nyuuzyou/wikis huggingface dataset:
Rounding to Even/Examples/72.465 to 2dp

== Example of Rounding to Even ==
$72 \cdotp 465$ rounded to $2$ decimal places using rounding to even is $72 \cdotp 46$.
== Proof ==
We have that $72 \cdotp 465$ is equally far away from $72 \cdotp 46$ as it is from $72 \cdotp 47$.
But $72 \cdotp 46$ ends in an even integer.
That is, $7246$ is even.
Hence the result.
== Sources ==
* : Chapter $1$: Rounding of Data
Category:Examples of Rounding
Row Equivalence/Examples/Arbitrary Example 1

== Example of Row Equivalence ==
Let $\mathbf A$ be the matrix defined as:
:$\mathbf A := \begin {bmatrix} 1 & 0 & -1 & 1 \\ 2 & 1 & 0 & 1 \\ -1 & 1 & 0 & -2 \end {bmatrix}$
Let $\mathbf B$ be the matrix defined as:
:$\mathbf B := \begin {bmatrix} 1 & 0 & -1 & 1 \\ -1 & 1 & 0 & 2 \\ 0 & 3 & 0 & 5 \end {bmatrix}$
Then $\mathbf A$ and $\mathbf B$ are row equivalent.
== Proof ==
$(1): \quad$ Use the elementary row operation on $\mathbf A$:
:$e_1 := r_2 \to r_2 + 2 r_3$
to make $\mathbf A_1$:
:$\mathbf A_1 := \begin {bmatrix} 1 & 0 & -1 & 1 \\ 0 & 3 & 0 & 5 \\ -1 & 1 & 0 & -2 \end {bmatrix}$
$(2): \quad$ Use the elementary row operation on $\mathbf A_1$:
:$e_2 := r_2 \leftrightarrow r_3$
to make $\mathbf A_2$:
:$\mathbf A_2 := \begin {bmatrix} 1 & 0 & -1 & 1 \\ -1 & 1 & 0 & -2 \\ 0 & 3 & 0 & 5 \end {bmatrix}$
which is $\mathbf B$.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Example
Category:Row Equivalence
Row Equivalence/Examples/Arbitrary Example 2

== Example of Row Equivalence ==
Let $\mathbf A$ be the matrix defined as:
:$\mathbf A := \begin {bmatrix} 1 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end {bmatrix}$
Let $\mathbf B$ be the matrix defined as:
:$\mathbf B := \begin {bmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \\ 1 & -1 & 1 \end {bmatrix}$
Then $\mathbf A$ and $\mathbf B$ are row equivalent.
== Proof ==
We use elementary row operations to convert both $\mathbf A$ and $\mathbf B$ to reduced echelon form.
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to either $\mathbf A$ or $\mathbf B$ accordingly.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$ or $\mathbf B_k$.
First we take $\mathbf A := \begin {bmatrix} 1 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end {bmatrix}$.
$e_1 := r_2 \to r_2 - 2 r_1$
$e_2 := r_3 \to r_3 - r_1$
:$\mathbf A_2 := \begin {bmatrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & -1 & 2
\end {bmatrix}$
$e_3 := r_3 \to r_3 + r_2$
:$\mathbf A_3 := \begin {bmatrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end {bmatrix}$
$e_4 := r_3 \to \dfrac {r_3} 4$
:$\mathbf A_4 := \begin {bmatrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 1
\end {bmatrix}$
$e_5 := r_1 \to r_1 + r_3$
$e_6 := r_2 \to r_2 - 2 r 3$
:$\mathbf A_6 := \begin {bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end {bmatrix}$
Then we take $\mathbf B := \begin {bmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \\ 1 & -1 & 1 \end {bmatrix}$.
$e_1 := r_1 \leftrightarrow r_3$
:$\mathbf B_1 := \begin {bmatrix}
1 & -1 & 1 \\
0 & 2 & 1 \\
3 & -1 & 1 \\
\end {bmatrix}$
$e_2 := r_3 \to r_3 - 3 r_1$
:$\mathbf B_2 := \begin {bmatrix}
1 & -1 & 1 \\
0 & 2 & 1 \\
0 & 2 & -2 \\
\end {bmatrix}$
$e_3 := r_3 \to \dfrac {r_3} 2$
:$\mathbf B_3 := \begin {bmatrix}
1 & -1 & 1 \\
0 & 2 & 1 \\
0 & 1 & -1 \\
\end {bmatrix}$
$e_4 := r_2 \leftrightarrow r_3$
:$\mathbf B_4 := \begin {bmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
0 & 2 & 1 \\
\end {bmatrix}$
$e_5 := r_1 \to r_1 + r_2$
$e_6 := r_1 \to r_1 - 2 r 1$
:$\mathbf B_6 := \begin {bmatrix}
1 & 0 & 0 \\
0 & 1 & -1 \\
0 & 0 & 3 \\
\end {bmatrix}$
$e_7 := r_3 \to \dfrac {r_3} 3$
:$\mathbf B_7 := \begin {bmatrix}
1 & 0 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \\
\end {bmatrix}$
$e_8 := r_2 \to r_2 + r 3$
:$\mathbf B_8 := \begin {bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end {bmatrix}$
It is seen that both $\mathbf A$ and $\mathbf B$ are row equivalent to the same reduced echelon matrix.
The result follows by Row Equivalence is Equivalence Relation. 
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $2 \ \text {(i)}$
Category:Row Equivalence
Row Equivalence/Examples/Arbitrary Example 3

