nyuuzyou/wikis huggingface dataset:
Gaussian Elimination/Examples/Arbitrary Matrix 1

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
0 & 0 & 5 & 35 & -24 & 1 \\
0 & 2 & 1 & -1 & 1 & 0 \\
0 & 3 & 2 & 2 & -1 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 5 & 3 & 1 & 0 & 1
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
0 & 1 & 0 & -4 & 0 & 26 \\
0 & 0 & 1 & 7 & 0 & -43 \\
0 & 0 & 0 & 0 & 1 & -9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0
\end {bmatrix}$
== Proof ==
$(1): \quad$ Use the elementary row operation $r_4 \leftrightarrow r_5$ to move the zero row to the bottom
$(2): \quad$ Use the elementary row operation $r_2 \to \dfrac {r_2} 2$ to put a leading $1$ in row $2$
$(3): \quad$ Use the elementary row operation $r_2 \leftrightarrow r_1$ to put the row with the leftmost leading $1$ at the top.
Hence we get matrix $\mathbf A_1$:
:$\mathbf A_1 = \begin {bmatrix}
0 & 1 & \dfrac 1 2 & -\dfrac 1 2 & \dfrac 1 2 & 0 \\
0 & 0 & 5 & 35 & -24 & 1 \\
0 & 3 & 2 & 2 & -1 & 1 \\
0 & 5 & 3 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 
\end {bmatrix}$
$(4): \quad$ Use the following elementary row operations:
:$r_3 \leftrightarrow r_3 - 3 r_1$
:$r_4 \leftrightarrow r_4 - 5 r_1$
to clear column $2$.
Hence we get matrix $\mathbf A_2$:
:$\mathbf A_2 = \begin {bmatrix}
0 & 1 & \dfrac 1 2 & -\dfrac 1 2 & \dfrac 1 2 & 0 \\
0 & 0 & 5 & 35 & -24 & 1 \\
0 & 0 & \dfrac 1 2 & \dfrac 7 2 & -\dfrac 5 2 & 1 \\
0 & 0 & \dfrac 1 2 & \dfrac 7 2 & -\dfrac 5 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 
\end {bmatrix}$
$(5): \quad$ Use the following elementary row operations:
:$r_2 \leftrightarrow r_3$
:$r_1 \to r_1 - r_2$
:$r_4 \to r_4 - r_2$
:$r_3 \to r_3 - 10 r_1$
:$r_2 \to 2 r_2$
Hence we get matrix $\mathbf A_3$:
:$\mathbf A_3 = \begin {bmatrix}
0 & 1 & 0 & -4 & 3 & 1 \\
0 & 0 & 1 & 7 & -5 & 2 \\
0 & 0 & 0 & 0 & 1 & -9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 
\end {bmatrix}$
$(6): \quad$ Use the following elementary row operations:
:$r_1 \to r_1 - 3 r_3$
:$r_2 \to r_2 + 5 r_3$
Hence we get matrix $\mathbf A_3$:
:$\mathbf A_3 = \begin {bmatrix}
0 & 1 & 0 & -4 & 0 & 26 \\
0 & 0 & 1 & 7 & 0 & 43 \\
0 & 0 & 0 & 0 & 1 & -9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 
\end {bmatrix}$
Different sequences of elementary row operations can be used, but it is necessary to be precise about the order in which they are applied.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 2

