# Solutions to Warmup Questions

## Linear Algebra

### Vectors

Define the vectors $$u = \begin{pmatrix} 1 \\2 \\3 \end{pmatrix}$$, $$v = \begin{pmatrix} 4\\5\\6 \end{pmatrix}$$, and the scalar $$c = 2$$.

1. $$u + v = \begin{pmatrix}5\\7\\9\end{pmatrix}$$
2. $$cv = \begin{pmatrix}8\\10\\12\end{pmatrix}$$
3. $$u \cdot v = 1(4) + 2(5) + 3(6) = 32$$

If you are having trouble with these problems, please review Section 6.1 "Working with Vectors" in Chapter 6.

Are the following sets of vectors linearly independent?

1. $$u = \begin{pmatrix} 1\\ 2\end{pmatrix}$$, $$v = \begin{pmatrix} 2\\4\end{pmatrix}$$

$$\leadsto$$ No: $2u = \begin{pmatrix} 2\\ 4\end{pmatrix}, v = \begin{pmatrix} 2\\ 4\end{pmatrix}$ so infinitely many linear combinations of $$u$$ and $$v$$ that amount to 0 exist.

1. $$u = \begin{pmatrix} 1\\ 2\\ 5 \end{pmatrix}$$, $$v = \begin{pmatrix} 3\\ 7\\ 9 \end{pmatrix}$$

$$\leadsto$$ Yes: we cannot find linear combination of these two vectors that would amount to zero.

1. $$a = \begin{pmatrix} 2\\ -1\\ 1 \end{pmatrix}$$, $$b = \begin{pmatrix} 3\\ -4\\ -2 \end{pmatrix}$$, $$c = \begin{pmatrix} 5\\ -10\\ -8 \end{pmatrix}$$

$$\leadsto$$ No: After playing around with some numbers, we can find that $-2a = \begin{pmatrix} -4\\ 2\\ -2 \end{pmatrix}, 3b = \begin{pmatrix} 9\\ -12\\ -6 \end{pmatrix}, -1c = \begin{pmatrix} -5\\ 10\\ 8 \end{pmatrix}$

So $-2a + 3b - c = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

i.e., a linear combination of these three vectors that would amount to zero exists.

If you are having trouble with these problems, please review Section 6.2.

### Matrices

${\bf A}=\begin{pmatrix} 7 & 5 & 1 \\ 11 & 9 & 3 \\ 2 & 14 & 21 \\ 4 & 1 & 5 \end{pmatrix}$

What is the dimensionality of matrix $${\bf A}$$? 4 $$\times$$ 3

What is the element $$a_{23}$$ of $${\bf A}$$? 3

Given that

${\bf B}=\begin{pmatrix} 1 & 2 & 8 \\ 3 & 9 & 11 \\ 4 & 7 & 5 \\ 5 & 1 & 9 \end{pmatrix}$

$\mathbf{A} + \mathbf{B} = \begin{pmatrix} 8 & 7 & 9 \\ 14 & 18 & 14 \\ 6 & 21 & 26 \\ 9 & 2 & 14 \end{pmatrix}$

Given that

${\bf C}=\begin{pmatrix} 1 & 2 & 8 \\ 3 & 9 & 11 \\ 4 & 7 & 5 \\ \end{pmatrix}$

$\mathbf{A} + \mathbf{C} = \text{No solution, matrices non-conformable}$

Given that

$c = 2$

$c\textbf{A} = \begin{pmatrix} 14 & 10 & 2 \\ 22 & 18 & 6 \\ 4 & 28 & 42 \\ 8 & 2 & 10 \end{pmatrix}$

If you are having trouble with these problems, please review Section 6.3.

## Operations

### Summation

Simplify the following

1. $$\sum\limits_{i = 1}^3 i = 1 + 2+ 3 = 6$$

2. $$\sum\limits_{k = 1}^3(3k + 2) = 3\sum\limits_{k=1}^3k + \sum\limits_{k=1}^3 2= 3\times 6 + 3\times 2 = 24$$

3. $$\sum\limits_{i= 1}^4 (3k + i + 2) = 3\sum\limits_{i= 1}^4k + \sum\limits_{i= 1}^4i + \sum\limits_{i= 1}^42 = 12k + 10 + 8 = 12k + 18$$

### Products

1. $$\prod\limits_{i= 1}^3 i = 1\cdot 2\cdot 3 = 6$$

2. $$\prod\limits_{k=1}^3(3k + 2) = (3 + 2)\cdot (6 + 2)\cdot (9 + 2) = 440$$

To review this material, please see Section 1.1.

