nyuuzyou/wikis huggingface dataset:
Simultaneous Homogeneous Linear First Order ODEs/Examples/y' - 3y + 2z = 0, y' + 4y - z = 0

== Theorem ==
Consider the system of linear first order ordinary differential equations with constant coefficients:
The general solution to $(1)$ and $(2)$ consists of the linear combinations of the following:
== Proof ==
Using the technique of Solution to Simultaneous Homogeneous Linear First Order ODEs with Constant Coefficients, we calculate the roots of the quadratic equation:
:$\paren {k + a} \paren {k + d} - b c = 0$
where:
That is:
:$\paren {k - 3} \paren {k - 1} - 8 = 0$
or:
:$k^2 - 4 k - 5 = 0$
This has roots:
We also obtain:
When $k = 5$ we get that $A + B = 0$.
When $k = -1$ we get that $2 A - B = 0$.
This provides us with the solutions:
or:
From these, the general solution is constructed:
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.2$ First order systems: Example $2$
Category:Examples of Systems of Differential Equations
Category:Examples of Linear First Order ODEs
Simultaneous Linear Equations/Examples

== Examples of Simultaneous Linear Equations ==
=== Arbitrary System $1$ ===
=== Arbitrary System $2$ ===
=== Arbitrary System $3$ ===
=== Arbitrary System $4$ ===
=== Arbitrary System $5$ ===
=== Arbitrary System $6$ ===
=== Arbitrary System $7$ ===
=== Arbitrary System $8$ ===
=== Arbitrary System $9$ ===
=== Arbitrary System $10$ ===
=== Arbitrary System $11$ ===
=== Arbitrary System $12$ ===
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 1

== Example of Simultaneous Linear Equations ==
The system of simultaneous linear equations:
has as its solution set:
== Proof ==
Subtract $2 \times$ equation $(1)$ from equation $(2)$.
Subtract $4 \times$ equation $(1)$ from equation $(3)$.
This gives us:
Divide equation $(2')$ by $3$ to get $(2)$.
Add $2 \times$ equation $(2)$ to equation $(1)$.
Subtract $9 \times$ equation $(2)$ from equation $(3')$.
This gives us:
From equation $(3)$ we have directly that $x_3 = \dfrac 3 2$.
Substituting for $x_3$ in equation $(1')$ and equation $(2)$ gives the single solution:
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction: Example $\text {(i)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 10

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ has as its solution set:
where $z$ and $w$ can be any numbers.
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & 2 & -1 & 1 \\ 2 & 1 & 1 & 0 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \\ w \end {pmatrix} = \begin {pmatrix} 1 \\ 2 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {cccc|c} 1 & 2 & -1 & 1 & 1 \\ 2 & 1 & 1 & 0 & 2 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 - 2 r_1$
Hence:
:$\begin {pmatrix} \mathbf A_1 & \mathbf b_1 \end {pmatrix} = \paren {\begin {array} {cccc|c}
1 & 2 & -1 & 1 & 1 \\
0 & -3 & 3 & -2 & 0 \\
\end {array} }$
$e_2 := r_2 \to -\dfrac 1 3 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {cccc|c}
1 & 2 & -1 & 1 & 1 \\
0 & 1 & -1 & 2/3 & 0 \\
\end {array} }$
$e_3 := r_1 \to r_1 - 2 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {cccc|c}
1 & 0 & 1 & -1/3 & 1 \\
0 & 1 & -1 & 2/3 & 0 \\
\end {array} }$
$e_4 := r_1 \to 3 r_1$
$e_5 := r_2 \to 3 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_5 & \mathbf b_5 \end {pmatrix} = \paren {\begin {array} {cccc|c}
3 & 0 & 1 & -1 & 3 \\
0 & 3 & -3 & 2 & 0 \\
\end {array} }$
Thus:
and it cannot be simplified further.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.12 \ \text {(a)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 11

