# 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 \(c = 2\).

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

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?

- \(u = \begin{pmatrix} 1\\ 2\end{pmatrix}\), \(v = \begin{pmatrix} 2\\4\end{pmatrix}\)

\(\leadsto\) 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.

- \(u = \begin{pmatrix} 1\\ 2\\ 5 \end{pmatrix}\), \(v = \begin{pmatrix} 3\\ 7\\ 9 \end{pmatrix}\)

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

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

\(\leadsto\) 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

\(\sum\limits_{i = 1}^3 i = 1 + 2+ 3 = 6\)

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

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

### Products

\(\prod\limits_{i= 1}^3 i = 1\cdot 2\cdot 3 = 6\)

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

To review this material, please see Section 1.1.

### Logs and exponents

Simplify the following

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

To review this material, please see Section 1.3

## Limits

Find the limit of the following.

- \(\lim\limits_{x \to 2} (x - 1) = 1\)
- \(\lim\limits_{x \to 2} \frac{(x - 2) (x - 1)}{(x - 2)} = 1\), 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 \(x - 1\).
- \(\lim\limits_{x \to 2}\frac{x^2 - 3x + 2}{x- 2} = 1\), 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 \(\frac{d}{dx}f(x)\)

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

## 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 \(x\)?

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

- 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) .
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 \(A^C = \{2, 4, 6, 9, 10\}\)

- If we roll two fair dice, what is the probability that their sum would be 11? \(\leadsto \frac{1}{18}\)
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).