Math Series 5
1. Characteristic roots of the matrix
$$ \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \\ \end{bmatrix} $$ are ______.
$$ \begin{bmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \\ \end{bmatrix} $$ are ______.
- $ \pm i $
- $ \pm 2 $
- $ \pm 1 $
- $ \pm 4 $
Answer: C
Solution:
Solution:
To find the characteristic roots or the eigenvalue ($\lambda$), subtract $\lambda$ from the diagonal entries of your matrix, and then taking the determinant.
$$ det(A -\lambda I) = 0 $$
$$ A - \lambda I = \begin{bmatrix} \cos \theta - \lambda & -\sin \theta \\ -\sin \theta & -\cos \theta - \lambda \\ \end{bmatrix} $$
Set up the characteristic equation by taking the determinant and setting it equal to zero: $$\begin{align} det(A-\lambda I) & = 0\\ & = (\cos \theta - \lambda)(-\cos \theta - \lambda) - (-\sin \theta)(-\sin \theta) \\ & = -\cos^2 \theta + \lambda \cos \theta - \lambda \cos \theta + \lambda^2 - \sin^2 \theta \\ \end{align}$$
Since $ \sin^2 \theta + \cos^2 \theta = 1$, $$\begin{align} \lambda^2 - 1 & = 0 \\ \lambda^2 & = 1 \\ \lambda & = \pm 1 \end{align}$$
$$ det(A -\lambda I) = 0 $$
$$ A - \lambda I = \begin{bmatrix} \cos \theta - \lambda & -\sin \theta \\ -\sin \theta & -\cos \theta - \lambda \\ \end{bmatrix} $$
Set up the characteristic equation by taking the determinant and setting it equal to zero: $$\begin{align} det(A-\lambda I) & = 0\\ & = (\cos \theta - \lambda)(-\cos \theta - \lambda) - (-\sin \theta)(-\sin \theta) \\ & = -\cos^2 \theta + \lambda \cos \theta - \lambda \cos \theta + \lambda^2 - \sin^2 \theta \\ \end{align}$$
Since $ \sin^2 \theta + \cos^2 \theta = 1$, $$\begin{align} \lambda^2 - 1 & = 0 \\ \lambda^2 & = 1 \\ \lambda & = \pm 1 \end{align}$$
2. The points (x1,y1), (x2,y2), (x3,y3) are collinear if the rank of the matrix
$$
\begin{bmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{bmatrix}
$$
is ________.
- 1 or 2
- 2 or 3
- 1 or 3
- 2
Answer: A
Solution:
Solution:
Rank is the order or dimension of the largest sub-matrix with non-zero determinant. In simpler terms, it is the number of linearly independent rows or columns.
$$ \text{Rank} + \text{Nullity} = \text{no. of columns} $$ where Nullity is the number of dependent rows or columns.
$$ \text{Rank} + \text{Nullity} = \text{no. of columns} $$ where Nullity is the number of dependent rows or columns.
Since the points are collinear, they have at least one relationship (ie. slope) or at least one dependent point (ie. row).
If one dependent point (row): Rank = 3 - 1 = 2.
If two dependent points (row): Rank = 3 - 2 = 1.
Therefore, the rank is 1 or 2.
3. If the matrix is
$$
A =
\begin{bmatrix}
-1 & 2 & 3 \\
0 & 3 & 5 \\
0 & 0 & -2 \\
\end{bmatrix}
$$
then the eigenvalues of $A^3 + 5A + 8I$ are ________.