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ has the single solution:
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & -1 & 5 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 1 \\ 2 \\ 1 \\ 1 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & 1 & 2 \\ 1 & 0 & 2 & 1 \\ 1 & -1 & 5 & 1 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_3 \to r_3 - r_1$
$e_2 := r_4 \to r_4 - r_1$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 1 & -1 & 1 \\
0 & 1 & 1 & 2 \\
0 & -1 & 3 & 0 \\
0 & -2 & 6 & 0
\end {array} }$
$e_3 := r_1 \to r_1 - r_2$
$e_4 := r_3 \to r_3 + r_2$
$e_5 := r_4 \to r_4 + 2 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_5 & \mathbf b_5 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & -2 & -1 \\
0 & 1 & 1 & 2 \\
0 & 0 & 4 & 2 \\
0 & 0 & 8 & 4
\end {array} }$
$e_6 := r_3 \to \dfrac 1 4 r_3$
Hence:
:$\begin {pmatrix} \mathbf A_6 & \mathbf b_6 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & -2 & -1 \\
0 & 1 & 1 & 2 \\
0 & 0 & 1 & 1/2 \\
0 & 0 & 8 & 4
\end {array} }$
$e_7 := r_1 \to r_1 + 2 r_3$
$e_8 := r_2 \to r_2 - r_3$
$e_9 := r_4 \to r_4 - 8 r_3$
Hence:
:$\begin {pmatrix} \mathbf A_9 & \mathbf b_9 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 3/2 \\
0 & 0 & 1 & 1/2 \\
0 & 0 & 0 & 0
\end {array} }$
Thus:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.12 \ \text {(b)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 12

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ is inconsistent and so has no solutions.
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & 2 & 3 & 1 \\ -1 & 1 & 1 & -1 \\ 1 & 5 & 7 & 1 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \\ w \end {pmatrix} = \begin {pmatrix} -1 \\ 2 \\ 1 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {cccc|c} 1 & 2 & 3 & 1 & -1 \\ -1 & 1 & 1 & -1 & 2 \\ 1 & 5 & 7 & 1 & 1 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 + r_1$
$e_2 := r_3 \to r_3 - r_1$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {cccc|c}
 1 & 2 & 3 & 1 & -1 \\
 0 & 3 & 4 & 0 & 1 \\
 0 & 3 & 4 & 0 & 2
\end {array} }$
$e_3 := r_3 \to r_3 - r_2$
Hence:
:$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {cccc|c}
 1 & 2 & 3 & 1 & -1 \\
 0 & 3 & 4 & 0 & 1 \\
 0 & 0 & 0 & 0 & 1
\end {array} }$
The bottom line of this augmented matrix leads to the false statement $0 = 1$.
It follows that this system of simultaneous linear equations is inconsistent.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.12 \ \text {(c)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 2

== Example of Simultaneous Linear Equations ==
The system of simultaneous linear equations:
has no solutions.
== Proof ==
 $(1)$ and $(2)$ together have a solution.
Subtract $2 \times$ equation $(1)$ from equation $(2)$.
which is an inconsistency.
Hence there is no such solution.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction: Example $\text {(ii)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 3

== Example of Simultaneous Linear Equations ==
The system of simultaneous linear equations:
has as its solution set:
where $t$ is any number.
== Proof ==
Subtract $2 \times$ equation $(1)$ from equation $(2)$.
This gives us:
Divide equation $(2')$ by $3$ to get $(2)$.
Add $2 \times$ equation $(2)$ to equation $(1)$.
This gives us:
That is:
Whatever value is assigned to $x_3$, the system of simultaneous linear equations will be fulfilled.
Thus we set $x_3$ to any arbitrary $t$ and the result follows.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction: Example $\text {(iii)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 6

