The general linear group of degree $n$ for $3$D matrices $GL(n,n,p;F)$

TL;DR

This paper defines the determinant and inverse for 3D matrices over a field, and introduces a new general linear group for 3D matrices, extending classical linear algebra concepts to higher dimensions.

Contribution

It constructs the general linear group of degree n for 3D matrices, expanding the algebraic framework for 3D matrix analysis.

Findings

Defined determinant for 3D matrices over a field

Introduced inverse for 3D matrices

Constructed the general linear group for 3D matrices

Abstract

In this article we give the meaning of the determinant for 3D matrices with elements from a field F, and the meaning of 3D inverse matrix. Based on my previous work titled '3D Matrix Rings', we want to constructed the 'general linear group of degree $n$ for 3D matrices, which i mark with $GL(n,n,p;F)$' for 3D-matrices, analog to 'general linear group of degree $n$' known.

Figures

Equations

$${\mathcal{M}}_{m\times n\times p}(\mathbf{F})=\left\{a_{i,j,k}|a_{i,j,k}\in F-\text{field }\forall i=\overline{1,m};j=\overline{1,n};k=\overline{1,p}\right\}$$

$${\mathbf{C}}_{m\times n\times p}(\mathbf{F})=\left\{c_{i,j,k}|c_{i,j,k}=a_{i,j,k}+b_{i,j,k},\forall i=\overline{1,m};j=\overline{1,n};k=\overline{1,p}\right\}$$

$$\mathbf{A}$$

$$\mathbf{B}$$

$${\mathbf{O}=\mathbf{O}}_{m\times n\times p}=\left\{{\left(0_{\mathbf{F}}\right)}_{i,j,k}|\forall i=\overline{1,m};j=\overline{1,n};k=\overline{1,p}\right\}\in {\mathcal{M}}_{m\times n\times p}(\mathbf{F})$$

$${\mathbf{A}}_{m\times n\times p}=\left\{\left(a_{i,j,k}\right)|\forall i=\overline{1,m};j=\overline{1,n};k=\overline{1,p}\right\}\in {\mathcal{M}}_{m\times n\times p}(\mathbf{F})$$

$$-{\mathbf{A}}_{m\times n\times p}=\left\{\left(-a_{i,j,k}\right)|\forall i=\overline{1,m};j=\overline{1,n};k=\overline{1,p}\right\}\in {\mathcal{M}}_{m\times n\times p}(\mathbf{F}).$$

$$a_{i,j,k}+\left(-a_{i,j,k}\right)=0_{\mathbf{F}},\forall i=\overline{1,m};j=\overline{1,n};k=\overline{1,p},$$

$${\mathbf{A}}_{m\times n\times p}+\left(-{\mathbf{A}}_{m\times n\times p}\right)$$

$$\forall\mathbf{A=}\left[\begin{array}[]{c}\left(\begin{array}[]{cccc}a_{11p}&a_{12p}&\cdots&a_{1np}\\ a_{21p}&a_{22p}&\cdots&a_{2np}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1p}&a_{n2p}&\cdots&a_{nnp}\end{array}\right)\\ \mathbf{\vdots}\\ \left(\begin{array}[]{cccc}a_{112}&a_{122}&\cdots&a_{1n2}\\ a_{212}&a_{222}&\cdots&a_{2n2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n12}&a_{n22}&\cdots&a_{nn2}\end{array}\right)\\ \left(\begin{array}[]{cccc}a_{111}&a_{121}&\cdots&a_{1n1}\\ a_{211}&a_{221}&\cdots&a_{2n1}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n11}&a_{n21}&\cdots&a_{nn1}\end{array}\right)\end{array}\right];\mathbf{B=}\left[\begin{array}[]{c}\left(\begin{array}[]{cccc}b_{11p}&b_{12p}&\cdots&b_{1np}\\ b_{21p}&b_{22p}&\cdots&b_{2np}\\ \vdots&\vdots&\ddots&\vdots\\ b_{n1p}&b_{n2p}&\cdots&b_{nnp}\end{array}\right)\\ \mathbf{\vdots}\\ \left(\begin{array}[]{cccc}b_{112}&b_{122}&\cdots&b_{1n2}\\ b_{212}&b_{222}&\cdots&b_{2n2}\\ \vdots&\vdots&\ddots&\vdots\\ b_{n12}&b_{n22}&\cdots&b_{nn2}\end{array}\right)\\ \left(\begin{array}[]{cccc}b_{111}&b_{121}&\cdots&b_{1n1}\\ b_{211}&b_{221}&\cdots&b_{2n1}\\ \vdots&\vdots&\ddots&\vdots\\ b_{n11}&b_{n21}&\cdots&b_{nn1}\end{array}\right)\end{array}\right]$$

