21_solutions-warmup.qmd

# Solutions to Warmup Questions {.unnumbered}

## Linear Algebra {.unnumbered}

### Vectors {.unnumbered}

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 @sec-vector-def "Working with Vectors" in Chapter @sec-linearalgebra.

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.

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

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

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

$\leadsto$ 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 @sec-linearindependence.

### Matrices {.unnumbered}

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

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 @sec-matrixbasics.

## Operations {.unnumbered}

### Summation {.unnumbered}

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 {.unnumbered}

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 @ref-sum-notation.

### Logs and exponents {.unnumbered}

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 @sec-logexponents

## Limits {.unnumbered}

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 @sec-limitsfun

## Calculus {.unnumbered}

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 @sec-derivintro - @sec-derivpoly

## Optimization {.unnumbered}

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 {.unnumbered}

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 @sec-setoper - @sec-probdef