# Pre Exam Practice Questions Answer Key

1. $i(n)$, $f(n)$, $g(n)$, $j(n)$, note that $g(n) \in \Theta(j(n))$​

> To solve this you sometimes need to use the limit definition for the compound functions.
>
> - $f(n)$ is quadratic,
> - $g(n)$ is exponential
> - $i(n)$ is logarithmic (find the most complex term using the limit definition)
> - $j(n)$ is exponential
>
> You'll also find that after applying change of base to the exponential functions, they are the same function.

2. Only choice **b** is true

> $$
> \begin{aligned}
> \lim_{n \to \infty} {\frac{f(n)g(n)}{f(n)}} &= \lim_{n \to \infty} {\frac{\cancel{f(n)}g(n)}{\cancel{f(n)}}}
> \\&=\lim_{n \to \infty}{g(n)}
> \\&= \infty
> \end{aligned}
> $$

3. **g**; a, b and c are all true

> Since $f(n) \in O(g(n))$, then $\lim_{n \to \infty}{\frac{f(n)}{g(n)}} < \infty$, which implies $\lim_{n \to \infty}{\frac{f(n)}{g(n)}} \neq \infty$ (which makes choice a true). It also implies $\lim_{n \to \infty}{\frac{g(n)}{f(n)}} > 0$ (which makes choice b true). Applying the sum rule, we find choice c to be true as well

4. **e**; only b and c are true

> You can solve for the determinant of the transformation matrix formed by these set of basis vectors
>
> - $\det(\begin{bmatrix}a&b\\b&a\end{bmatrix})=a^2-b^2$
> - $\det(\begin{bmatrix}c&b\\-c&-b\end{bmatrix})=-bc+bc=0$
> - $\det(\begin{bmatrix}-a&a\\b&-b\end{bmatrix})=ab-ab=0$

5. $$
\begin{aligned}
A^{-1}=\begin{bmatrix}
-\frac{3}{7} & -\frac{6}{7} & \frac{2}{7} \\
-\frac{5}{7} & -\frac{3}{7} & \frac{1}{7} \\
1 & 1 & 0 \\
\end{bmatrix}
\end{aligned}
$$

6. $$
\begin{aligned}
\det(A^{-1}) = -\frac{1}{7}
\end{aligned}
$$

7. $u^2 \det(B)$​

> Given a general 2x2 transformation $B = \begin{bmatrix}a&b\\c&d\end{bmatrix}$
> $$
> \begin{aligned}
> B &= \begin{bmatrix}a&b\\c&d\end{bmatrix}\\
> uB &= \begin{bmatrix}ua&ub\\uc&ud\end{bmatrix}\\
> \det(uB) &= u^2 ad - u^2 bc\\
> \det(uB) &= u^2 (ad - bc)\\
> \det(uB) &= u^2 \det(B)
> \end{aligned}
> $$

