nyuuzyou/wikis huggingface dataset:
Matrix Product with Adjugate Matrix

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Let $\adj {\mathbf A}$ be its adjugate matrix.
Then:
where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$, and $\mathbf I_n$ denotes the unit matrix of order $n$.
== Proof ==
Let $\mathbf A = \paren {a_{i j} }$.
Let $A_{i j}$ denote the cofactor of $a_{i j} \in \mathbf A$.
=== Right Multiplication ===
We show that $\mathbf A \cdot \adj {\mathbf A} = \map \det {\mathbf A} \cdot \mathbf I_n$.
Let $i, j \in \set {1, \ldots, n}$. 
If $i = j$, expanding $\map \det {\mathbf A}$ along row $i$ shows that:
:$\ds \map \det {\mathbf A} = \sum_{k \mathop = 1}^n a_{i k} A_{i k}$
If $i \ne j$, define $\mathbf A'$ as the matrix obtained by replacing row $j$ of $\mathbf A$ with row $i$ of $\mathbf A$.
Then $\mathbf A' = \begin {bmatrix} a' \end {bmatrix}_n$ has two identical rows, so:
By definition of matrix product, element $\tuple {i, j}$ of $\mathbf A \cdot \adj {\mathbf A}$ is:
:$\ds \sum_{k \mathop = 1}^n a_{i k} A_{j k} = \begin {cases}
0_R & \text {for} & i \ne j \\
\map \det {\mathbf A} & \text {for} & i = j
\end {cases}$
Hence:
:$\mathbf A \cdot \adj {\mathbf A} = \map \det {\mathbf A} \cdot \mathbf I_n$
=== Left Multiplication ===
We show that $\adj {\mathbf A} \cdot \mathbf A = \map \det {\mathbf A} \cdot \mathbf I_n$.
Let $i, j \in \set {1, \ldots, n}$.
If $i = j$, expanding $\map \det {\mathbf A}$ along column $j$ shows that:
:$\ds \map \det {\mathbf A} = \sum_{k \mathop = 1}^n a_{k j} A_{k j}$
If $i \ne j$, define $\mathbf A'$ as the matrix obtained by replacing column $i$ of $\mathbf A$ with column $j$ of $\mathbf A$.
Then $\mathbf A' = \begin{bmatrix} a' \end{bmatrix}_n$ has two identical columns, so:
By definition of matrix product, element $\tuple {i, j}$ of $\adj {\mathbf A} \cdot \mathbf A$ is:
:$\ds \sum_{k \mathop = 1}^n A_{k i} a_{k j} = \begin {cases}
0_R & \text {for} & i \ne j \\
\map \det {\mathbf A} & \text {for} & i = j
\end {cases}$
Hence:
:$\adj {\mathbf A} \cdot \mathbf A = \map \det {\mathbf A} \cdot \mathbf I_n$
== Also see ==
* Matrix is Nonsingular iff Determinant has Multiplicative Inverse
== Sources ==
* : $\S 7.4$
* : $\S 1.2.3$: Sums and Products: Exercises -- Second Set
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.7$ Minors and cofactors: $(2)$
* 
* 
Category:Inverse Matrices
Category:Conventional Matrix Multiplication
Category:Adjugate Matrices
Category:Determinants
Matrix Scalar Product/Examples/Arbitrary Example 1

== Examples of Matrix Scalar Product ==
The matrix scalar product of the matrix $\begin {pmatrix} a & b \\ c & d \end {pmatrix}$ by the scalar $k$ works as follows:
:$k \begin {pmatrix} a & b \\ c & d \end {pmatrix} = \begin {pmatrix} k a & k b \\ k c & k d \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Matrix Scalar Product
Matrix Scalar Product Distributes over Matrix Entrywise Addition

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be matrices both of order $m \times n$.
Let $k$ be a scalar.
Then:
:$k \paren {\mathbf A + \mathbf B} = k \mathbf A + k \mathbf B$
where:
:$+$ denotes matrix entrywise addition
:$k \mathbf A$ etc. denotes matrix scalar product.
== Proof ==
<onlyinclude>
Let $a_{i j}$ and $b_{i j}$ denote the $\tuple {i, j}$th entry in $\mathbf A$ and $\mathbf B$ respectively.
== Sources ==
* 
* 
Category:Matrix Scalar Product
Category:Matrix Entrywise Addition
Category:Examples of Distributive Operations
Matrix Scalar Product Distributes over Number Addition