== Example of Row Equivalence ==
Let $\mathbf A$ be the matrix defined as:
:$\mathbf A := \begin {bmatrix}
 1 & -1 & 1 & 2 \\
-2 & 3 & 0 & 1 \\
 1 & 0 & -1 & 3 \\
\end {bmatrix}$
Let $\mathbf B$ be the matrix defined as:
:$\mathbf B := \begin {bmatrix}
 0 & -1 & 2 & 3 \\
 1 & 2 & -1 & 0 \\
-2 & -5 & 4 & 3 \\
\end {bmatrix}$
Then $\mathbf A$ and $\mathbf B$ are not row equivalent.
== Proof ==
We use elementary row operations to convert both $\mathbf A$ and $\mathbf B$ to reduced echelon form.
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to either $\mathbf A$ or $\mathbf B$ accordingly.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$ or $\mathbf B_k$.
First we take $\mathbf A := \begin {bmatrix}
 1 & -1 & 1 & 2 \\
-2 & 3 & 0 & 1 \\
 1 & 0 & -1 & 3 \\
\end {bmatrix}$.
$e_1 := r_2 \to r_2 + 2 r_1$
$e_2 := r_3 \to r_3 - r_1$
:$\mathbf A_2 := \begin {bmatrix}
1 & -1 & 1 & 2 \\
0 & 1 & 2 & 5 \\
0 & 1 & -2 & 1 \\
\end {bmatrix}$
$e_3 := r_3 \to r_3 - r_2$
$e_4 := r_1 \to r_1 + r_2$
:$\mathbf A_4 := \begin {bmatrix}
1 & 0 & 3 & 7 \\
0 & 1 & 2 & 5 \\
0 & 0 & -4 & -4 \\
\end {bmatrix}$
$e_5 := r_3 \to -\dfrac {r_3} 4$
:$\mathbf A_5 := \begin {bmatrix}
1 & 0 & 3 & 7 \\
0 & 1 & 2 & 5 \\
0 & 0 & 1 & 1 \\
\end {bmatrix}$
$e_6 := r_1 \to r_1 - 3 r 3$
$e_7 := r_2 \to r_1 - 2 r 3$
:$\mathbf A_7 := \begin {bmatrix}
1 & 0 & 0 & 4 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & 1 \\
\end {bmatrix}$
Then we take $\mathbf B := \begin {bmatrix} 
 0 & -1 & 2 & 3 \\
 1 & 2 & -1 & 0 \\
-2 & -5 & 4 & 3 \\
\end {bmatrix}$.
$e_1 := r_1 \leftrightarrow r_2$
:$\mathbf B_1 := \begin {bmatrix}
 1 & 2 & -1 & 0 \\
 0 & -1 & 2 & 3 \\
-2 & -5 & 4 & 3 \\
\end {bmatrix}$
$e_2 := r_3 \to r_3 + 2 r_1$
:$\mathbf B_2 := \begin {bmatrix}
 1 & 2 & -1 & 0 \\
 0 & -1 & 2 & 3 \\
 0 & -1 & 2 & 3 \\
\end {bmatrix}$
$e_3 := r_3 \to r_3 - r_2$
:$\mathbf B_3 := \begin {bmatrix}
 1 & 2 & -1 & 0 \\
 0 & -1 & 2 & 3 \\
 0 & 0 & 0 & 0 \\
\end {bmatrix}$
A zero row has appeared.
There is no need to proceed further.
The reduced echelon matrices for $\mathbf A$ and $\mathbf B$ are different.
The result follows by Row Equivalence is Equivalence Relation. 
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Exercises $1.2$: $2 \ \text {(ii)}$
Category:Row Equivalence
Row Equivalence is Equivalence Relation