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
1 & -1 & 2 & 1 \\
2 & 1 & -1 & 1 \\
1 & -2 & 1 & 1 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
1 & 0 & 0 & \dfrac 5 8 \\
0 & 1 & 0 & -\dfrac 1 8 \\
0 & 0 & 1 & \dfrac 1 8 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \to r_2 - 2 r_1$
$e_2 := r_3 \to r_3 - r_1$
Hence:
:$\mathbf A_2 = \begin {bmatrix}
1 & -1 & 2 & 1 \\
0 & 3 & -5 & -1 \\
0 & -1 & -1 & 0 \\
\end {bmatrix}$
$e_3 := r_2 \to \dfrac {r_2} 3$
:$\mathbf A_3 = \begin {bmatrix}
1 & -1 & 2 & 1 \\
0 & 1 & -\dfrac 5 3 & -\dfrac 1 3 \\
0 & -1 & -1 & 0 \\
\end {bmatrix}$
$e_4 := r_3 \to r_3 + r_2$
:$\mathbf A_4 = \begin {bmatrix}
1 & -1 & 2 & 1 \\
0 & 1 & -\dfrac 5 3 & -\dfrac 1 3 \\
0 & 0 & -\dfrac 8 3 & -\dfrac 1 3 \\
\end {bmatrix}$
$e_5 := r_3 \to -\dfrac 3 8 r_3$
:$\mathbf A_5 = \begin {bmatrix}
1 & -1 & 2 & 1 \\
0 & 1 & -\dfrac 5 3 & -\dfrac 1 3 \\
0 & 0 & 1 & \dfrac 1 8 \\
\end {bmatrix}$
It may be noted that $\mathbf A_5$ is in echelon form.
It remains to convert it to reduced echelon form.
$e_6 := r_1 \to r_1 + r_2$
:$\mathbf A_6 = \begin {bmatrix}
1 & 0 & \dfrac 1 3 & \dfrac 2 3 \\
0 & 1 & -\dfrac 5 3 & -\dfrac 1 3 \\
0 & 0 & 1 & \dfrac 1 8 \\
\end {bmatrix}$
$e_7 := r_2 \to r_2 + \dfrac 5 3 r_3$
:$\mathbf A_7 = \begin {bmatrix}
1 & 0 & \dfrac 1 3 & \dfrac 2 3 \\
0 & 1 & 0 & -\dfrac 1 8 \\
0 & 0 & 1 & \dfrac 1 8 \\
\end {bmatrix}$
$e_8 := r_1 \to r_1 - \dfrac 1 3 r_3$
:$\mathbf A_8 = \begin {bmatrix}
1 & 0 & 0 & \dfrac 5 8 \\
0 & 1 & 0 & -\dfrac 1 8 \\
0 & 0 & 1 & \dfrac 1 8 \\
\end {bmatrix}$
and it is seen that $\mathbf A_8$ is the required reduced echelon form.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(i)}$
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 3

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
1 & 1 & -1 \\
1 & -1 & 2 \\
2 & 0 & 2 \\
2 & 1 & -1 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \to r_2 - r_1$
$e_2 := r_3 \to r_3 - 2 r_1$
$e_3 := r_4 \to r_4 - 2 r_1$
Hence:
:$\mathbf A_3 = \begin {bmatrix}
1 & 1 & -1 \\
0 & -2 & 1 \\
0 & -2 & 4 \\
0 & -1 & 1 \\
\end {bmatrix}$
$e_4 := r_2 \to r_2 - r_4$
:$\mathbf A_4 = \begin {bmatrix}
1 & 1 & -1 \\
0 & -1 & 0 \\
0 & -2 & 4 \\
0 & -1 & 1 \\
\end {bmatrix}$
$e_5 := r_2 \to -r_2$
:$\mathbf A_5 = \begin {bmatrix}
1 & 1 & -1 \\
0 & 1 & 0 \\
0 & -2 & 4 \\
0 & -1 & 1 \\
\end {bmatrix}$
$e_6 := r_3 \to r_3 + 2 r_2$
$e_7 := r_4 \to r_4 + r_2$
:$\mathbf A_7 = \begin {bmatrix}
1 & 1 & -1 \\
0 & 1 & 0 \\
0 & 0 & 4 \\
0 & 0 & 1 \\
\end {bmatrix}$
$e_8 := r_3 \to \dfrac {r_3} 4$
:$\mathbf A_8 = \begin {bmatrix}
1 & 1 & -1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 1 \\
\end {bmatrix}$
$e_9 := r_4 \to r_4 - r_3$
:$\mathbf A_9 = \begin {bmatrix}
1 & 1 & -1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end {bmatrix}$
It may be noted that $\mathbf A_9$ is in echelon form.
It remains to convert it to reduced echelon form.
$e_{10} := r_1 \to r_1 - r_2$
:$\mathbf A_{10} = \begin {bmatrix}
1 & 0 & -1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end {bmatrix}$
$e_{11} := r_1 \to r_1 + r_3$
:$\mathbf A_{11} = \begin {bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end {bmatrix}$
and it is seen that $\mathbf A_{11}$ is the required reduced echelon form.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(ii)}$
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 4