== Theorem ==
Let $\GF$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\GF$.
For $\mathbf A \in \map \MM {m, n}$ and $\lambda \in \GF$, let $\lambda \mathbf A$ be defined as the matrix scalar product of $\lambda$ and $\mathbf A$.
The matrix scalar product is distributive on $\map \MM {m, n}$, in the following sense:
For all $\mathbf A$ in $\map \MM {m, n}$ and $\lambda, \mu \in \GF$:
:$\paren {\lambda + \mu} \mathbf A = \lambda \mathbf A + \mu \mathbf A$
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $5$
Category:Matrix Scalar Product Distributes over Number Addition
Category:Matrix Scalar Product
Category:Addition
Category:Examples of Distributive Operations
Matrix Scalar Product is Associative

== Theorem ==
Let $\GF$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\GF$.
For $\mathbf A \in \map \MM {m, n}$ and $\lambda$ \in $\GF$, let $\lambda \mathbf A$ be defined as the matrix scalar product of $\lambda$ and $\mathbf A$.
The matrix scalar product is associative on $\map \MM {m, n}$, in the following sense:
For all $\mathbf A$ in $\map \MM {m, n}$ and $\lambda, \mu \in \GF$:
:$\lambda \paren {\mu \mathbf A} = \paren {\lambda \mu} \mathbf A$
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $4$
Category:Matrix Scalar Product is Associative
Category:Matrix Scalar Product
Category:Examples of Associative Operations‎
Matrix Scalar Product with Zero gives Zero Matrix

== Theorem ==
Let $\Bbb F$ denote one of the standard number systems.
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over $\Bbb F$.
For $\mathbf A \in \map \MM {m, n}$ and $\lambda$ \in $\Bbb F$, let $\lambda \mathbf A$ be defined as the matrix scalar product of $\lambda$ and $\mathbf A$.
When $\lambda = 0$, we have for all $\mathbf A$ in $\map \MM {m, n}$:
:$0 \mathbf A = \mathbf 0$
where $\mathbf 0$ denotes the zero matrix.
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $6$
Category:Matrix Scalar Product
Category:Addition
Category:Examples of Distributive Operations
Matrix Similarity is Equivalence Relation/Proof 2

== Theorem ==
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ square matrices $\mathbf A$.
So matrix similarity is reflexive.
=== Symmetric ===
Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
As $\mathbf P$ is nonsingular, we have:
So matrix similarity is symmetric.
=== Transitive ===
Let $\mathbf B = \mathbf P_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf P_2^{-1} \mathbf B \mathbf P_2$.
Then:
By Product of Matrices is Nonsingular iff Matrices are Nonsingular, $\paren {\mathbf P_1 \mathbf P_2}$ is nonsingular.
So matrix similarity is transitive.
So, by definition, matrix similarity is an equivalence relation.
== Sources ==
* : $\S 16$: Equivalence relations (Worked Example)
Category:Matrix Similarity is Equivalence Relation
Matrix Space Semigroup under Hadamard Product

== Theorem ==
Let $\map {\MM_S} {m, n}$ be the matrix space over a semigroup $\struct {S, \cdot}$.
Then the algebraic structure $\struct {\map {\MM_S} {m, n}, \circ}$, where $\circ$ is the Hadamard product, is also a semigroup.
If $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
If $\struct {S, \cdot}$ is a monoid then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
== Proof ==
$\struct {S, \cdot}$ is a semigroup and is therefore closed and associative.
As $\struct {S, \cdot}$ is closed, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Closure of Hadamard Product‎.
As $\struct {S, \cdot}$ is associative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Associativity of Hadamard Product.
Thus if $\struct {S, \cdot}$ is a semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
If $\struct {S, \cdot}$ is commutative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Commutativity of Hadamard Product.
Thus if $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
Let $\struct {S, \cdot}$ be a monoid, with identity $e$.
Then from Zero Matrix is Identity for Hadamard Product, $\struct {\map {\MM_S} {m, n}, \circ}$ also has an identity and is therefore also a monoid.
Category:Hadamard Product
Matrix Space is Module