$$\mathbf{A=}\left[\begin{array}[]{c}\mathbf{A}_{p}\\ \vdots\\ \mathbf{A}_{2}\\ \mathbf{A}_{1}\end{array}\right],\mathbf{B=}\left[\begin{array}[]{c}\mathbf{B}_{p}\\ \vdots\\ \mathbf{B}_{2}\\ \mathbf{B}_{1}\end{array}\right]\in\mathcal{M}_{n\times n\times p}(\mathbf{F})$$

$$\mathbf{C=A\odot B=\left[\begin{array}[]{c}\mathbf{A}_{p}\\ \vdots\\ \mathbf{A}_{2}\\ \mathbf{A}_{1}\end{array}\right]\odot}\left[\begin{array}[]{c}\mathbf{B}_{p}\\ \vdots\\ \mathbf{B}_{2}\\ \mathbf{B}_{1}\end{array}\right]=\mathbf{\left[\begin{array}[]{c}\mathbf{A}_{p}\ast\mathbf{B}_{p}\\ \vdots\\ \mathbf{A}_{2}\ast\mathbf{B}_{2}\\ \mathbf{A}_{1}\ast\mathbf{B}_{1}\end{array}\right]}$$

$$a_{1\times 1\times p}=\left[\begin{array}[]{c}\left(\alpha_{11p}\right)\\ \vdots\\ \left(\alpha_{112}\right)\\ \left(\alpha_{111}\right)\end{array}\right],$$

$$\mathbf{A}_{m\times n\times p}\mathbf{=}\left[\begin{array}[]{c}\left(\begin{array}[]{cccc}a_{11p}&a_{12p}&\cdots&a_{1np}\\ a_{21p}&a_{22p}&\cdots&a_{2np}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1p}&a_{m2p}&\cdots&a_{mnp}\end{array}\right)\\ \vdots\\ \left(\begin{array}[]{cccc}a_{112}&a_{122}&\cdots&a_{1n2}\\ a_{212}&a_{222}&\cdots&a_{2n2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m12}&a_{m22}&\cdots&a_{mn2}\end{array}\right)\\ \left(\begin{array}[]{cccc}a_{111}&a_{121}&\cdots&a_{1n1}\\ a_{211}&a_{221}&\cdots&a_{2n1}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m11}&a_{m21}&\cdots&a_{mn1}\end{array}\right)\end{array}\right],$$