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
2 & 2 & 5 & 3 \\
6 & 1 & 5 & 4 \\
4 & -1 & 0 & 1 \\
2 & 0 & 1 & 1 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
1 & 0 & \dfrac 1 2 & \dfrac 1 2 \\
0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \to r_2 - 3 r_1$
$e_2 := r_3 \to r_3 - 2 r_1$
$e_3 := r_4 \to r_4 - r_1$
Hence:
:$\mathbf A_3 = \begin {bmatrix}
2 & 2 & 5 & 3 \\
0 & -5 & -10 & -5 \\
0 & -5 & -10 & -5 \\
0 & -2 & -4 & -2 \\
\end {bmatrix}$
$e_4 := r_2 \to -\dfrac {r_2} 5$
$e_5 := r_3 \to -\dfrac {r_3} 5$
$e_6 := r_4 \to -\dfrac {r_4} 2$
:$\mathbf A_6 = \begin {bmatrix}
2 & 2 & 5 & 3 \\
0 & 1 & 2 & 1 \\
0 & 1 & 2 & 1 \\
0 & 1 & 2 & 1 \\
\end {bmatrix}$
$e_7 := r_3 - r_2$
$e_8 := r_4 - r_2$
:$\mathbf A_8 = \begin {bmatrix}
2 & 2 & 5 & 3 \\
0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
$e_9 := r_1 \to \dfrac {r_1} 2$
:$\mathbf A_9 = \begin {bmatrix}
1 & 1 & \dfrac 5 2 & \dfrac 3 2 \\
0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
It may be noted that $\mathbf A_9$ is in echelon form.
It remains to convert it to reduced echelon form.
$e_{10} := r_1 \to r_1 - r_2$
:$\mathbf A_{10} = \begin {bmatrix}
1 & 0 & \dfrac 1 2 & \dfrac 1 2 \\
0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
and it is seen that $\mathbf A_{10}$ is the required reduced echelon form.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(iii)}$
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 5