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over $\struct {R, +, \circ}$.
Then the matrix space $\map {\MM_R} {m, n}$ of all $m \times n$ matrices over $R$ is a module.
== Proof ==
This follows as $\map {\MM_R} {m, n}$ is a direct instance of the module given in the module of all mappings, where $\map {\MM_R} {m, n}$ is the $R$-module $R^{\closedint 1 m \times \closedint 1 n}$.
The $S$ of that example is the set $\closedint 1 m \times \closedint 1 n$, while the $G$ of that example is the $R$-module $R$.
Let $\sqbrk a_{i j}$ and $\sqbrk b_{i j}$ be the $\tuple{i, j}$th element of $m \times n$ matrices $\mathbf A$ and $\mathbf B$ respectively.
Let $\lambda$ and $\mu$ be arbitrary elements of $R$.
We have for all $i \in \closedint 1 m$, $j \in \closedint 1 n$, by egregious abuse of notation:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Algebra
Category:Matrix Spaces
Matrix is Nonsingular iff Determinant has Multiplicative Inverse

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Then $\mathbf A$ is nonsingular its determinant is a invertible in $R$.
If $R$ is one of the standard number fields $\Q$, $\R$ or $\C$, this translates into:
:$\mathbf A$ is nonsingular its determinant is non-zero.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Also see ==
* Determinant of Inverse Matrix
* Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant: $\mathbf A^{-1} = \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}$
== Sources ==
* : $2.1$: Matrices
* : $\S 7.4$
* : $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.9 \ (2)$
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
Category:Matrix is Nonsingular iff Determinant has Multiplicative Inverse
Category:Inverse Matrices
Category:Determinants
Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be an nonsingular square matrix of order $n$.
Let $\mathbf B = \mathbf A^{-1}$ be the inverse of $\mathbf A$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Then:
:$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$
== Proof ==
Let $\mathbf A$ be nonsingular with $\mathbf B = \mathbf A^{-1}$.
Let $1_R$ denote the unity of $R$.
Let $\mathbf I_n$ denote the unit matrix of order $n$.
Then:
 
This shows that:
:$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$
== Sources ==
* : $\S 7.4$
Category:Matrix is Nonsingular iff Determinant has Multiplicative Inverse
Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Sufficient Condition

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Let the determinant of $\mathbf A$ be invertible in $R$.
Then $\mathbf A$ is a nonsingular matrix.
== Proof ==
Let $\map \det {\mathbf A}$ be invertible in $R$.
From Matrix Product with Adjugate Matrix:
Thus:
Thus $\mathbf A$ is nonsingular, and:
:$\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$
== Also see ==
* Determinant of Inverse Matrix
Category:Matrix is Nonsingular iff Determinant has Multiplicative Inverse
Matrix is Nonsingular iff Rank equals Order

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Then $\mathbf A$ is nonsingular its rank also equals $n$.
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
Category:Inverse Matrices
Category:Rank of Matrix
Category:Nonsingular Matrices
Matrix is Row Equivalent to Echelon Matrix

== Theorem ==
Let $\mathbf A = \sqbrk a_{m n}$ be a matrix of order $m \times n$ over a field $F$.
Then $A$ is row equivalent to an echelon matrix of order $m \times n$.
== Proof ==
Using the operation Row Operation to Clear First Column of Matrix, $\mathbf A$ is converted to $\mathbf B$, which will be in the form:
:$\begin{bmatrix}
0 & \cdots & 0 & 1 & b_{1, j + 1} & \cdots & b_{1 n} \\
0 & \cdots & 0 & 0 & b_{2, j + 1} & \cdots & b_{2 n} \\
\vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & 0 & b_{m, j + 1} & \cdots & b_{m n} \\
\end{bmatrix}$
If some zero rows have appeared, do some further elementary row operations, that is row interchanges, to put them at the bottom.
We then address our attention to the submatrix:
:$\begin{bmatrix}
b_{2, j + 1} & b_{2, j + 2} & \cdots & b_{2 n} \\
b_{3, j + 1} & b_{3, j + 2} & \cdots & b_{3 n} \\
 \vdots & \vdots & \ddots & \vdots \\
b_{m, j + 1} & b_{m, j + 2} & \cdots & b_{m n} \\
\end{bmatrix}$
and perform the same operation on that.
This results in the submatrix being transformed into the form:
:$\begin{bmatrix}
 1 & c_{2, j + 2} & \cdots & c_{2 n} \\
 0 & c_{3, j + 2} & \cdots & c_{3 n} \\
\vdots & \vdots & \ddots & \vdots \\
 0 & c_{m, j + 2} & \cdots & c_{m n} \\
\end{bmatrix}$
Again, we process the submatrix:
:$\begin{bmatrix}
c_{3, j + 2} & \cdots & c_{3 n} \\
 \vdots & \ddots & \vdots \\
c_{m, j + 2} & \cdots & c_{m n} \\
\end{bmatrix}$
Thus we progress, until the entire matrix is in echelon form.
== Examples ==
== Also presented as ==
Some sources use the non-unity variant of the echelon matrix, and so do not seek to ensure that the leading coefficients are equal to $1$.
However, it is often left as an optional step to perform the necessary elementary row operation to multiply each row by the reciprocal of its leading coefficients so as to make it so.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations
* 
* 
Category:Echelon Matrices
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Echelon Matrix/Examples