== Theorem ==
Row equivalence is an equivalence relation.
== Proof ==
In the following, $\mathbf A$, $\mathbf B$ and $\mathbf C$ denote arbitrary matrices in a given matrix space $\map \MM {m, n}$ for $m, n \in \Z{>0}$.
We check in turn each of the conditions for equivalence:
=== Reflexive ===
Let $r_i$ denote an arbitrary row of $\mathbf A$.
Let $e$ denote the elementary row operation $r_i \to 1 r_i$ applied to $\mathbf A$.
Then trivially:
:$\map e {\mathbf A} = \mathbf A$
and so $\mathbf A$ is trivially row equivalent to itself.
So row equivalence has been shown to be reflexive.
=== Symmetric ===
Let $\mathbf A$ be row equivalent to $\mathbf B$.
Let $\Gamma$ be the row operation that transforms $\mathbf A$ into $\mathbf B$.
From Row Operation has Inverse there exists a row operation $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$.
Thus $\mathbf B$ is row equivalent to $\mathbf A$.
So row equivalence has been shown to be symmetric.
=== Transitive ===
Let $\mathbf A$ be row equivalent to $\mathbf B$, and let $\mathbf B$ be row equivalent to $\mathbf C$.
Let $\Gamma_1$ be the row operation that transforms $\mathbf A$ into $\mathbf B$.
Let $\Gamma_2$ be the row operation that transforms $\mathbf B$ into $\mathbf C$.
From Sequence of Row Operations is Row Operation, $\mathbf C$ is row equivalent to $\mathbf A$.
So row equivalence has been shown to be transitive.
Row equivalence has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Also see ==
* Column Equivalence is Equivalence Relation
Category:Examples of Equivalence Relations
Category:Row Operations
Category:Row Equivalence
Row Equivalent Matrix for Homogeneous System has same Solutions

== Theorem ==
Let $\mathbf A$ be a matrix in the matrix space $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a homogeneous system of linear equations.
Let $\mathbf H$ be row equivalent to $\mathbf A$.
Then the solution set of $\mathbf H \mathbf x = \mathbf 0$ equals the solution set of $\mathbf A \mathbf x = \mathbf 0$.
That is:
:$\mathbf A \sim \mathbf H \implies \set {\mathbf x: \mathbf A \mathbf x = \mathbf 0} = \set {\mathbf x: \mathbf H \mathbf x = \mathbf 0}$
where $\sim$ represents row equivalence.
=== Corollary ===
== Proof ==
Let:
be the system of equations to be solved.
Suppose the elementary row operation of multiplying one row $i$ by a non-zero scalar $\lambda$ is performed.
Recall, the $i$th row of the matrix represents the $i$th equation of the system to be solved.
Then this is logically equivalent to multiplying the $i$th equation on both sides by the scalar $\lambda$:
which clearly has the same solutions as the original equation.
Suppose the elementary row operation of adding a scalar multiple of row $i$ to another row $j$ is performed.
Recall that the $i$th and $j$th row of the matrix represent the $i$th and $j$th equation in the system to be solved.
Thus this is logically equivalent to manipulating the $i$th and $j$th equations as such:
As both sides of equation $i$ are equal to each other, this operation is simply performing the same act on both sides of equation $j$.
This clearly will have no effect on the solution set of the system of equations.
Suppose the elementary row operation of interchanging row $i$ and row $j$ is performed.
Recall that the $i$th and $j$th row of the matrix represent the $i$th and $j$th equation in the system to be solved.
Then, interchanging row $i$ and row $j$ is logically equivalent to switching the $i$th equation and the $j$th equation of the system to be solved.
But clearly the system containing the following two equations:
has the same solution set as a system instead containing the following two equations:
Hence the result, by the definition of row equivalence.
Category:Linear Algebra
Row Equivalent Matrix for Homogeneous System has same Solutions/Corollary