$${\mathbf{B}}_{m\times n\times p}$$

$$=\left[\begin{array}[]{c}\left(\begin{array}[]{cccc}\alpha_{11p}\cdot a_{11p}&\alpha_{11p}\cdot a_{12p}&\cdots&\alpha_{11p}\cdot a_{1np}\\ \alpha_{11p}\cdot a_{21p}&\alpha_{11p}\cdot a_{22p}&\cdots&\alpha_{11p}\cdot a_{2np}\\ \vdots&\vdots&\ddots&\vdots\\ \alpha_{11p}\cdot a_{m1p}&\alpha_{11p}\cdot a_{m2p}&\cdots&\alpha_{11p}\cdot a_{mnp}\end{array}\right)\\ \mathbf{\vdots}\\ \left(\begin{array}[]{cccc}\alpha_{112}\cdot a_{112}&\alpha_{112}\cdot a_{122}&\cdots&\alpha_{112}\cdot a_{1n2}\\ \alpha_{112}\cdot a_{212}&\alpha_{112}\cdot a_{222}&\cdots&\alpha_{112}\cdot a_{2n2}\\ \vdots&\vdots&\ddots&\vdots\\ \alpha_{112}\cdot a_{m12}&\alpha_{112}\cdot a_{m22}&\cdots&\alpha_{112}\cdot a_{mn2}\end{array}\right)\\ \left(\begin{array}[]{cccc}\alpha_{111}\cdot a_{111}&\alpha_{111}\cdot a_{121}&\cdots&\alpha_{111}\cdot a_{1n1}\\ \alpha_{111}\cdot a_{211}&\alpha_{111}\cdot a_{221}&\cdots&\alpha_{111}\cdot a_{2n1}\\ \vdots&\vdots&\ddots&\vdots\\ \alpha_{111}\cdot a_{m11}&\alpha_{111}\cdot a_{m21}&\cdots&\alpha_{111}\cdot a_{mn1}\end{array}\right)\end{array}\right]$$

$$\ast :{\mathcal{M}}_{1\times 1\times p}(\mathbf{F})\times {\mathcal{M}}_{m\times n\times p}(\mathbf{F})⟶{\mathcal{M}}_{m\times n\times p}(\mathbf{F}).$$

$$a_{1\times 1\times 3}=\left[\begin{array}[]{c}\left(2\right)\\ \left(5\right)\\ \left(3\right)\end{array}\right]\ \$$

$$\mathbf{A}_{2\times 2\times 3}=\left[\begin{array}[]{c}\left(\begin{array}[]{cc}2&3\\ 4&5\end{array}\right)\\ \left(\begin{array}[]{cc}5&0\\ 9&1\end{array}\right)\\ \left(\begin{array}[]{cc}1&4\\ 5&3\end{array}\right)\end{array}\right].$$

$${\mathbf{B}}_{2\times 2\times 3}$$

$$\det\left(\mathbf{A}_{n\times n\times p}\right)=\left[\begin{array}[]{c}\det\left(\mathbf{A}_{n\times n,p}\right)\\ \mathbf{\vdots}\\ \det\left(\mathbf{A}_{n\times n,2}\right)\\ \det\left(\mathbf{A}_{n\times n,1}\right)\end{array}\right]$$

$$\det\left(\mathbf{A}_{n\times n,1}\right)=\det\left(\begin{array}[]{cccc}a_{111}&a_{121}&\cdots&a_{1n1}\\ a_{211}&a_{221}&\cdots&a_{2n1}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n11}&a_{n21}&\cdots&a_{nn1}\end{array}\right);$$

$$\det\left(\mathbf{A}_{n\times n,2}\right)=\det\left(\begin{array}[]{cccc}a_{112}&a_{122}&\cdots&a_{1n2}\\ a_{212}&a_{222}&\cdots&a_{2n2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n12}&a_{n22}&\cdots&a_{nn2}\end{array}\right);...;$$

$$\det\left(\mathbf{A}_{n\times n,p}\right)=\det\left(\begin{array}[]{cccc}a_{11p}&a_{12p}&\cdots&a_{1np}\\ a_{21p}&a_{22p}&\cdots&a_{2np}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1p}&a_{n2p}&\cdots&a_{nnp}\end{array}\right)$$

$$\det \left({\mathbf{A}}_{n\times n\times p}\right)\ne \widetilde{{}^a{0}_{\mathbf{F}}}⟺\det \left({\mathbf{A}}_{n\times n,i}\right)\ne 0_{\mathbf{F}},\forall i=\overline{1,p}.$$