== Examples of Use of Matrix is Row Equivalent to Echelon Matrix ==
=== Arbitrary Matrix $1$ ===
=== Arbitrary Matrix $2$ ===
=== Arbitrary Matrix $3$ ===
=== Arbitrary Matrix $4$ ===
=== Arbitrary Matrix $5$ ===
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4$
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 1

== Examples of Use of Matrix is Row Equivalent to Echelon Matrix ==
Let $\mathbf A = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end {bmatrix}$
This can be converted into the echelon form:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$
== Proof ==
Using Row Operation to Clear First Column of Matrix we obtain:
:$\mathbf B = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end {bmatrix}$
which happens on the first step, exchanging row $1$ with row $3$.
Then we investigate the submatrix:
:$\mathbf B' = \begin {bmatrix} 1 & 0 \\ 1 & 1 \\ \end {bmatrix}$
Using Row Operation to Clear First Column of Matrix we obtain:
:$\mathbf C' = \begin {bmatrix} 1 & 0 \\ 0 & 1 \\ \end {bmatrix}$
which is obtained by adding $-1$ of row $1$ of $\mathbf B'$ to row $2$ of $\mathbf B'$.
This is the same as adding $-1$ of row $2$ of $\mathbf B$ to row $3$ of $\mathbf B$.
Thus we are left with:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(a)}$
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 2

== Examples of Use of Matrix is Row Equivalent to Echelon Matrix ==
Let $\mathbf A = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ \end {bmatrix}$
This can be converted into the echelon form:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
== Proof ==
Using Row Operation to Clear First Column of Matrix we obtain:
:$\mathbf B = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \\ \end {bmatrix}$
which is obtained by:
:adding $-2$ of row $1$ to row $2$
:adding $-3$ of row $1$ to row $3$.
Then we investigate the submatrix:
:$\mathbf B' = \begin {bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end {bmatrix}$
Using Row Operation to Clear First Column of Matrix we obtain:
:$\mathbf C' = \begin {bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ \end {bmatrix}$
which is obtained by adding $-1$ of row $1$ of $\mathbf B'$ to row $2$ of $\mathbf B'$.
Thus we are left with:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(b)}$
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 3

== Examples of Use of Matrix is Row Equivalent to Echelon Matrix ==
Let $\mathbf A = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$
This can be converted into the echelon form:
:$\mathbf E = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end {bmatrix}$
== Proof ==
Using Row Operation to Clear First Column of Matrix we obtain:
:$\mathbf B = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$
which is obtained by adding $-2$ of row $1$ to row $2$.
Then we investigate the submatrix:
:$\mathbf B' = \begin {bmatrix} 0 & 0 & -1 \\ 0 & 1 & 1 \\ \end {bmatrix}$
Using Row Operation to Clear First Column of Matrix we obtain:
:$\mathbf C' = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {bmatrix}$
which is obtained by:
:$(1): \quad$ exchanging row $1$ of $\mathbf B'$ with row $2$ of $\mathbf B'$.
:$(2): \quad$ multiplying row $2$ of $\mathbf B'$ by $-1$.
Thus we are left with:
:$\mathbf E = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end {bmatrix}$
== Also presented as ==
Some sources use the non-unity variant of the echelon matrix.
Such sources do not require that the leading coefficients necessarily have to equal to $1$.
Hence they consider the final step to convert row $3$ of $\mathbf E$ from $\begin {bmatrix} 0 & 0 & 0 & -1 \end {bmatrix}$ to $\begin {bmatrix} 0 & 0 & 0 & 1 \end {bmatrix}$ as optional.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(c)}$
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 4