== Theorem ==
Let $\mathbf A$ be a matrix in the matrix space $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a homogeneous system of linear equations.
Then:
:$\set {\mathbf x: \mathbf A \mathbf x = \mathbf 0} = \set {\mathbf x: \map {\mathrm {ref} } {\mathbf A} \mathbf x = \mathbf 0}$
where $\map {\mathrm {ref} } {\mathbf A}$ is the reduced echelon form of $\mathbf A$.
== Proof ==
Follows from Row Equivalent Matrix for Homogeneous System has same Solutions and from Matrix is Row Equivalent to Reduced Echelon Matrix.
Category:Matrix Theory
Category:Linear Algebra
Row Operation has Inverse

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\Gamma$ be a row operation which transforms $\mathbf A$ to a new matrix $\mathbf B \in \map \MM {m, n}$.
Then there exists another row operation $\Gamma'$ which transforms $\mathbf B$ back to $\mathbf A$.
== Proof ==
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary row operations that compose $\Gamma$.
Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary row matrices.
From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, we have:
:$\mathbf R \mathbf A = \mathbf B$
where $\mathbf R$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
:$\mathbf R = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
By Elementary Row Matrix is Nonsingular, each of $\mathbf E_i$ is nonsingular.
By Product of Matrices is Nonsingular iff Matrices are Nonsingular, it follows that $\mathbf R$ is likewise nonsingular.
Thus $\mathbf R$ has an inverse $\mathbf R^{-1}$.
Hence:
We have:
From Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse, each of ${\mathbf E_i}^{-1}$ is the elementary row matrix corresponding to the inverse $e'_i$ of the corresponding elementary row operation $e_i$.
Let $\Gamma'$ be the row operation composed of the finite sequence of elementary row operations $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$.
Thus $\Gamma'$ is a row operation which transforms $\mathbf B$ into $\mathbf A$.
Hence the result.
== Also see ==
* Column Operation has Inverse
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.15 \ \text {(a)}$
Category:Row Operations
Row Operation to Clear First Column of Matrix

== Theorem ==
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over a field $K$.
Then there exists a row operation to convert $\mathbf A$ into another $m \times n$ matrix $\mathbf B = \sqbrk b_{m n}$ with the following properties:
:$(1): \quad$ Except possibly for element $b_{1 1}$, all the elements of column $1$ are $0$
:$(2): \quad$ If $b_{1 1} \ne 0$, then $b_{1 1} = 1$.
This process is referred to as clearing the first column.
== Proof ==
The following algorithm generates a sequence of elementary row operations which convert $\mathbf A$ to $\mathbf B$.
Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the state of $\mathbf A$ after having processed the latest step.
After each step, an implicit step can be included that requires that the form of $\mathbf A'$ is inspected to see if it is in the form $\mathbf B$, and if so, terminating the algorithm, but this is not essential.
:$(1): \quad$ Are all elements in the first column of $\mathbf A$ equal to $0$?
:::If so, there is nothing to do, and the required row operation is the unit matrix $\mathbf I_m$.
:::Otherwise, move on to step $(2)$.
:$(2): \quad$ Is element $a_{1 1}$ equal to $0$?
:::If so:
::::$\text (a): \quad$ find the smallest $k$ such that row $k$ of $\mathbf A$ such that $a_{k 1} \ne 0$
::::$\text (b): \quad$ use the elementary row operation $r_1 \leftrightarrow r_k$ which will result $a'_{1 1} = a_{k 1}$ and $a'_{k 1} = 0$.
:Move on to step $(3)$.
:$(3): \quad$ Is element $a'_{1 1}$ equal to $1$?
:::If so, use the elementary row operation $r_1 \to \lambda r_1$ where $\lambda = \dfrac 1 {a'_{1 1} }$, which will result $a'_{1 1} = 1$.
:Move on to step $4$
:$(4): \quad$ For each row $j$ from $2$ to $m$, do the following:
:::Is $a_{j 1} \ne 0$?
::::If so, use the elementary row operation $r_j \leftrightarrow r_j + \mu r_1$, where $\mu = -\dfrac {a'_{j 1} } {a'{1 1} }$, which will result in $a'_{j 1} = 0$.
This will result in an $m \times n$ matrix in the required form.
Exercising the above algorithm will have generated a sequence of elementary row operations $e_1, e_2, \ldots, e_t$.
For each $e_k$ we create the elementary row matrix $\mathbf E_k$.
We then assemble the matrix product:
:$\mathbf R := \mathbf E_t \mathbf E_{t - 1} \mathbf E_{t - 2} \dotsm \mathbf E_2 \mathbf E_1$
From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, $\mathbf R$ is the resulting $m \times m$ matrix corresponding to the row operation which is used to convert $\mathbf A$ to $\mathbf B$.
== Examples ==
== Also presented as ==
Some sources are not concerned about making $b_{1 1}$ equal to $1$.
Hence they do not use step $3$ of the algorithm.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
Category:Row Operations
Category:Row Operation to Clear First Column of Matrix
Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 1