$$\det\left(\mathbf{A}_{2\times 2\times 3}\right)=\left[\begin{array}[]{c}\det\left(\begin{array}[]{cc}2&3\\ 4&5\end{array}\right)\\ \det\left(\begin{array}[]{cc}5&0\\ 9&1\end{array}\right)\\ \det\left(\begin{array}[]{cc}1&4\\ 5&3\end{array}\right)\end{array}\right]=\left[\begin{array}[]{c}\left(-2\right)\\ \left(5\right)\\ \left(-17\right)\end{array}\right]$$

$$\widehat{a_{1\times 1\times p}}=\left[\begin{array}[]{c}\left(\frac{1}{\alpha_{11p}}\right)\\ \vdots\\ \left(\frac{1}{\alpha_{112}}\right)\\ \left(\frac{1}{\alpha_{111}}\right)\end{array}\right].$$

$$\widehat{a_{1\times 1\times 3}}=\left[\begin{array}[]{c}\left(\frac{1}{2}\right)\\ \left(\frac{1}{5}\right)\\ \left(\frac{1}{3}\right)\end{array}\right].$$

$$\forall \mathbf{A},\mathbf{B},\mathbf{C}\in {\mathcal{M}}_{n\times n\times p}(\mathbf{F}),\left[\mathbf{A}\odot \mathbf{B}\right]\odot \mathbf{C}=\mathbf{A}\odot \left[\mathbf{B}\odot \mathbf{C}\right]$$

$$\ \vskip 12.0pt plus 4.0pt minus 4.0pt\mathbf{I}_{n\times n\times p}=\left[\begin{array}[]{c}\mathbf{I}_{n\times n,p}\\ \vdots\\ \mathbf{I}_{n\times n,2}\\ \mathbf{I}_{n\times n,1}\end{array}\right],$$

$$\odot$$

Full text

1. Introduction, 3D matrix

This paper comes as a continuation of the ideas that arise based on my previous work, of the 3D matrix ring with element from an whatever field $\mathbf{F}$ (see [1]). At this point we are making a brief summary associated with 3D matrices and the 3D matrix presented in [1]. Our objective is to constructed the ’general linear group of degree $n$ for 3D matrices, which i mark with $GL(n,n,p;F)$’ for 3D-matrices, analog to ’general linear group of degree $n$’ known. For this we need new notions, which we will give in the following points.

Definition 1**.**

[1]** 3-dimensional $m\times n\times p$ matrix will call, a matrix which has: m-horizontal layers (analogous to m-rows), n-vertical page (analogue with n-columns in the usual matrices) and p-vertical layers (p-1 of which are hidden).

The set of these matrixes the write how:

[TABLE]

1.1. ADDITION OF 3D MATRIX

Definition 2**.**

[1]** The addition of two matrices $\mathbf{A,B}\in\mathcal{M}_{m\times n\times p}(\mathbf{F})$ we will call the matrix:

[TABLE]

The appearance of the addition of $m\times n\times p,$ 3D matrices will be as in Figure 1, where matrices $\mathbf{A}$ and $\mathbf{B}$ have the following appearance,

[TABLE]

Definition 3**.**

[1*]*The 3-D, Zero matrix $m\times n\times p,$ we will called the matrix that has all its elements zero.

[TABLE]

Definition 4**.**

[1]** The opposite matric of an matrice

[TABLE]

will, called matrix

[TABLE]

Where $\left(-a_{i,j,k}\right)$ is a opposite element of element $a_{i,j,k}\in\mathbf{F,}$ so

[TABLE]

and $\left(\mathbf{F,+,\cdot}\right)$ is field, which satisfies the condition

[TABLE]

Theorem 1**.**

[1]$\left(\mathcal{M}_{m\times n\times p}(\mathbf{F}),+\right)$* is abelian grup.*

1.2. THE MULTIPLICATION OF $n\times n\times p$ MATRICES

In the same way, as have the meaning of a 3D $3\times 3\times 3$ matrix multiplication to [1], we give the definition of 3D matrix multiplication for  $\mathbf{A,B}\in\mathcal{M}_{n\times n\times p}(\mathbf{F})$.