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
 0 & 1 & 1 & 1 & 2 & 2 \\
-1 & 4 & 3 & 3 & 4 & 7 \\
 2 & 1 & 3 & 2 & 8 & 3 \\
 3 & 1 & 4 & -1 & 4 & 0 \\
 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
 1 & 0 & 1 & 0 & 2 & 0 \\
 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 0 & 0 & 1 & 2 & 1 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \leftrightarrow r_1$
Hence:
:$\mathbf A_1 = \begin {bmatrix}
-1 & 4 & 3 & 3 & 4 & 7 \\
 0 & 1 & 1 & 1 & 2 & 2 \\
 2 & 1 & 3 & 2 & 8 & 3 \\
 3 & 1 & 4 & -1 & 4 & 0 \\
 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
$e_2 := r_1 \to -r_1$
:$\mathbf A_2 = \begin {bmatrix}
 1 & -4 & -3 & -3 & -4 & -7 \\
 0 & 1 & 1 & 1 & 2 & 2 \\
 2 & 1 & 3 & 2 & 8 & 3 \\
 3 & 1 & 4 & -1 & 4 & 0 \\
 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
$e_3 := r_1 \to r_1 + 4 r_2$
:$\mathbf A_3 = \begin {bmatrix}
 1 & 0 & 1 & 1 & 4 & 1 \\
 0 & 1 & 1 & 1 & 2 & 2 \\
 2 & 1 & 3 & 2 & 8 & 3 \\
 3 & 1 & 4 & -1 & 4 & 0 \\
 5 & 2 & 7 & 0 & 10 & 2 \\
\end {bmatrix}$
$e_4 := r_3 \to r_3 - 2 r_1$
$e_5 := r_4 \to r_4 - 3 r_1$
$e_6 := r_5 \to r_5 - 5 r_1$
Hence:
:$\mathbf A_6 = \begin {bmatrix}
 1 & 0 & 1 & 1 & 4 & 1 \\
 0 & 1 & 1 & 1 & 2 & 2 \\
 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 1 & 1 & -4 & -8 & -3 \\
 0 & 2 & 2 & -5 & -10 & -3 \\
\end {bmatrix}$
$e_7 := r_3 \to r_3 - r_2$
$e_8 := r_4 \to r_4 - r_2$
:$\mathbf A_8 = \begin {bmatrix}
 1 & 0 & 1 & 1 & 4 & 1 \\
 0 & 1 & 1 & 1 & 2 & 2 \\
 0 & 0 & 0 & -1 & -2 & -1 \\
 0 & 0 & 0 & -5 & -10 & -5 \\
 0 & 2 & 2 & -5 & -10 & -3 \\
\end {bmatrix}$
$e_9 := r_3 \to -r_3$
$e_{10} := r_5 \to r_5 - r_4$
:$\mathbf A_{10} = \begin {bmatrix}
 1 & 0 & 1 & 1 & 4 & 1 \\
 0 & 1 & 1 & 1 & 2 & 2 \\
 0 & 0 & 0 & 1 & 2 & 1 \\
 0 & 0 & 0 & -5 & -10 & -5 \\
 0 & 2 & 2 & 0 & 0 & 2 \\
\end {bmatrix}$
$e_{12} := r_4 \to r_4 + 5 r_3$
$e_{13} := r_2 \to r_2 - r_3$
$e_{14} := r_1 \to r_1 - r_3$
:$\mathbf A_{14} = \begin {bmatrix}
 1 & 0 & 1 & 0 & 2 & 0 \\
 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 0 & 0 & 1 & 2 & 1 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 2 & 2 & 0 & 0 & 2 \\
\end {bmatrix}$
$e_{15} := r_5 \to r_5 - 2 r_2$
:$\mathbf A_{15} = \begin {bmatrix}
 1 & 0 & 1 & 0 & 2 & 0 \\
 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 0 & 0 & 1 & 2 & 1 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 \\
\end {bmatrix}$
and it is seen that $\mathbf A_{16}$ is the required reduced echelon form.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(iv)}$
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 6

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
 i & 1 - i & i & 0 \\
1 & -2 & 0 & i \\
1 - i & -1 + i & 1 & 1 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
1 & 0 & 1 + i & 1 \\
0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_2 \leftrightarrow r_1$
Hence:
:$\mathbf A_1 = \begin {bmatrix}
1 & -2 & 0 & i \\
 i & 1 - i & i & 0 \\
1 - i & -1 + i & 1 & 1 \\
\end {bmatrix}$
$e_2 := r_3 \to r_3 - r_1$
$e_3 := r_3 \to r_3 + r_2$
:$\mathbf A_3 = \begin {bmatrix}
1 & -2 & 0 & i \\
 i & 1 - i & i & 0 \\
0 & 2 & 1 + i & 1 - i \\
\end {bmatrix}$
$e_4 := r_2 \to r_2 - i r_1$
:$\mathbf A_4 = \begin {bmatrix}
1 & -2 & 0 & i \\
0 & 1 + i & i & 1 \\
0 & 2 & 1 + i & 1 - i \\
\end {bmatrix}$
$e_5 := r_1 \to r_1 + r_3$
:$\mathbf A_5 = \begin {bmatrix}
1 & 0 & 1 + i & 1 \\
0 & 1 + i & i & 1 \\
0 & 2 & 1 + i & 1 - i \\
\end {bmatrix}$
$e_6 := r_3 \to \dfrac {r_3} 2$
:$\mathbf A_6 = \begin {bmatrix}
1 & 0 & 1 + i & 1 \\
0 & 1 + i & i & 1 \\
0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\
\end {bmatrix}$
$e_7 := r_2 \to r_2 - \paren {1 + i} r_3$
:$\mathbf A_7 = \begin {bmatrix}
1 & 0 & 1 + i & 1 \\
0 & 0 & 0 & 0 \\
0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\
\end {bmatrix}$
$e_8 := r_2 \leftrightarrow r_3$
:$\mathbf A_8 = \begin {bmatrix}
1 & 0 & 1 + i & 1 \\
0 & 1 & \dfrac {1 + i} 2 & \dfrac {1 - i} 2 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
and it is seen that $\mathbf A_8$ is the required reduced echelon form.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(v)}$
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 7

