Solutions to Warmup Questions

Linear Algebra

Vectors

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

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

If you are having trouble with these problems, please review Section 6.1 Working with Vectors “Working with Vectors” in Chapter 6 Linear Algebra.

Are the following sets of vectors linearly independent?

$u = (\begin{matrix} 1 \\ 2 \end{matrix})$ , $v = (\begin{matrix} 2 \\ 4 \end{matrix})$

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

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

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

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

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

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

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 Linear Independence.

Matrices

$A = (\begin{matrix} 7 & 5 & 1 \\ 11 & 9 & 3 \\ 2 & 14 & 21 \\ 4 & 1 & 5 \end{matrix})$

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

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

Given that

$B = (\begin{matrix} 1 & 2 & 8 \\ 3 & 9 & 11 \\ 4 & 7 & 5 \\ 5 & 1 & 9 \end{matrix})$

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

Given that

$C = (\begin{matrix} 1 & 2 & 8 \\ 3 & 9 & 11 \\ 4 & 7 & 5 \end{matrix})$

$A + C = No solution, matrices non-conformable$

Given that

$c = 2$

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

If you are having trouble with these problems, please review Section 6.3 Basics of Matrix Algebra.

Operations

Summation

Simplify the following

$\sum_{i = 1}^{3} i = 1 + 2 + 3 = 6$
$\sum_{k = 1}^{3} (3 k + 2) = 3 \sum_{k = 1}^{3} k + \sum_{k = 1}^{3} 2 = 3 \times 6 + 3 \times 2 = 24$
$\sum_{i = 1}^{4} (3 k + i + 2) = 3 \sum_{i = 1}^{4} k + \sum_{i = 1}^{4} i + \sum_{i = 1}^{4} 2 = 12 k + 10 + 8 = 12 k + 18$

Products

$\prod_{i = 1}^{3} i = 1 \cdot 2 \cdot 3 = 6$
$\prod_{k = 1}^{3} (3 k + 2) = (3 + 2) \cdot (6 + 2) \cdot (9 + 2) = 440$

To review this material, please see Section @ref-sum-notation.

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, $e^{a} e^{b} e^{c} = e^{a + b + c}$ ,
$\log 0 = undefined$ – 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 \log and \exp

Limits

Find the limit of the following.

$lim_{x \to 2} (x - 1) = 1$
$lim_{x \to 2} \frac{(x - 2) (x - 1)}{(x - 2)} = 1$ , 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 $x - 1$ .
$lim_{x \to 2} \frac{x^{2} - 3 x + 2}{x - 2} = 1$ , same as above.

To review this material please see Section 2.3 Limits of a Function

Calculus

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

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

For a review, please see Section 3.1 Derivatives - 3.2 Higher-Order Derivatives (Derivatives of Derivatives of Derivatives)

Optimization

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

$f (x) = x$ $⇝$ neither exists.
$f (x) = x^{2}$ $⇝$ a minimum $f (x) = 0$ exists at $x = 0$ , but not a maximum.
$f (x) = - (x - 2)^{2}$ $⇝$ 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? $⇝ \frac{1}{18}$
If we roll two fair dice, what is the probability that their sum would be 12? $⇝ \frac{1}{36}$ . 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 Sets - 5.3 Probability