== Example of Use of Row Operation to Clear First Column of Matrix ==
Let $\mathbf A$ be the matrix:
:$\mathbf A = \begin {pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$
The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:
:$\mathbf R = \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end {pmatrix}$
and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:
:$\mathbf B = \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end {pmatrix}$
== Proof ==
We use Row Operation to Clear First Column of Matrix as follows:
:$(1): \quad$ Apply the elementary row operation $r_1 \leftrightarrow r_2$ to exchange row $1$ with row $2$.
From Elementary Matrix corresponding to Elementary Row Operation: Exchange Rows, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_1 := \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A$ to $\mathbf A_1$ as follows:
:$(2): \quad$ Apply the elementary row operation $r_3 \to r_3 + \paren {-1} r_1$ to add $-1$ times row $1$ to row $3$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_2 := \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A_1$ to $\mathbf A_2$ as follows:
The matrix is in the correct form, and so:
:$\mathbf B = \mathbf A_2$
Hence we can calculate $\mathbf R$:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.2 \ \text {(a)}$
Category:Row Operation to Clear First Column of Matrix
Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 2

== Example of Use of Row Operation to Clear First Column of Matrix ==
Let $\mathbf A$ be the matrix:
:$\mathbf A = \begin {pmatrix} 1 & 0 & 1 & 1 \\ -1 & 0 & 2 & 1 \\ -1 & 1 & 5 & 9 \end {pmatrix}$
The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:
:$\mathbf R = \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$
and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:
:$\mathbf B = \begin {pmatrix} 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & 2 \\ 0 & 1 & 6 & 10 \end {pmatrix}$
== Proof ==
We use Row Operation to Clear First Column of Matrix as follows:
:$(1): \quad$ Apply the elementary row operation $r_2 \to r_2 + r_1$ to to add $1$ times row $1$ to row $2$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_1 := \begin {pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A$ to $\mathbf A_1$ as follows:
:$(2): \quad$ Apply the elementary row operation $r_3 \to r_3 + r_1$ to add $1$ times row $1$ to row $3$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_2 := \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A_1$ to $\mathbf A_2$ as follows:
The matrix is in the correct form, and so:
:$\mathbf B = \mathbf A_2$
Hence we can calculate $\mathbf R$:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.2 \ \text {(b)}$
Category:Row Operation to Clear First Column of Matrix
Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 3