- -1 ,27, 8
- -1, 3, -2
- 2, 50, -10
- 2, 50, 10
Answer: C
Solution:
Solution:
$$
\text{Identity Matrix, I = }
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
$$
A^3 + 5A + 8I =
\begin{bmatrix}
2 & 24 & 36 \\
0 & 50 & 60 \\
0 & 0 & -10 \\
\end{bmatrix}
$$
Getting the eigenvalues,
$$ A^3 + 5A + 8I = \begin{bmatrix} 2-\lambda & 24 & 36 \\ 0 & 50-\lambda & 60 \\ 0 & 0 & -10-\lambda \\ \end{bmatrix} $$ $$\begin{align} det & = 0 \\ & = (2-\lambda)((50-\lambda)(-10-\lambda)-(60)(0)) - 24 ((0)(-10-\lambda)-(60)(0))+36((0)(0)-(50-\lambda)(0))\\ & = (2-\lambda)(50-\lambda)(-10-\lambda) \\ \end{align}$$ $$\begin{align} \text{Let } -10-\lambda & = 0 \\ \lambda & = -10 \end{align}$$ $$\begin{align} \text{Let } 2-\lambda & = 0 \\ \lambda & = 2 \end{align}$$ $$\begin{align} \text{Let } 50-\lambda & = 0 \\ \lambda & = 50 \end{align}$$
Getting the eigenvalues,
$$ A^3 + 5A + 8I = \begin{bmatrix} 2-\lambda & 24 & 36 \\ 0 & 50-\lambda & 60 \\ 0 & 0 & -10-\lambda \\ \end{bmatrix} $$ $$\begin{align} det & = 0 \\ & = (2-\lambda)((50-\lambda)(-10-\lambda)-(60)(0)) - 24 ((0)(-10-\lambda)-(60)(0))+36((0)(0)-(50-\lambda)(0))\\ & = (2-\lambda)(50-\lambda)(-10-\lambda) \\ \end{align}$$ $$\begin{align} \text{Let } -10-\lambda & = 0 \\ \lambda & = -10 \end{align}$$ $$\begin{align} \text{Let } 2-\lambda & = 0 \\ \lambda & = 2 \end{align}$$ $$\begin{align} \text{Let } 50-\lambda & = 0 \\ \lambda & = 50 \end{align}$$
4. If
$$
\mathbf{u} =
\begin{vmatrix}
x^2 & y^2 & z^2 \\
x & y & z \\
1 & 1 & 1 \\
\end{vmatrix}
$$
then $\mathbf{u}_x + \mathbf{u}_y + \mathbf{u}_z$ is equal to ________.
- 0
- x + y + z
- 1
- 2(x + y + z)
Answer: A
Solution:
Solution:
$$
\mathbf{u}_x = \frac{\partial \mathbf{u}}{\partial x} \text{, }
\mathbf{u}_y = \frac{\partial \mathbf{u}}{\partial y} \text{, }
\mathbf{u}_z = \frac{\partial \mathbf{u}}{\partial z}
$$
$$
\mathbf{u}_x = \frac{\partial \mathbf{u}}{\partial x} =
\begin{vmatrix}
2x & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{vmatrix}
= 0
$$
$$
\mathbf{u}_y = \frac{\partial \mathbf{u}}{\partial y} =
\begin{vmatrix}
0 & 2y & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{vmatrix}
= 0
$$
$$
\mathbf{u}_z = \frac{\partial \mathbf{u}}{\partial z} =
\begin{vmatrix}
0 & 0 & 2z \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{vmatrix}
= 0
$$
Therefore, $\mathbf{u}_x + \mathbf{u}_y + \mathbf{u}_z = 0$.
5. On average, a baseball player gets a hit in one out of three attempts. Assume that the attempts are independent, what is the probability that he gets exactly three hits in six attempts?
- 0.2195
- 0.1097
- 0.01646
- 0.0412
Answer: A
Solution:
Solution:
This is a Binomial Distribution problem where:
- It has repeated trials
- There are two possible outcomes in every trial
- The trials are independent or has constant probability
6. The random variable, X is discrete and uniformly distributed with values 1, 2, 3, 4, 5. Find the variance of X.
- 1
- 3
- 2
- 4
Answer: C
Solution:
Solution:
Variance can be calculated by:
$$\begin{align}
\lambda = \sigma^2 & = \frac{\sum |x-\bar x|^2}{n} \\
\sigma & - \text{population standard deviation}\\
\bar x & - \text{mean}\\
n & - \text{no. of terms}\\
\end{align}$$
Using Casio FX-ES plus scientific calculator, input the following:
[MODE] > [3: STAT] > [1: 1-VAR] $$ \left[ \begin{array}{c|c} &x\\ 1&1\\ 2&2\\ 3&3\\ 4&4\\ 5&5\\ \end{array} \right]$$ [AC] > [SHIFT] > [1 (STAT)] > [4: Var] > [3: $\sigma x$] > [$x^2$]
Variance, $\lambda = \sigma x^2 = 2$.