Definition 5**.**

[1]** The multiplication of two matrices $\mathbf{A,B}\in\mathcal{M}_{n\times n\times p}(\mathbf{F})$ we will call the matrix $\mathbf{C=A\odot B}\in\mathcal{M}_{n\times n\times p}(\mathbf{F}),$ calculated as follows:

Matrices will normally have the appearance:

[TABLE]

If we write more briefly

[TABLE]

where $\mathbf{A}_{i}$ and $\mathbf{B}_{i}$ are the $n\times n$ matrices $\forall i=\overline{1,p}.$ Then

[TABLE]

where action $"\ast"$ is the usual multiplication of matrices.

2. Multi-Scalars and Multiplication of 3D matrices with multi-scalar

In this section we will introduce the concept of ’multi-scalar’, and we will give a clear idea of the multiplication of 3D matrices with multi-scalar.

Definition 6**.**

Multi-scalar will call one $1\times 1\times p,$ 3D matrix.

Remark 1**.**

A multi-scalar \ a_{1\times 1\times p}=\left[\begin{array}[]{c}\left(\alpha_{11p}\right)\\ \vdots\\ \left(\alpha_{112}\right)\\ \left(\alpha_{111}\right)\end{array}\right], we will call ”absolutely” different from zero only if $\alpha_{11i}\neq 0_{\mathbf{F}},\forall i=\overline{1,p}.$ For the ”absolutely zero” multi-scalar we will use the note $\widetilde{{}^{a}0_{\mathbf{F}}},$ wich is \ \widetilde{{}^{a}0_{\mathbf{F}}}=\left[\begin{array}[]{c}\left(0_{\mathbf{F},p}\right)\\ \vdots\\ \left(0_{\mathbf{F},2}\right)\\ \left(0_{\mathbf{F,}1}\right)\end{array}\right].

Let’s have a **multi-scalar  **

[TABLE]

and 3D matrix

[TABLE]

Definition 7**.**

Multiplication of 3D-matrix $\mathbf{A}_{m\times n\times p}\in\mathcal{M}_{m\times n\times p}(\mathbf{F})$  with multi-scalar $a_{1\times 1\times p}$ will we call the 3D matrix $\mathbf{B}_{m\times n\times p}\in\mathcal{M}_{m\times n\times p}(\mathbf{F}),$ calculated as follows:

[TABLE]

[TABLE]

So multiplication of the 3D matrix with a multi-scalar is a function

[TABLE]

Example 1**.**

Let’s have a multi-scalar

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**

and 3D matrix

[TABLE]

Matrix obtained by multiplying the 3D matrix $A_{2\times 2\times 3}$ with multi-scalar $a_{1\times 1\times 3},$ it is the matrix:

[TABLE]

2.1. DETERMINANTS OF 3D-MATRICES

Regarding the determinant we will only talk about form matrices $\mathbf{A}_{n\times n\times p}\in\mathcal{M}_{n\times n\times p}(\mathbf{F}),$ ie for matrices that vertical layers have square matrices.

Definition 8**.**

Determinant of the matrix $\mathbf{A}_{n\times n\times p}\in\mathcal{M}_{n\times n\times p}(\mathbf{F}),$ we will call the multi-scalar

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

Referring to the multi-scalar note ’absolutely different from zero’, we say that for a 3D matrix $\mathbf{A}_{n\times n\times p},$

[TABLE]

Example 2**.**

Determinant of the matrix $\mathbf{A}_{2\times 2\times 3},$ of example 1, is multi-scalar:

[TABLE]

Definition 9**.**

Inverted* of the multi-scalar a_{1\times 1\times p}=\left[\begin{array}[]{c}\left(\alpha_{11p}\right)\\ \vdots\\ \left(\alpha_{112}\right)\\ \left(\alpha_{111}\right)\end{array}\right], will called the multi-scalar:*

[TABLE]