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ has as its solution set:
where $z$ can be any number.
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & 1 & 2 \\ -1 & 0 & 1 \\ -1 & 1 & 4 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} -1 \\ -1 \\ -3 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 1 & 2 & -1 \\ -1 & 0 & 1 & -1 \\ -1 & 1 & 4 & -3 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 + r_1$
$e_2 := r_3 \to r_3 + r_1$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
 1 & 1 & 2 & -1 \\
 0 & 1 & 3 & -2 \\
 0 & 2 & 6 & -4 \\
\end {array} }$
$e_3 := r_3 \to r_3 - 2 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
 1 & 1 & 2 & -1 \\
 0 & 1 & 3 & -2 \\
 0 & 0 & 0 & 0 \\
\end {array} }$
$e_4 := r_1 \to r_1 - r_2$
Hence:
:$\begin {pmatrix} \mathbf A_4 & \mathbf b_4 \end {pmatrix} = \paren {\begin {array} {ccc|ccc}
 1 & 0 & -1 & 1 \\
 0 & 1 & 3 & -2 \\
 0 & 0 & 0 & 0 \\
\end {array} }$
Thus $z$ is not restricted, and can be anything, leaving us with:
as the solution set:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 6/Mistake

== Source Work ==
:
:Part $\text I$: Matrices and vector spaces
::$1$ Matrices
:::$1.5$ Row and column operations
::::Solving linear equations
== Mistake ==
To solve:
:$\begin {array} {rcrcrcr}
 x & + & y & + & 2 z & = & -1 \\
-x & + & & & z & = & -1 \\
-x & + & y & + & 4 z & = & 3 \\
\end {array}$,
first put the equation in matrix form
:$\paren {\begin {array} {rrr} 1 & 1 & 2 \\ -1 & 0 & 1 \\ -1 & 1 & 4 \end {array} } \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \paren {\begin {array} {r} -1 \\ -1 \\ 3 \end {array} }$
and then put the augmented matrix formed from the matrix on the left with the column vector on the right into echelon form:
:$\paren {\begin {array} {rrr|r}
 1 & 1 & 2 & -1 \\
-1 & 0 & 1 & -1 \\
-1 & 1 & 4 & 3 \end {array} } \to \paren {\begin {array} {rrr|r}
 1 & 1 & 2 & -1 \\
 0 & 1 & 3 & -2 \\
 0 & 2 & 6 & -4 \end {array} } \to \paren {\begin {array} {rrr|r}
 1 & 1 & 2 & -1 \\
 0 & 1 & 3 & -2 \\
 0 & 0 & 0 & 0 \end {array} }$.
== Correction ==
That first transformation should be:
$\paren {\begin {array} {rrr|r}
 1 & 1 & 2 & -1 \\
-1 & 0 & 1 & -1 \\
-1 & 1 & 4 & 3 \end {array} } \to \paren {\begin {array} {rrr|r}
 1 & 1 & 2 & -1 \\
 0 & 1 & 3 & -2 \\
 0 & 2 & 6 & 2 \end {array} }$
which leads to:
:$\paren {\begin {array} {rrr|r}
 1 & 1 & 2 & -1 \\
 0 & 1 & 3 & -2 \\
 0 & 0 & 0 & 6 \end {array} }$
Hence from the resulting $0 = 6$ it is seen that the initial system of simultaneous linear equations has no solutions.
The original system of simultaneous linear equations can be amended to:
:$\begin {array} {rcrcrcr}
 x & + & y & + & 2 z & = & -1 \\
-x & + & & & z & = & -1 \\
-x & + & y & + & 4 z & = & -3 \\
\end {array}$
from which the solution is derived in Simultaneous Linear Equations: Arbitrary System $6$.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Mistakes/Linear Algebra
Simultaneous Linear Equations/Examples/Arbitrary System 7