[MODE] > [3: STAT] > [1: 1-VAR] $$ \left[ \begin{array}{c|c} &x\\ 1&1\\ 2&2\\ 3&3\\ 4&4\\ 5&5\\ \end{array} \right]$$ [AC] > [SHIFT] > [1 (STAT)] > [4: Var] > [3: $\sigma x$] > [$x^2$]
Variance, $\lambda = \sigma x^2 = 2$.
7. The solution set for the inequality
$$
\frac{1}{x-2} \lt \frac{1}{x+3}
$$
- (-3, -2)
- [-3, -2]
- (-3, 2)
- (2, 3)
Answer: C
Solution:
Solution:
For the inequality to be true, it can be assumed that $\frac{1}{x-2}$ equates to a negative value and $\frac{1}{x+3}$ equates to a positive value.
$$\begin{align}
\frac{1}{x-2} & = (-) \\
x-2 & = (-) \\
x-2 & \lt 0 \\
x & \lt 2 \\
\frac{1}{x+3} & = (+) \\
x+3 & = (+) \\
x+3 & \gt 0 \\
x & \gt -3
\end{align}$$
Since $x\lt2$ and $x\gt-3$, then $-3 \lt x \lt 2$ or $(-3,2)$
8. What is the indicial equation for the ordinary differential equation $x^2y"+4xy'+4y=0$?
- $r^2+2r+4=0$
- $r^2+4r+4$
- $r^2+3r+4=0$
- $r^2+4xr+4=0$
Answer: C
Solution:
Solution:
The differential equation is an Euler-Cauchy Equation:
$$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$$
To get the auxiliary (indicial) equation of the Euler-Cauchy equation, let:
$$\begin{align}
y & = x^m \\
y' & = mx^{m-1} \\
y" & = m(m-1)x^{m-2} \\
\end{align}$$
Solving,
$$
x^2y"+4xy'+4y = 0 \\
x^2(m)(m-1)(x^{m-2})+4xmx^{m-1}+4x^m = 0\\
m^2-m+4m+4 = 0\\
m^2+3m+4 = 0\\
$$
9.
$$ \lim_{n \to \infty} \prod_{m=2}^n \left( 1 - \frac1m \right) $$ is equal to
- 1
- e-1
- e
- 0
Answer: D
Solution:
Solution:
$$\begin{align}
& \lim_{n \to \infty} \prod_{m=2}^n \left( 1 - \frac1m \right) \\
& = (1-\frac12)(1-\frac13)(1-\frac14)(1-\frac15) \ldots (1-\frac 1 \infty) \\
& = 0
\end{align}$$
10. Which of the following complex numbers is equal to $(1+i)^{4/3}$?
- $ 2^{2/3}(-\frac12 + i\frac{\sqrt3}{2}) $
- $ 2^{2/3}(\sqrt{\frac23} + i\frac1{\sqrt3}) $
- $ 2^{2/3}(\sqrt{\frac32} - i \frac12) $
- $ 2^{2/3}(\frac12 + i \frac{\sqrt3}2 ) $
Answer: D
Solution:
Solution:
First, simplify:
$$\begin{align}
(1+i)^{4/3} & = (1+i)^{4 \cdot 1/3} \\
& = [ (1+i)^2(1+1)^2]^{1/3} \\
& = (-4)^{1/3}
\end{align}$$
To solve for the nth root of complex numbers $(r \lt \theta)^{a/b}$:
$$
\text{nth root } = r^{a/b} \lt \frac{ a\theta + 2\pi k}b
$$
Solving,
$$\begin{align}
(-4)^{1/3} & \to r \lt \theta \\
& = ( 4 \lt \pi )^{1/3} \\
\text{nth root} & = 4^{1/3} \lt \frac{ \pi + 2 \pi k}3
\end{align}$$
@ k = 0,
principal root = 0.7937 + 1.3747i
@ k = 1,
2nd root = -1.5874
@ k = 2,
3rd root = 0.7937 - 1.3747i
From the principal root,
$$ 0.7937 + 1.3747i = 2^{2/3} \left(\frac12 + \frac {\sqrt3}2 i \right)$$
principal root = 0.7937 + 1.3747i
@ k = 1,
2nd root = -1.5874
@ k = 2,
3rd root = 0.7937 - 1.3747i
From the principal root,
$$ 0.7937 + 1.3747i = 2^{2/3} \left(\frac12 + \frac {\sqrt3}2 i \right)$$
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