== Example of Use of Gaussian Elimination ==
Let $\mathbf A$ denote the matrix:
:$\mathbf A = \begin {bmatrix}
 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\
\sqrt 2 & -3 & 1 + \sqrt 2 & -1 - 2 \sqrt 2 \\
 -1 & \sqrt 2 & -1 & 1 \\
\sqrt 2 - 2 & -2 + 4 \sqrt 2 & -2 - \sqrt 2 & 3 + \sqrt 2 \\
\end {bmatrix}$
The reduced echelon form of $\mathbf A$ is:
:$\mathbf E = \begin {bmatrix}
1 & 0 & 1 - \sqrt 2 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
== Proof ==
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.
$e_1 := r_3 \to r_3 + r_1$
$e_2 := r_4 \to r_4 - r_2$
$e_3 := r_4 \to r_4 + 2 r_1$
Hence:
:$\mathbf A_3 = \begin {bmatrix}
 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\
\sqrt 2 & -3 & 1 + \sqrt 2 & -1 - 2 \sqrt 2 \\
 0 & 1 & -1 & 1 + \sqrt 2 \\
 0 & 3 + 2 \sqrt 2 & -3 - 2 \sqrt 2 & 4 + 5 \sqrt 2 \\
\end {bmatrix}$
$e_4 := r_2 \to r_2 - \sqrt 2 r_1$
:$\mathbf A_4 = \begin {bmatrix}
1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\
0 & -1 - \sqrt 2 & 1 + \sqrt 2 & -3 - 2 \sqrt 2 \\
0 & 1 & -1 & 1 + \sqrt 2 \\
0 & 3 + 2 \sqrt 2 & -3 - 2 \sqrt 2 & 4 + 5 \sqrt 2 \\
\end {bmatrix}$
$e_5 := r_2 \leftrightarrow r_3$
:$\mathbf A_5 = \begin {bmatrix}
1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\
0 & 1 & -1 & 1 + \sqrt 2 \\
0 & -1 - \sqrt 2 & 1 + \sqrt 2 & -3 - 2 \sqrt 2 \\
0 & 3 + 2 \sqrt 2 & -3 - 2 \sqrt 2 & 4 + 5 \sqrt 2 \\
\end {bmatrix}$
$e_6 := r_1 \to r_1 - r_3$
$e_7 := r_4 \to r_4 + 2 r_3$
:$\mathbf A_7 = \begin {bmatrix}
1 & 2 & -1 - \sqrt 2 & 3 + 3 \sqrt 2 \\
0 & 1 & -1 & 1 + \sqrt 2 \\
0 & -1 - \sqrt 2 & 1 + \sqrt 2 & -3 - 2 \sqrt 2 \\
0 & 1 & -1 & -2 + \sqrt 2 \\
\end {bmatrix}$
$e_8 := r_1 \to r_1 - 2 r_2$
$e_9 := r_3 \to r_3 + \paren {1 + \sqrt 2} r_2$
$e_{10} := r_4 \to r_4 - r_2$
:$\mathbf A_{10} = \begin {bmatrix}
1 & 0 & 1 - \sqrt 2 & 1 + \sqrt 2 \\
0 & 1 & -1 & 1 + \sqrt 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -3 \\
\end {bmatrix}$
$e_{11} := r_4 \to -\dfrac 1 3 r_4$
$e_{12} := r_3 \leftrightarrow r_4$
:$\mathbf A_{12} = \begin {bmatrix}
1 & 0 & 1 - \sqrt 2 & 1 + \sqrt 2 \\
0 & 1 & -1 & 1 + \sqrt 2 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
$e_{13} := r_1 \to r_1 - \paren {1 + \sqrt 2} r_3$
$e_{14} := r_2 \to r_2 - \paren {1 + \sqrt 2} r_3$
:$\mathbf A_{12} = \begin {bmatrix}
1 & 0 & 1 - \sqrt 2 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
and it is seen that $\mathbf A_{12}$ is the required reduced echelon form.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(vi)}$
Category:Examples of Gaussian Elimination
Gaussian Elimination/Examples/Arbitrary Matrix 7/Mistake