Example 3**.**

Inverted* of the multi-scalar  a_{1\times 1\times 3}=\left[\begin{array}[]{c}\left(2\right)\\ \left(5\right)\\ \left(3\right)\end{array}\right], is the multi-scalar:*

[TABLE]

3. THE MULTIPLICATION GROUP OF 3D MATRICES

Referring to [1], we have a 3D Matrix Ring, and in that paper we have shown the possibility of unitary ring. The rest of the assertions that lead us from unitary ring to a Skew-Field are summing up in this

Theorem 2**.**

The structure $\left(\mathcal{M}_{n\times n\times p}(\mathbf{F}),\odot\right)$ is a unitary semi-group.

Proof.

We show first that:

[TABLE]

truly,

\left[\mathbf{A}\odot\mathbf{B}\right]\odot\mathbf{C=}\left(\left[\begin{array}[]{c}\mathbf{A}_{p}\\ \vdots\\ \mathbf{A}_{2}\\ \mathbf{A}_{1}\end{array}\right]\odot\left[\begin{array}[]{c}\mathbf{B}_{p}\\ \vdots\\ \mathbf{B}_{2}\\ \mathbf{B}_{1}\end{array}\right]\right)\odot\left[\begin{array}[]{c}\mathbf{C}_{p}\\ \vdots\\ \mathbf{C}_{2}\\ \mathbf{C}_{1}\end{array}\right]=\left[\begin{array}[]{c}\mathbf{A}_{p}\cdot\mathbf{B}_{p}\\ \vdots\\ \mathbf{A}_{2}\cdot\mathbf{B}_{2}\\ \mathbf{A}_{1}\cdot\mathbf{B}_{1}\end{array}\right]\odot\left[\begin{array}[]{c}\mathbf{C}_{p}\\ \vdots\\ \mathbf{C}_{2}\\ \mathbf{C}_{1}\end{array}\right]=\left[\begin{array}[]{c}\left(\mathbf{A}_{p}\cdot\mathbf{B}_{p}\right)\cdot\mathbf{C}_{p}\\ \vdots\\ \left(\mathbf{A}_{2}\cdot\mathbf{B}_{2}\right)\cdot\mathbf{C}_{2}\\ \left(\mathbf{A}_{1}\cdot\mathbf{B}_{1}\right)\cdot\mathbf{C}_{1}\end{array}\right]=\left[\begin{array}[]{c}\mathbf{A}_{p}\cdot\left(\mathbf{B}_{p}\cdot\mathbf{C}_{p}\right)\\ \vdots\\ \mathbf{A}_{2}\cdot\left(\mathbf{B}_{2}\cdot\mathbf{C}_{2}\right)\\ \mathbf{A}_{1}\cdot\left(\mathbf{B}_{1}\cdot\mathbf{C}_{1}\right)\end{array}\right]=\left[\begin{array}[]{c}\mathbf{A}_{p}\\ \vdots\\ \mathbf{A}_{2}\\ \mathbf{A}_{1}\end{array}\right]\odot\left[\begin{array}[]{c}\mathbf{B}_{p}\cdot\mathbf{C}_{p}\\ \vdots\\ \mathbf{B}_{2}\cdot\mathbf{C}_{2}\\ \mathbf{B}_{1}\cdot\mathbf{C}_{1}\end{array}\right]

=\left[\begin{array}[]{c}\mathbf{A}_{p}\\ \vdots\\ \mathbf{A}_{2}\\ \mathbf{A}_{1}\end{array}\right]\odot\left(\left[\begin{array}[]{c}\mathbf{B}_{p}\\ \vdots\\ \mathbf{B}_{2}\\ \mathbf{B}_{1}\end{array}\right]\odot\left[\begin{array}[]{c}\mathbf{C}_{p}\\ \vdots\\ \mathbf{C}_{2}\\ \mathbf{C}_{1}\end{array}\right]\right)=\mathbf{A}\odot\left[\mathbf{B}\odot\mathbf{C}\right].