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ has as its solution set:
where $z$ can be any number.
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & 2 & 1 \\ -1 & 2 & 0 \\ 5 & -2 & 2 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 2 \\ -1 \\ 7 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 2 \\ -1 & 2 & 0 & -1 \\ 5 & -2 & 2 & 7 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 + r_1$
$e_2 := r_3 \to r_3 - 5 r_1$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 2 & 1 & 2 \\
0 & 4 & 1 & 1 \\
0 & -12 & -3 & -3 \\
\end {array} }$
$e_3 := r_3 \to r_3 + 3 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_3 & \mathbf b_3 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 2 & 1 & 2 \\
0 & 4 & 1 & 1 \\
0 & 0 & 0 & 0 \\
\end {array} }$
$e_4 := r_2 \to \dfrac 1 4 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_4 & \mathbf b_4 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 2 & 1 & 2 \\
0 & 1 & 1/4 & 1/4 \\
0 & 0 & 0 & 0 \\
\end {array} }$
$e_5 := r_1 \to r_1 - 2 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_5 & \mathbf b_5 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & 1/2 & 3/2 \\
0 & 1 & 1/4 & 1/4 \\
 0 & 0 & 0 & 0 \\
\end {array} }$
$e_6 := r_1 \to 2 r_1$
$e_7 := r_5 \to 4 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_7 & \mathbf b_7 \end {pmatrix} = \paren {\begin {array} {ccc|c}
2 & 0 & 1 & 3 \\
0 & 4 & 1 & 1 \\
0 & 0 & 0 & 0 \\
\end {array} }$
Thus $z$ is not restricted, and can be anything, leaving us with:
as the solution set:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.11 \ \text {(a)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 8

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ is inconsistent and so has no solutions.
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & -1 & -1 \\ 2 & -1 & 0 \\ 2 & 0 & 2 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 1 \\ 1 \\ 1 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & -1 & -1 & 1 \\ 2 & -1 & 0 & 1 \\ 2 & 0 & 2 & 1 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 - 2 r_1$
$e_2 := r_3 \to r_3 - 2 r_1$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & -1 & -1 & 1 \\
0 & 1 & 2 & -1 \\
0 & 2 & 4 & -1 \\
\end {array} }$
$e_3 := r_1 \to r_1 + r_2$
$e_4 := r_3 \to r_3 - 2 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_4 & \mathbf b_4 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & 1 & 0 \\
0 & 1 & 2 & -1 \\
0 & 0 & 0 & 1 \\
\end {array} }$
The bottom line of this augmented matrix leads to the false statement $0 = 1$.
It follows that this system of simultaneous linear equations is inconsistent.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.11 \ \text {(b)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations/Examples/Arbitrary System 9

== Example of Simultaneous Linear Equations ==
Let $S$ denote the system of simultaneous linear equations:
$S$ has the single solution:
== Proof ==
We express $S$ in matrix representation:
:$\begin {pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ -1 & 0 & 1 \end {pmatrix} \begin {pmatrix} x \\ y \\ z \end {pmatrix} = \begin {pmatrix} 1 \\ 1 \\ 1 \end {pmatrix}$
and consider the augmented matrix:
:$\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix} = \paren {\begin {array} {ccc|c} 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ -1 & 0 & 1 & 1 \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf b_k \end {pmatrix}$.
$e_1 := r_2 \to r_2 - r_1$
$e_2 := r_3 \to r_3 + r_1$
Hence:
:$\begin {pmatrix} \mathbf A_2 & \mathbf b_2 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 2 & 1 & 1 \\
0 & -1 & 0 & 0 \\
0 & 2 & 2 & 2 \\
\end {array} }$
$e_3 := r_1 \to r_1 + 2 r_2$
$e_4 := r_3 \to r_3 + 2 r_2$
Hence:
:$\begin {pmatrix} \mathbf A_4 & \mathbf b_4 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & 1 & 1 \\
0 & -1 & 0 & 0 \\
0 & 0 & 2 & 2 \\
\end {array} }$
$e_5 := r_2 \to -r_2$
$e_6 := r_3 \to \dfrac 1 2 r_3$
Hence:
:$\begin {pmatrix} \mathbf A_6 & \mathbf b_6 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
\end {array} }$
$e_7 := r_1 \to r_1 - r_3$
Hence:
:$\begin {pmatrix} \mathbf A_7 & \mathbf b_7 \end {pmatrix} = \paren {\begin {array} {ccc|c}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
\end {array} }$
Thus:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.11 \ \text {(c)}$
Category:Examples of Simultaneous Linear Equations
Simultaneous Linear Equations has Unique Solution iff Rank of Matrix of Coefficients equals Number of Columns

