The problem consists of three parts: a. Given a quadratic equation $2x^2 - 4x + 5 = 0$ with roots $\alpha$ and $\beta$, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$. b. Express the complex numbers $\frac{1+i}{1-2i}$ and $\frac{2+3i}{5+i}$ in the form $a+ib$, where $a$ and $b$ are real numbers. c. Given two matrices $A = \begin{pmatrix} 1 & 2 & 1 \\ 4 & 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 1 & -2 \\ -4 & 5 & 2 \end{pmatrix}$, find $A-B$ and $B^T B$.

1. Problem Description

The problem consists of three parts:

a. Given a quadratic equation $2x^2 - 4x + 5 = 0$ with roots $\alpha$ and $\beta$, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

b. Express the complex numbers $\frac{1+i}{1-2i}$ and $\frac{2+3i}{5+i}$ in the form $a+ib$, where $a$ and $b$ are real numbers.

c. Given two matrices $A = \begin{pmatrix} 1 & 2 & 1 \\ 4 & 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 1 & -2 \\ -4 & 5 & 2 \end{pmatrix}$, find $A-B$ and $B^T B$.

2. Solution Steps

a.

We are given the quadratic equation $2x^2 - 4x + 5 = 0$ with roots $\alpha$ and $\beta$. We want to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

We can rewrite $\frac{1}{\alpha} + \frac{1}{\beta}$ as $\frac{\alpha + \beta}{\alpha \beta}$.

From the quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}$.

In our case, $a=2$, $b=-4$, and $c=5$.

Therefore, $\alpha + \beta = -\frac{-4}{2} = 2$ and $\alpha \beta = \frac{5}{2}$.

So, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{2}{\frac{5}{2}} = \frac{2 \cdot 2}{5} = \frac{4}{5}$.

b.

i. We want to express $\frac{1+i}{1-2i}$ in the form $a+ib$.

Multiply the numerator and denominator by the conjugate of the denominator:

$\frac{1+i}{1-2i} = \frac{(1+i)(1+2i)}{(1-2i)(1+2i)} = \frac{1 + 2i + i + 2i^2}{1 - (2i)^2} = \frac{1 + 3i - 2}{1 - (-4)} = \frac{-1 + 3i}{5} = -\frac{1}{5} + \frac{3}{5}i$.

ii. We want to express $\frac{2+3i}{5+i}$ in the form $a+ib$.

Multiply the numerator and denominator by the conjugate of the denominator:

$\frac{2+3i}{5+i} = \frac{(2+3i)(5-i)}{(5+i)(5-i)} = \frac{10 - 2i + 15i - 3i^2}{25 - i^2} = \frac{10 + 13i + 3}{25 + 1} = \frac{13 + 13i}{26} = \frac{13}{26} + \frac{13}{26}i = \frac{1}{2} + \frac{1}{2}i$.

c.

i. We want to find $A - B$, where $A = \begin{pmatrix} 1 & 2 & 1 \\ 4 & 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 1 & -2 \\ -4 & 5 & 2 \end{pmatrix}$.

$A - B = \begin{pmatrix} 1-3 & 2-1 & 1-(-2) \\ 4-(-4) & 0-5 & 2-2 \end{pmatrix} = \begin{pmatrix} -2 & 1 & 3 \\ 8 & -5 & 0 \end{pmatrix}$.

ii. We want to find $B^T B$.

First, find $B^T$, the transpose of $B$:

$B^T = \begin{pmatrix} 3 & -4 \\ 1 & 5 \\ -2 & 2 \end{pmatrix}$.

Now, compute $B^T B$:

$B^T B = \begin{pmatrix} 3 & -4 \\ 1 & 5 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 & -2 \\ -4 & 5 & 2 \end{pmatrix} = \begin{pmatrix} 3(3) + (-4)(-4) & 3(1) + (-4)(5) & 3(-2) + (-4)(2) \\ 1(3) + 5(-4) & 1(1) + 5(5) & 1(-2) + 5(2) \\ -2(3) + 2(-4) & -2(1) + 2(5) & -2(-2) + 2(2) \end{pmatrix} = \begin{pmatrix} 9 + 16 & 3 - 20 & -6 - 8 \\ 3 - 20 & 1 + 25 & -2 + 10 \\ -6 - 8 & -2 + 10 & 4 + 4 \end{pmatrix} = \begin{pmatrix} 25 & -17 & -14 \\ -17 & 26 & 8 \\ -14 & 8 & 8 \end{pmatrix}$.

3. Final Answer

a. $\frac{4}{5}$

b. i. $-\frac{1}{5} + \frac{3}{5}i$

ii. $\frac{1}{2} + \frac{1}{2}i$

c. i. $\begin{pmatrix} -2 & 1 & 3 \\ 8 & -5 & 0 \end{pmatrix}$

ii. $\begin{pmatrix} 25 & -17 & -14 \\ -17 & 26 & 8 \\ -14 & 8 & 8 \end{pmatrix}$