== Source Work ==
:
Chapter $1$: Linear Equations and Matrices:
:$1.2$ Elementary Row Operations on Matrices:
::Exercises $1.2$:
:::$1 \ \text {(vi)}$
:Solutions to Exercises
== Mistake ==
:Find the reduced echelon matrices of the following matrices
::(vi) $\begin {bmatrix}
 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\
\sqrt 2 & -3 & 1 + \sqrt 2 & -1 - 2 \sqrt 2 \\
 -1 & \sqrt 2 & -1 & 1 \\
\sqrt 2 - 2 & -2 + 4 \sqrt 2 & -2 - \sqrt 2 & 3 + \sqrt 2 \\
\end {bmatrix}$
:Solutions to Exercises: Exercises $1.2 \ \text {(vi)}$: $\begin {bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end {bmatrix}$
== Correction ==
Evaluation according to Gaussian Elimination: Arbitrary Matrix $7$ gives the answer as:
:$\begin {bmatrix}
1 & 0 & 1 - \sqrt 2 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $1 \ \text {(vi)}$
Category:Mistakes/Linear Algebra: An Introduction
Gaussian Hypergeometric Function Difference Equation

== Theorem ==
Let $a, b, c \in \C$.
Let $c \notin \Z_{\le 0}$.
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
=== Formulation 4 ===
== Also see ==
* Gauss's Continued Fraction
* Hypergeometric Function Difference Equation
== Sources ==
* : Chapter Disquisitiones generales circa seriem infinitam $1 + \dfrac {\alpha \beta} \gamma x + $ etc.
Category:Gaussian Hypergeometric Function Difference Equation
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Series
Gaussian Hypergeometric Function Difference Equation/Formulation 1

== Theorem ==
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
== Proof ==
== Also see ==
* Gaussian Hypergeometric Function Difference Equation/Formulation 2
* Gaussian Hypergeometric Function Difference Equation/Formulation 3
* Gaussian Hypergeometric Function Difference Equation/Formulation 4
Category:Gaussian Hypergeometric Function Difference Equation
Gaussian Hypergeometric Function Difference Equation/Formulation 2

== Theorem ==
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
== Proof ==
== Also see ==
* Gaussian Hypergeometric Function Difference Equation/Formulation 1
* Gaussian Hypergeometric Function Difference Equation/Formulation 3
* Gaussian Hypergeometric Function Difference Equation/Formulation 4
Category:Gaussian Hypergeometric Function Difference Equation
Gaussian Hypergeometric Function Difference Equation/Formulation 3

== Theorem ==
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
== Proof ==
== Also see ==
* Gaussian Hypergeometric Function Difference Equation/Formulation 1
* Gaussian Hypergeometric Function Difference Equation/Formulation 2
* Gaussian Hypergeometric Function Difference Equation/Formulation 4
Category:Gaussian Hypergeometric Function Difference Equation
Gaussian Hypergeometric Function Difference Equation/Formulation 4

== Theorem ==
where:
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
:$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
== Proof ==
== Also see ==
* Gaussian Hypergeometric Function Difference Equation/Formulation 1
* Gaussian Hypergeometric Function Difference Equation/Formulation 2
* Gaussian Hypergeometric Function Difference Equation/Formulation 3
Category:Gaussian Hypergeometric Function Difference Equation
Gaussian Integer Units are 4th Roots of Unity