From the definition of multiplication to 3D matrices it is clear that:

[TABLE]

is unit element of $\mathcal{M}_{n\times n\times p}(\mathbf{F})$, related to multiplication. ∎

Remark 2**.**

We mark with $\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),$ 3D matrix with determinants ’absolutely different from zero’, ie $\det\left(\mathbf{A}_{n\times n\times p}\right)\neq\widetilde{{}^{a}0_{\mathbf{F}}}$, (So vertical layers of 3D matrix, are 2D non-singular matrices.).

Proposition 1**.**

The set of  $\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),$ is closed regarding multiplication.

Well!

[TABLE]

Proof.

Let’s have

[TABLE]

[TABLE]

So all 2D matrices $\mathbf{A}_{n\times n,i}$ and $\mathbf{B}_{n\times n,i}$, $\forall i=\overline{1,p}$ , are non-singular matrices. A well-known result in Linear algebra, but also mentioned in [2], in pp.230, and [5] in pp317, which is

[TABLE]

We use this result as the vertical layers of 3D matrix are 2D-matries, and so we have that:

\det\left(\mathbf{A}\odot\mathbf{B}\right)\mathbf{=}\det\left(\left[\begin{array}[]{c}\mathbf{A}_{n\times n,p}\\ \vdots\\ \mathbf{A}_{n\times n,2}\\ \mathbf{A}_{n\times n,1}\end{array}\right]\odot\left[\begin{array}[]{c}\mathbf{B}_{n\times n,p}\\ \vdots\\ \mathbf{B}_{n\times n,2}\\ \mathbf{B}_{n\times n,1}\end{array}\right]\right)=\det\left[\begin{array}[]{c}\mathbf{A}_{n\times n,p}\cdot\mathbf{B}_{n\times n,p}\\ \vdots\\ \mathbf{A}_{n\times n,2}\cdot\mathbf{B}_{n\times n,2}\\ \mathbf{A}_{n\times n,1}\cdot\mathbf{B}_{n\times n,1}\end{array}\right]=\left[\begin{array}[]{c}\det\left(\mathbf{A}_{n\times n,p}\cdot\mathbf{B}_{n\times n,p}\right)\\ \vdots\\ \det\left(\mathbf{A}_{n\times n,2}\cdot\mathbf{B}_{n\times n,2}\right)\\ \det\left(\mathbf{A}_{n\times n,1}\cdot\mathbf{B}_{n\times n,1}\right)\end{array}\right]\overset{[2]}{=}\left[\begin{array}[]{c}\det\left(\mathbf{A}_{n\times n,p}\right)\cdot\det\left(\mathbf{B}_{n\times n,p}\right)\\ \vdots\\ \det\left(\mathbf{A}_{n\times n,2}\right)\cdot\det\left(\mathbf{B}_{n\times n,2}\right)\\ \det\left(\mathbf{A}_{n\times n,1}\right)\cdot\det\left(\mathbf{B}_{n\times n,1}\right)\end{array}\right]=

=\left[\begin{array}[]{c}\det\left(\mathbf{A}_{n\times n,p}\right)\\ \vdots\\ \det\left(\mathbf{A}_{n\times n,2}\right)\\ \det\left(\mathbf{A}_{n\times n,1}\right)\end{array}\right]\odot\left[\begin{array}[]{c}\det\left(\mathbf{B}_{n\times n,p}\right)\\ \vdots\\ \det\left(\mathbf{B}_{n\times n,2}\right)\\ \det\left(\mathbf{B}_{n\times n,1}\right)\end{array}\right]=\det\left(\mathbf{A}\right)\odot\det\left(\mathbf{B}\right)\neq\widetilde{{}^{a}0_{\mathbf{F}}}\Rightarrow

$\Rightarrow\mathbf{A\odot B}\in\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}).$ ∎

Theorem 3**.**

The structure $\left(\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),\odot\right)$ is a Group.