== Example of Use of Row Operation to Clear First Column of Matrix ==
Let $\mathbf A$ be the matrix:
:$\mathbf A = \begin {pmatrix} 0 & 4 & 7 & 10 \\ 2 & 5 & 8 & 11 \\ 3 & 6 & 9 & 12 \end {pmatrix}$
The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:
:$\mathbf R = \begin {pmatrix} 0 & \dfrac 1 2 & 0 \\ 1 & 0 & 0 \\ 0 & -\dfrac 3 2 & 1 \end {pmatrix}$
and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:
:$\mathbf B = \begin {pmatrix} 1 & \dfrac 5 2 & 4 & \dfrac {11} 2 \\ 0 & 4 & 7 & 10 \\ 0 & -\dfrac 3 2 & -3 & -\dfrac 9 2 \end {pmatrix}$
== Proof ==
We use Row Operation to Clear First Column of Matrix as follows:
:$(1): \quad$ Apply the elementary row operation $r_1 \leftrightarrow r_2$ to exchange row $1$ with row $2$.
From Elementary Matrix corresponding to Elementary Row Operation: Exchange Rows, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_1 := \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A$ to $\mathbf A_1$ as follows:
:$(2): \quad$ Apply the elementary row operation $r_1 \to \dfrac 1 2 r_1$ to exchange row $1$ with row $2$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_2 := \begin {pmatrix} \dfrac 1 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A_1$ to $\mathbf A_2$ as follows:
We calculate $\mathbf E_2 \mathbf E_1$:
:$(3): \quad$ Apply the elementary row operation $r_3 \to r_3 + \paren {-3} r_1$ to add $-3$ times row $1$ to row $3$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
:$\mathbf E_3 := \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A_2$ to $\mathbf A_3$ as follows:
The matrix is in the correct form, and so:
:$\mathbf B = \mathbf A_3$
Hence we can calculate $\mathbf R$:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.2 \ \text {(c)}$
Category:Row Operation to Clear First Column of Matrix
Row Rank of Matrix equals Rank of Matrix

== Theorem ==
Let $\mathbf A$ be a matrix.
The row rank of $\mathbf A$ is equal to the rank of $\mathbf A$.
== Proof ==
The rank of $\mathbf A$ is defined as the dimension of the column space of $\mathbf A$.
That is, the rank of $\mathbf A$ is the column rank of $\mathbf A$.
The result follows from Column Rank of Matrix equals Row Rank.
== Sources ==
* 
Category:Row Rank
Category:Rank of Matrix
Row and Column Sums of Adjacency Matrix of Undirected Graph equals Vertex Degree

== Theorem ==
Let $G$ be an undirected graph.
Let $\mathbf M$ be the adjacency matrix for $G$.
The row sums and column sums of the rows and columns of $\mathbf M$ are equal to the degrees of the corresponding vertices of $G$.
== Proof ==
== Sources ==
* 
* 
Category:Adjacency Matrices
Category:Undirected Graphs
Row in Pascal's Triangle forms Palindromic Sequence

== Theorem ==
Each of the rows of Pascal's triangle forms a palindromic sequence.
== Proof ==
The $n$th row of Pascal's triangle consists of the finite sequence:
:$\dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n$
By the Symmetry Rule for Binomial Coefficients:
:$\dbinom n m = \dbinom n {n - m}$
Hence we can write the $n$th row in reverse order:
and the sequences are seen to be the same.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $8$
Category:Pascal's Triangle
Rowing With and Against the Tide

== Problem ==
:If, during ebb tide, a wherry should set out from Haverhill, to come down the river,
:and, at the same time, another should set out from Newburyport, to go up the river,
:allowing the difference to be $18$ miles;
:suppose the current forwards one and retards the other $1 \frac 1 2$ miles per hour;
:the boats are equally laden, the rowers equally good,
:and, in the common way of working in still water, would proceed at the rate of $4$ miles per hour;
:when, in the river, will the two boats meet?
== Solution ==
:$5 \frac 5 8$ miles upstream from Newburyport.
== Proof ==
The rowers are moving at $8$ miles per hour relative to each other.
Hence they will meet in $\dfrac {18} 8 = \dfrac 9 4$ hours.
During that time, the rower from Newburyport is rowing at a speed of $4 - 1 \frac 1 2 = 2 \frac 1 2 = \dfrac 5 2$ miles per hour relative to land.
Thus he gets as far is it takes to travel for $\dfrac 9 4$ hours, that is: $\dfrac 9 4 \times \dfrac 5 2 = \dfrac {45} 8 = 5 \frac 5 8$ miles from Newburyport.
== Historical Note ==
== Sources ==
* : Rowing with and against the Tide: $145$
Category:Travelling With and Against the Flow Puzzles