== Examples of Use of Matrix is Row Equivalent to Echelon Matrix ==
Let $\mathbf A = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 3 & 3 \\ \end {bmatrix}$
This can be converted into the echelon form:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
== Proof ==
It is noted that $\mathbf A$ is already most of the way there.
It remains to use the elementary row operation:
:$e := r_3 \to r_3 - 3 r_2$
to convert $\mathbf A$ to the form:
:$\mathbf E = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(d)}$
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Echelon Matrix/Examples/Arbitrary Matrix 5

== Examples of Use of Matrix is Row Equivalent to Echelon Matrix ==
Let $\mathbf A = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ -1 & -2 & -3 & -4 & -5 \\ \end {bmatrix}$
This can be converted into the echelon form:
:$\mathbf E = \begin {bmatrix} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$
== Proof ==
This matrix can more easily be handled by direct application of elementary row operations, as follows.
Let $e_1$ be the elementary row operation:
:$e_1 := r_3 \to r_3 - r_1$
which leaves:
$\mathbf A_1 = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & -2 & -4 & -6 & -8 \\ \end {bmatrix}$
Let $e_2$ be the elementary row operation:
:$e_2 := r_3 \to r_3 + 2 r_2$
which leaves:
:$\mathbf A_2 = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$
$\mathbf A_2$ is in non-unity echelon form.
It remains to perform the elementary row operation $e_3$ to convert it into echelon form:
:$e_3 := r_1 \to -r_1$
which leaves:
:$\mathbf E = \begin {bmatrix} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$
== Also presented as ==
Some sources use the non-unity variant of the echelon matrix.
Such sources do not require that the leading coefficients necessarily have to equal to $1$.
Hence they consider the final step to convert row $1$ of $\mathbf E$ from $\begin {bmatrix} -1 & 0 & 1 & 2 & 3 \end {bmatrix}$ to $\begin {bmatrix} 1 & 0 & -1 & -2 & -3 \end {bmatrix}$ as optional.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.4 \ \text {(e)}$
Category:Matrix is Row Equivalent to Echelon Matrix
Matrix is Row Equivalent to Reduced Echelon Matrix

== Theorem ==
Let $\mathbf A = \sqbrk a_{m n}$ be a matrix of order $m \times n$ over a field $F$.
Then $A$ is row equivalent to a reduced echelon matrix of order $m \times n$.
== Proof ==
== Also see ==
* Definition:Gaussian Elimination
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.2$ Elementary Row Operations on Matrices: Theorem $1.5$
* 
* 
Category:Echelon Matrices
Matrix is Singular iff Product with Non-Zero Vector is Zero

== Theorem ==
Let $\mathbf A$ be a square matrix of order $n$.
Then $\mathbf A$ is singular if there exists a vector $\mathbf v$ of $n$ such that:
:$\mathbf v \ne \mathbf 0$
:$\mathbf A \mathbf v = \mathbf 0$
where $\mathbf 0$ is the zero vector.
== Proof ==
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.5$: Determinants: Lemma $2.2.1$
Category:Singular Matrices
Matrix of Bilinear Form Under Change of Basis

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\AA$ and $\BB$ be ordered bases of $M$.
Let $\mathbf M_{\AA, \BB}$ be the change of basis matrix from $\AA$ to $\BB$.
Let $f : M \times M \to R$ be a bilinear form.
Let $\mathbf M_{f, \AA}$ be its matrix relative to $\AA$.
Then its matrix relative to $\BB$ equals:
:$\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$
== Proof ==
Let $m \in M$, and let $\sqbrk m_\AA$ and $\sqbrk m_\BB$ denote its coordinate vectors relative to $\AA$ and $\BB$.
We have:
Thus $\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$.
Category:Bilinear Forms (Linear Algebra)
Category:Change of Basis