== Theorem ==
The units of the ring of Gaussian integers:
:$\set {1, i, -1, -i}$
are the (complex) $4$th roots of $1$.
== Proof ==
We have that $i = \sqrt {-1}$ is the imaginary unit.
Thus:
So $\set {1, i, -1, -i}$ constitutes the set of the $4$th roots of unity.
== Sources ==
* 
* 
Category:Gaussian Integers
Category:Complex Roots of Unity
Gaussian Integer Units form Multiplicative Subgroup of Complex Numbers

== Theorem ==
The group of Gaussian integer units under complex multiplication:
:$\struct {U_\C, \times} = \struct {\set {1, i, -1, -i}, \times}$
forms a subgroup of the multiplicative group of complex numbers.
== Proof ==
By Units of Gaussian Integers form Group, $\struct {U_\C, \times}$ forms a group.
Each of the elements of $U_\C$ is a complex number, and non-zero, and therefore $U_\C \subseteq \C \setminus \set 0$.
The result follows by definition of subgroup.
== Sources ==
* : $\S 36$: Subgroups: Simple illustrations: $(2)$
* 
Category:Multiplicative Group of Complex Numbers
Category:Gaussian Integers
Category:Examples of Subgroups
Gaussian Integers are Closed under Addition

== Theorem ==
The set of Gaussian integers $\Z \sqbrk i$ is closed under addition:
:$\forall x, y \in \Z \sqbrk i: x + y \in \Z \sqbrk i$
== Proof ==
Let $x$ and $y$ be Gaussian integers.
Then:
== Sources ==
* 
* 
Category:Gaussian Integers
Category:Complex Addition
Category:Algebraic Closure
Gaussian Integers are Closed under Multiplication

== Theorem ==
The set of Gaussian integers $\Z \sqbrk i$ is closed under multiplication:
:$\forall x, y \in \Z \sqbrk i: x \times y \in \Z \sqbrk i$
== Proof ==
Let $x$ and $y$ be Gaussian integers.
Then:
== Sources ==
* 
* 
Category:Gaussian Integers
Category:Complex Multiplication
Category:Algebraic Closure
Gaussian Integers are Closed under Negation

== Theorem ==
The set of Gaussian integers $\Z \sqbrk i$ is closed under negation:
:$\forall x \in \Z \sqbrk i: -x \in \Z \sqbrk i$
== Proof ==
Let $x$ be a Gaussian integer.
Then:
Category:Gaussian Integers
Category:Negation Functions
Category:Algebraic Closure
Gaussian Integers are Closed under Subtraction

== Theorem ==
The set of Gaussian integers $\Z \sqbrk i$ is closed under subtraction:
:$\forall x, y \in \Z \sqbrk i: x - y \in \Z \sqbrk i$
== Proof ==
Let $x$ and $y$ be Gaussian integers.
Then:
== Sources ==
* 
* 
Category:Gaussian Integers
Category:Complex Subtraction
Category:Algebraic Closure
Gaussian Integers does not form Subfield of Complex Numbers

== Theorem ==
The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
is not a subfield of $\C$.
== Proof ==
Proof by Counterexample:
We have that:
:$2 + 0 i \in \Z \sqbrk i$
However there is no $z \in \Z \sqbrk i$ such that:
:$x \paren {2 + 0 i} = 1 + 0 i$
So, by definition, $\Z \sqbrk i$ is not a field.
Thus $\Z \sqbrk i$ is not a subfield of $\C$.
== Also see ==
* Gaussian Integers form Subring of Complex Numbers
Category:Subfields
Category:Gaussian Integers
Gaussian Integers form Euclidean Domain

== Theorem ==
Let $\struct {\Z \sqbrk i, +, \times}$ be the integral domain of Gaussian Integers.
Let $\nu: \Z \sqbrk i \to \R$ be the real-valued function defined as: 
:$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$ 
where $\cmod a$ is the (complex) modulus of $a$.
Then $\nu$ is a Euclidean valuation on $\Z \sqbrk i$.
Hence $\struct {\Z \sqbrk i, +, \times}$ with $\nu: \Z \sqbrk i \to \Z$ forms a Euclidean domain.
== Proof 1 ==
== Proof 2 ==
Category:Examples of Euclidean Domains
Category:Gaussian Integers
Category:Gaussian Integers form Euclidean Domain