### Logs and exponents

Simplify the following

1. $$4^2 = 16$$
2. $$4^2 2^3 = 2^{2\cdot 2}2^{3} = 2^{4 + 3} = 128$$
3. $$\log_{10}100 = \log_{10}10^2 = 2$$
4. $$\log_{2}4 = \log_{2}2^2 = 2$$
5. when $$\log$$ is the natural log, $$\log e = \log_{e} e^1 = 1$$
6. when $$a, b, c$$ are each constants, $$e^a e^b e^c = e^{a + b + c}$$,
7. $$\log 0 = \text{undefined}$$ -- no exponentiation of anything will result in a 0.
8. $$e^0 = 1$$ -- any number raised to the 0 is always 1.
9. $$e^1 = e$$ -- any number raised to the 1 is always itself
10. $$\log e^2 = \log_e e^2 = 2$$

To review this material, please see Section 1.3

## Limits

Find the limit of the following.

1. $$\lim\limits_{x \to 2} (x - 1) = 1$$
2. $$\lim\limits_{x \to 2} \frac{(x - 2) (x - 1)}{(x - 2)} = 1$$, though note that the original function $$\frac{(x - 2) (x - 1)}{(x - 2)}$$ would have been undefined at $$x = 2$$ because of a divide by zero problem; otherwise it would have been equal to $$x - 1$$.
3. $$\lim\limits_{x \to 2}\frac{x^2 - 3x + 2}{x- 2} = 1$$, same as above.

To review this material please see Section 2.3

## Calculus

For each of the following functions $$f(x)$$, find the derivative $$f'(x)$$ or $$\frac{d}{dx}f(x)$$

1. $$f(x)=c$$, $$f'(x) = 0$$
2. $$f(x)=x$$, $$f'(x) = 1$$
3. $$f(x)=x^2$$, $$f'(x) = 2x$$
4. $$f(x)=x^3$$, $$f'(x) = 3x^2$$
5. $$f(x)=3x^2+2x^{1/3}$$, $$f'(x) = 6x + \frac{2}{3}x^{-2/3}$$
6. $$f(x)=(x^3)(2x^4)$$, $$f'(x) = \frac{d}{dx}2x^7 = 14x^6$$

For a review, please see Section 3.1 - 3.2

## Optimization

For each of the followng functions $$f(x)$$, does a maximum and minimum exist in the domain $$x \in \mathbf{R}$$? If so, for what are those values and for which values of $$x$$?

1. $$f(x) = x$$ $$\leadsto$$ neither exists.
2. $$f(x) = x^2$$ $$\leadsto$$ a minimum $$f(x) = 0$$ exists at $$x = 0$$, but not a maximum.
3. $$f(x) = -(x - 2)^2$$ $$\leadsto$$ a maximum $$f(x) = 0$$ exists at $$x = 2$$, but not a minimum.

If you are stuck, please try sketching out a picture of each of the functions.

## Probability

1. If there are 12 cards, numbered 1 to 12, and 4 cards are chosen, $$\binom{12}{4} = \frac{12\cdot 11\cdot 10\cdot 9}{4!} = 495$$ possible hands exist (unordered, without replacement) .
2. Let $$A = \{1,3,5,7,8\}$$ and $$B = \{2,4,7,8,12,13\}$$. Then $$A \cup B = \{1, 2, 3, 4, 5, 7, 8, 12, 13\}$$, $$A \cap B = \{7, 8\}$$? If $$A$$ is a subset of the Sample Space $$S = \{1,2,3,4,5,6,7,8,9,10\}$$, then the complement $$A^C = \{2, 4, 6, 9, 10\}$$

3. If we roll two fair dice, what is the probability that their sum would be 11? $$\leadsto \frac{1}{18}$$
4. If we roll two fair dice, what is the probability that their sum would be 12? $$\leadsto \frac{1}{36}$$. There are two independent dice, so $$6^2 = 36$$ options in total. While the previous question had two possibilities for a sum of 11 (5,6 and 6,5), there is only one possibility out of 36 for a sum of 12 (6,6).

For a review, please see Sections 5.2 - 5.3