Set of 3D matrix $\mathbf{A}_{n\times n\times p}\in\mathcal{M}_{n\times n\times p}(\mathbf{F}),$ with determinants ’absolutely different from zero’, ie $\det\left(\mathbf{A}_{n\times n\times p}\right)\neq\widetilde{{}^{a}0_{\mathbf{F}}}$, associated with ordinary multiplication is a Group.

Proof.

1. It is clear that the set $\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F})$ is sub-set of $\mathcal{M}_{n\times n\times p}(\mathbf{F})$, and in the foregoing assertion we showed that the multiplication is closed in this set, so we have that $\left(\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),\odot\right)$ is subsemigroup of semigroup $\left(\mathcal{M}_{n\times n\times p}(\mathbf{F}),\odot\right),$ see [3], [4].

2. It is clear that the $\mathbf{I}_{n\times n\times p}\in\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F})$, after

[TABLE]

3. Show that: $\forall\mathbf{A}_{n\times n\times p}\in\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),($such that $\det\left(\mathbf{A}_{n\times n\times p}\right)\neq\widetilde{{}^{a}0_{\mathbf{F}}})$, $\exists\mathbf{A}_{n\times n\times p}^{-1}\in\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),$  such that

[TABLE]

Let’s have

[TABLE]

So we have, that: $\forall\mathbf{A}_{n\times n,i}$ (as a 2D matrix), $\exists\mathbf{A}_{n\times n,i}^{-1}$ such that $\mathbf{A}_{n\times n,i}\cdot\mathbf{A}_{n\times n,i}^{-1}=\mathbf{I}_{n\times n,i},\forall i=\overline{1,p}.$ From this, we can write that, the inverse of 3D matrix \mathbf{A}_{n\times n\times p}=\left[\begin{array}[]{c}\mathbf{A}_{n\times n,p}\\ \vdots\\ \mathbf{A}_{n\times n,2}\\ \mathbf{A}_{n\times n,1}\end{array}\right] is a 3D matrix \mathbf{A}_{n\times n\times p}^{-1}=\left[\begin{array}[]{c}\mathbf{A}_{n\times n,p}^{-1}\\ \vdots\\ \mathbf{A}_{n\times n,2}^{-1}\\ \mathbf{A}_{n\times n,1}^{-1}\end{array}\right], because the

[TABLE]

.

Where

[TABLE]

∎

4. FINDING OF 3D-INVERSE MATRIX

In this section, we provide a way to find the 3D reverse matrix of a 3D matrix.

Proposition 2**.**

3D inverse matrix of the matrix $\mathbf{A}_{n\times n\times p}\in\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),$ we called the 3D-matrix $\mathbf{A}_{n\times n\times p}^{-1}\in\mathcal{M}_{n\times n\times p}^{\ast}(\mathbf{F}),$ which has the following form:

[TABLE]

where $\overline{\overline{\mathbf{A}_{n\times n\times p}}}$ is the adjugate matrix of $\mathbf{A}_{n\times n\times p}$ (exactly page by page)and has the form:

[TABLE]

I give a clearer view of matrices $\mathbf{A}_{n\times n\times p}^{-1}$, as follows:

[TABLE]

Proof.

Verified with ease that:

[TABLE]

Where

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Let

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and

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to prove that it is true:

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really:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

∎

Example 4**.**

Let’s have  A_{3\times 3\times 2}=\left[\begin{array}[]{c}\left(\begin{array}[]{ccc}3&1&5\\ 0&2&1\\ 1&7&4\end{array}\right)\\ \left(\begin{array}[]{ccc}1&2&4\\ 8&1&1\\ 3&1&0\end{array}\right)\end{array}\right]\in\mathcal{M}_{3\times 3\times 2}^{\ast}(\mathbb{R}). Find its inverse matrix?

Solution 1**.**

The inverse matrix has form:

[TABLE]

we see that

[TABLE]

And

[TABLE]

Then,

[TABLE]

[TABLE]