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Consider the following vectors with entries in $\R$: $\mathbf x=\xvector{n}$ Consider the following vectors with entries in $\Z$: $\mathbf w=\avector{w}{8}$ $\Intersect{1}{n}$ $\summation{1}{n}$ $\product{1}{n}$ $\trans$ $\Abs{A}{b}$ $\VofF{a}{F}$ $\Vfgpoly$ $\Vfghpoly$ $\Vfpoly{f}{a}$ $\matAn{A}{a}$ $\matAn{B}{b}$ $\matABn$ $A = B =\matn{a}$ $\mataplhaAn{\alpha}{a}$ $\mataplhaAn{\beta}{b}$ $\polya{b}$ $\polyab$ $\polyabc$ $\polyabcc$\\ $\polyzero$ $\polyazero$ $\polyazeroo$ $\polyaalpha{\alpha}{a}$ $\polyaalpha{\beta}{b}$ $\polyabalpha{\alpha}$ $\polyabalpha{\beta}$ $\polyabalphaa{\eta}$ $\polyaalpha{(\alpha+\beta)}{a}$ $\polyax{(a(b\alpha_k))}$

System of linear equations

Two equations in two unknowns

The following is a system of two linear equations in the two unknowns $x$ and $y$:

$x − y = 1$

$3x + 4y = 6$.

A solution to the system is a pair $(x, y)$ of numbers that satisfy both equations.

Each of these equations represents a line in the xy-plane, so a solution is a point in the intersection of the lines.

A linear equation is an equation of the form $a_1x_1 + a_2x_2 + · · · a_nx_n = b$, where $a_1, a_2, . . . , a_n$  and $b$ are numbers.

A system of linear equation has one of the following:

  • a unique solution,
  • infinitely many solutions,
  • no solution
Problem 1: (Unique solution) Solve the following system

$x − y = 1$

$3x + 4y = 6$.

Solution:

Given equations:

$\int_{a}^{b} \! f(x)\,\mathrm{d}x$

$n=\underbrace{1+1+\cdots+1}_{\text{$n$ times}},$

$\left\|\frac{a}{b}\right \|$

$\|\mathbf{v}\|$

generates
\[\begin{array}{c}
a_1x+b_1y+c_1z=d_1 \\
a_2x+b_2y+c_2z=d_2 \\
a_3x+b_3y+c_3z=d_3
\end{array}.
\]

\[
A=\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{2}{3} &\frac{-1}{3} &\frac{-1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{-2}{3}
\end{bmatrix}
\]

\[
A=\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{2}{3} &\frac{-1}{3} &\frac{-1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{-2}{3}
\end{bmatrix}
\]

$f(x)=\sum _{}^{}{\alpha }_{k}{x}^{k}\:and\:g(x)=\sum _{}^{}{\beta }_{k}{x}^{k}$

\begin{array}{ |c|c|c| }\hline a & a^2 \pmod{5} & 2a^2 \pmod{5}\\ \hline 0 & 0 & 0\\ 1 & 1 & 2\\ 2 & 4 & 3\\ 3 & 4 & 3\\ 4 & 1 & 2\\ \hline \end{array}

\begin{align*} f(ab)&=(ab)^2 && (\text{by definition of $f$})\\ &=(ab)(ab)\\ &=a^2 b^2 && (\text{since $G$ is abelian})\\ &=f(a)f(b) && (\text{by definition of $f$}). \end{align*}

\begin{align*} \det(A)&=1+(-1)^{n+1} \\ &= \begin{cases} 2 & \text{ if } n \text{ is odd}\\ 0 & \text{ if } n \text{ is even}. \end{cases} \end{align*}

\[ \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 &1 \\ 0 & 1 & 2 & 3 & 5 \\ 0 & -2 & 0 & -2 & 2 \\ 0 & 1 & -2 & 3 & 1 \\ \end{array}\right] \xrightarrow{\substack{R_1-R_2 \ R_3-R_2 \ R_4-R_2}} \left[\begin{array}{rrrr|r} 1 & 0& -1 & -2 &-4 \\ 0 & 1 & 2 & 3 & 5 \\ 0 & 0 & 4 & 4 & 12 \\ 0 & 0 & -4 & 0 & -4 \\ \end{array}\right] \xrightarrow[\frac{-1}{4}R_4]{\frac{1}{4}R_3} \left[\begin{array}{rrrr|r} 1 & 0& -1 & -2 &-4 \\ 0 & 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 & 3 \\ 0 & 0 & 1 & 0& 1 \\ \end{array}\right] \]

\begin{align*} \left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 1 \\ 0 & 1 & 3 & 0 & -1 & 2 \\ 2 & 0 & -2 & 1 & -3 & 0 \\ \end{array}\right] \xrightarrow{R_3-2R_1} \left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 1 \\ 0 & 1 & 3 & 0 & -1 & 2 \\ 0 & 0 & 0 & 1 & 1 & -2 \\ \end{array}\right]. \end{align*}

\begin{align*} \mathbf{x}&=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}=\begin{bmatrix} x_3+2x_5+1 \\ -3x_3+x_5+2 \\ x_3 \\ -x_5-2 \\ x_5 \end{bmatrix}\\[10pt] &=x_3\begin{bmatrix} 1 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} +\begin{bmatrix} 1 \\ 2 \\ 0 \\ -2 \\ 0 \end{bmatrix}. \end{align*}

1. What is the full form of the RAM?





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Consider the following vectors with entries in $\R$: $\mathbf x=\xvector{n}$ Consider the following vectors with entries in $\Z$: $\mathbf w=\avector{w}{8}$ $\Intersect{1}{n}$ $\summation{1}{n}$ $\product{1}{n}$ $\trans$ $\Abs{A}{b}$

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