Consider the vectors $\bold{a}=\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}$ and $\bold{b}=\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$. What is the acute angle between $\bold{a}$ and $\bold{b}$ in radians?

$0$

No. The vectors are not parallel since one is not a multiple of the other.

$\pi/4$

No. Using the formula $\bold{a}\cdot\bold{b} = \|\bold{a}\||\bold{b}\|cos(\theta)$, where $\theta$ is the angle between the vectors and $\|\cdot\|$ denotes the magnitude function, you find that the angle is not $\pi/4$.

$\pi/3$

Yes. You can derive the answer by applying the formula $cos(\theta) = (\bold{a}\cdot\bold{b}) / (\|\bold{a}\||\bold{b}\|)$. Note that $\|\bold{a}\| = \|\bold{b}\| =\sqrt{1^2 + 1^2} = \sqrt{2}$ and $\bold{a}\cdot\bold{b} = 1\times1 + 1\times0 + 0\times1 = 1$. Thus, the cosine of the angle between $\bold{a}$ and $\bold{b}$ is $cos(\theta) = 1/2$ and $\theta = \pi/3$.

$\pi/2$

No. Using the formula $\bold{a}\cdot\bold{b} = \|\bold{a}\||\bold{b}\|cos(\theta)$, where $\theta$ is the angle between the vectors and $\|\cdot\|$ denotes the magnitude function, you find that the angle is not $\pi/2$.

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For which of the following use cases would you compute the determinant of a matrix?

To verify that a matrix is invertible.

Yes. A matrix is invertible if and only if its determinant is not $0$.

To check if the matrix is diagonalizable.

No. An $n \times n$ real matrix $A$ is diagonalizable if and only if there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. This happens if and only if there is a basis of $\mathbb{R}^n$ made from eigenvectors of A. Calculating the determinant of a matrix is not enough to conclude whether it is diagonalizable.

To check if the matrix is positive definite.

No. A symmetric $n \times n$ real matrix $M$ is positive definite if $x^TMx > 0$ for all non-zero $x \in \mathbb{R}^n$. This happens if and only if all its eigenvalues are positive. Calculating the determinant of a matrix is not enough to conclude whether it is positive definite.

None of the above.

No. One of the above answers is correct.

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The softmax function is used to convert an $n$-dimensional vector $x = (x_1, x_2, \cdots, x_n)^T$ into probabilities. It is defined as $softmax(x) = (\frac{e^{x_1}}{\textstyle\sum_{i=1}^ne^{x_i}}, \frac{e^{x_2}}{\textstyle\sum_{i=1}^ne^{x_i}}, \cdots, \frac{e^{x_n}}{\textstyle\sum_{i=1}^ne^{x_i}})^T$. What is the softmax of $x = \begin{pmatrix} \ln(2) \\ 0 \\ 0 \end{pmatrix}$?

$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

No. Remember that if $x \in \mathcal{R}^n$, then $softmax(x)_i = \frac{e^{x_i}}{\sum_{k=1}^n e^{x_k}}$.

$\frac{1}{6}\begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$

No. Remember that if $x \in \mathcal{R}^n$, then $softmax(x)_i = \frac{e^{x_i}}{\sum_{k=1}^n e^{x_k}}$.

$\frac{1}{4}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$

Yes. Note that $e^{\ln(2)} = 2$ and $e^0 = 1$. Thus, $softmax(x) = \begin{pmatrix} 2/4 \\ 1/4 \\ 1/4 \end{pmatrix} = \frac{1}{4}\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$.

$\frac{1}{3}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

No. Remember that if $x \in \mathcal{R}^n$, then $softmax(x)_i = \frac{e^{x_i}}{\sum_{k=1}^n e^{x_k}}$.

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Consider the slope function $f'(x) = \begin{cases} 0 &\text{if } x < 0 \\ 1 &\text{if } x > 0 \end{cases}$. Which of the following is a possible function for $f$?

$f(x) = x \quad \text{if } x > 0$

No. This function is not defined for non-positive numbers, so its derivative is also not defined for non-positive numbers.

$f(x) = \begin{cases} x &\text{if } x < 0 \\ x^2/2 &\text{if } x > 0 \end{cases}$

No. The derivative of this function is $f'(x) = \begin{cases} 1 &\text{if } x < 0 \\ x &\text{if } x > 0 \end{cases}$

$f(x) = \begin{cases} 1 &\text{if } x < 0 \\ x + 1 &\text{if } x > 0 \end{cases}$

Yes. The derivative of this function is $f'(x) = \begin{cases} 0 &\text{if } x < 0 \\ 1 &\text{if } x > 0 \end{cases}$

$f(x) = \begin{cases} 0 &\text{if } x < 0 \\ 1 &\text{if } x > 0 \end{cases}$

No. The derivative of this function is $f'(x) = 0 \quad \text{if } x \neq 0$

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