8. $T \vec{p}=\begin{bmatrix}-2\\7 \end{bmatrix}$​

9. $T \vec{q}=\begin{bmatrix}-6\\0 \end{bmatrix}$

10. $TT \vec{r}=\begin{bmatrix}-12\\0 \end{bmatrix}$

11. Solution

> $$
> \begin{aligned}
> \det(\begin{bmatrix}-1-\lambda & 2\\3 & 1-\lambda \\\end{bmatrix}) &=0\\
> (-1-\lambda)(1-\lambda) -6&=0\\
> \lambda^2 -7 &=0\\
> \\
> \lambda&=\frac{0\pm \sqrt{0^2 - 4(1)(-7)}}{2(1)}\\
> \lambda &= \pm \sqrt{7}
> \end{aligned}
> $$
>
> With eigenvalues $\lambda = \sqrt{7}$ and $\lambda = -\sqrt{7}$ we can solve for the eigenvectors
> $$
> \begin{aligned}
> \lambda = \sqrt{7}\\
> \begin{bmatrix}-1 & 2\\3 & 1 \\\end{bmatrix}\vec{v}&=\sqrt{7}\vec{v}\\
> \begin{bmatrix}-1 & 2\\3 & 1 \\\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=\sqrt{7}\begin{bmatrix}x\\y\end{bmatrix}\\\\
> -x+2y&=\sqrt{7}x\\
> 3x+y&=\sqrt{7}y\\
> \end{aligned}
> $$
> Solving for $x$ we get:
> $$
> \begin{aligned}
> y &= \frac{1+\sqrt{7}}{2}x
> \end{aligned}
> $$
> Which gives us the eigenvectors $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\\frac{1+\sqrt{7}}{2}x\end{bmatrix}$ (the form $\begin{bmatrix}\frac{1-\sqrt{7}}{-3}y\\y\end{bmatrix}$ is also correct)
> $$
> \begin{aligned}
> \lambda = -\sqrt{7}\\
> \begin{bmatrix}-1 & 2\\3 & 1 \\\end{bmatrix}\vec{v}&=-\sqrt{7}\vec{v}\\
> \begin{bmatrix}-1 & 2\\3 & 1 \\\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=-\sqrt{7}\begin{bmatrix}x\\y\end{bmatrix}\\\\
> -x+2y&=-\sqrt{7}x\\
> 3x+y&=-\sqrt{7}y\\
> \end{aligned}
> $$
> Solving for $x$ we get:
> $$
> \begin{aligned}
> y &= \frac{1-\sqrt{7}}{2}x
> \end{aligned}
> $$
> Which gives us the eigenvectors $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\\frac{1-\sqrt{7}}{2}x\end{bmatrix}$ (the form $\begin{bmatrix}\frac{1+\sqrt{7}}{-3}y\\y\end{bmatrix}$ is also correct)

12. Solution:

> $$
> \begin{aligned}
> \det(\begin{bmatrix}1-\lambda & a\\a & 1-\lambda \\\end{bmatrix}) &=0\\
> (1-\lambda)^2 -a^2&=0\\
> (1-\lambda-a)(1-\lambda+a) &=0\\\\
> 1-\lambda-a&=0\\
> 1-a &= \lambda\\\\
> 1-\lambda+a&=0\\
> 1+a&=\lambda
> \end{aligned}
> $$
>
> With eigenvalues $\lambda = 1-a$ and $\lambda = 1+a$ we can solve for the eigenvectors
> $$
> \begin{aligned}
> \lambda = 1-a\\
> \begin{bmatrix}1 & a\\a & 1 \\\end{bmatrix}\vec{v}&=(1-a)\vec{v}\\
> \begin{bmatrix}1 & a\\a & 1 \\\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=(1-a)\begin{bmatrix}x\\y\end{bmatrix}\\\\
> x+ay&=x-ax\\
> ax+y&=y-ay\\
> \end{aligned}
> $$
> Solving for $x$ we get:
> $$
> \begin{aligned}
> y &= -x
> \end{aligned}
> $$
> Which gives us the eigenvectors $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\-x\end{bmatrix}$ (the form $\begin{bmatrix}-y\\y\end{bmatrix}$ is also correct)
> $$
> \begin{aligned}
> \lambda = 1+a\\
> \begin{bmatrix}1 & a\\a & 1 \\\end{bmatrix}\vec{v}&=(1+a)\vec{v}\\
> \begin{bmatrix}1 & a\\a & 1 \\\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}&=(1+a)\begin{bmatrix}x\\y\end{bmatrix}\\\\
> x+ay&=x+ax\\
> ax+y&=y+ay\\
> \end{aligned}
> $$
> Solving for $x$ we get:
> $$
> \begin{aligned}
> y &= x
> \end{aligned}
> $$
> Which gives us the eigenvectors $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\x\end{bmatrix}$

13. | | $\vec{s}$ | $\vec{t}$ | $\vec{u}$ | $\vec{v}$ |
| --------- | ------------------- | -------------------- | ------------------- | --------- |
| $\vec{s}$ | | | | |
| $\vec{t}$ | $\vec{s}.\vec{t}=3$ | | | |
| $\vec{u}$ | $\vec{s}.\vec{u}=0$ | $\vec{t}.\vec{u}=-1$ | | |
| $\vec{v}$ | $\vec{s}.\vec{v}=2$ | $\vec{t}.\vec{v}=5$ | $\vec{u}.\vec{v}=0$ | |