== Theorem ==
Let $S$ be a system of $m$ simultaneous linear equations in $n$ variables:
:$\ds \forall i \in \set {1, 2, \ldots, m} : \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
Let $S$ be expressed in matrix form as:
:$\mathbf A \mathbf x = \mathbf b$
where:
:$\mathbf A = \begin {pmatrix}
\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \\
\end {pmatrix}$, $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$, $\mathbf b = \begin {pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {pmatrix}$
Then $S$ has exactly one solution :
:$\map \rho {\mathbf A} = n$
where $\map \rho {\mathbf A}$ denotes the rank of $\mathbf A$.
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Simultaneous Linear Equations
Category:Rank of Matrix
Simultaneous Linear Equations have Solution iff Ranks of Matrix of Coefficients and Augmented Matrix are Equal

== Theorem ==
Let $S$ be a system of simultaneous linear equations:
:$\ds \forall i \in \set {1, 2, \ldots, m} : \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
Let $S$ be expressed in matrix form as:
:$\mathbf {A x} = \mathbf b$
where:
:$\mathbf A = \begin {pmatrix}
\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \\
\end {pmatrix}$, $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$, $\mathbf b = \begin {pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {pmatrix}$
Then $S$ has at least one solution :
:$\map \rho {\mathbf A} = \map \rho {\begin {array} {c|c} \mathbf A & \mathbf b \end {array} }$
where:
:$\map \rho {\mathbf A}$ denotes the rank of $\mathbf A$
:$\paren {\begin {array} {c|c} \mathbf A & \mathbf b \end {array} }$ denotes the augmented matrix of $S$.
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Simultaneous Linear Equations
Category:Rank of Matrix
Sine Function in terms of Hypergeometric Function

== Theorem ==
:$\sin x = x \paren {\map { {}_0 \operatorname F_1} {\dfrac 3 2; \dfrac {-x^2} 4} }$
where:
:$x$ is a real number with $x \in \R$
:$\sin$ denotes the sine function
:$\map { {}_0 \operatorname F_1} {a; x}$ is the generalized hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac 1 { a^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Cosine Function in terms of Hypergeometric Function
* Hyperbolic Cosine Function in terms of Hypergeometric Function
* Hyperbolic Sine Function in terms of Hypergeometric Function
Category:Hypergeometric Functions
Category:Sine Function
Sine Function is Absolutely Convergent

== Theorem ==
The real sine function $\sin: \R \to \R$ is absolutely convergent.
=== Complex Case ===
== Proof ==
Recall the definition of the sine function:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
For:
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
to be absolutely convergent we want:
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } = \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!}$
to be convergent.
But:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!}$
is just the terms of:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!}$
for odd $n$.
Thus:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n + 1} } {\paren {2 n + 1}!} < \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!}$
But:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!} = \exp \size x$
from the Taylor Series Expansion for Exponential Function of $\size x$, which converges for all $x \in \R$.
The result follows from the Squeeze Theorem.
== Also see ==
* Cosine Function is Absolutely Convergent
== Sources ==
* : $\S 16.2$
Category:Sine Function
Category:Absolute Convergence
Sine Function is Absolutely Convergent/Complex Case/Proof 1

== Theorem ==
== Proof ==
The definition of the complex sine function is:
:$\ds \forall z \in \C: \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$
By definition of absolutely convergent complex series, we must show that the power series
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} }$
is convergent.
We have
The result follows from Squeeze Theorem for Complex Sequences.
== Sources ==
* $\S 1.5$
Category:Complex Sine Function is Absolutely Convergent