqrt{1^2 + 6^2} = \sqrt{37}$! C: **Incorrect:** Option C (

) does not satisfy the conditions derived in the step-by-step analysis for Unit Direction of Maximum Directional Derivative (correct choice: B). D: **Incorrect:** Option D (

) does not satisfy the conditions derived in the step-by-step analysis for Unit Direction of Maximum Directional Derivative (correct choice: B).

**Answer: 1.5** **Topic:** Multivariate Calculus — *First-Order Taylor Linear Approximation in Two Variables* ### Step-by-Step Solution **Step 1:** Write the linear approximation $L(x, y)$ of $f(x, y) = e^x \cos(y)$ around $(x_0, y_0) = (0, 0)$: $$L(x, y) = f(0, 0) + f_x(0, 0)(x - 0) + f_y(0, 0)(y - 0)$$ **Step 2:** Compute $f(0, 0)$ and the partial derivatives at $(0, 0)$: - $f(0, 0) = e^0 \cos(0) = 1 \times 1 = 1$ - $f_x(x, y) = e^x \cos(y) \implies f_x(0, 0) = 1$ - $f_y(x, y) = -e^x \sin(y) \implies f_y(0, 0) = -1 \times 0 = 0$ Thus, $L(x, y) = 1 + x$. **Step 3:** Evaluate $L(0.5, 0)$: $$L(0.5, 0) = 1 + 0.5 = 1.5$$ **Key Takeaway / Tip:** Since $\cos(y) = 1 - \frac{y^2}{2!} + \dots$ has zero first derivative at $y = 0$ and $e^x \approx 1 + x$, the linearization around $(0, 0)$ is simply $L(x, y) = 1 + x$.