Warmup Questions Solutions

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

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}

-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.

Matrices

\mathbf{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

\mathbf{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}

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

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

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.

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).

# Warmup Questions Solutions {-} ## 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. $u \cdot v = 1(4) + 2(5) + 3(6) = 32$ 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. ### Matrices {-} $$\mathbf{A}=\begin{pmatrix} 7 & 5 & 1 \\ 11 & 9 & 3 \\ 2 & 14 & 21 \\ 4 & 1 & 5 \end{pmatrix}$$ What is the dimensionality of matrix ${\bf A}$? 4 $\times$ 3 What is the element $a_{23}$ of ${\bf A}$? 3 Given that $$\mathbf{B}=\begin{pmatrix} 1 & 2 & 8 \\ 3 & 9 & 11 \\ 4 & 7 & 5 \\ 5 & 1 & 9 \end{pmatrix}$$ $$\mathbf{A} + \mathbf{B} = \begin{pmatrix} 8 & 7 & 9 \\ 14 & 18 & 14 \\ 6 & 21 & 26 \\ 9 & 2 & 14 \end{pmatrix}$$ 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 $$c = 2$$ $$c\textbf{A} = \begin{pmatrix} 14 & 10 & 2 \\ 22 & 18 & 6 \\ 4 & 28 & 42 \\ 8 & 2 & 10 \end{pmatrix}$$ ## 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. $\prod\limits_{k=1}^3(3k + 2) = (3 + 2)\cdot (6 + 2)\cdot (9 + 2) = 440$ ### 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. $\log e^2 = \log_e e^2 = 2$ ## 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. ## 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. $f(x)=(x^3)(2x^4)$, $f'(x) = \frac{d}{dx}2x^7 = 14x^6$ ## 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).