(
1
2
3
4
)
⊗
(
5
6
7
8
)
=
(
1
⋅
(
5
6
7
8
)
2
⋅
(
5
6
7
8
)
3
⋅
(
5
6
7
8
)
4
⋅
(
5
6
7
8
)
)
=
(
5
6
10
12
7
8
14
16
15
18
20
24
21
24
28
32
)
{\displaystyle {\begin{pmatrix}1&2\\3&4\end{pmatrix}}\otimes {\begin{pmatrix}5&6\\7&8\end{pmatrix}}={\begin{pmatrix}1\cdot {\begin{pmatrix}5&6\\7&8\end{pmatrix}}&2\cdot {\begin{pmatrix}5&6\\7&8\end{pmatrix}}\\\\3\cdot {\begin{pmatrix}5&6\\7&8\end{pmatrix}}&4\cdot {\begin{pmatrix}5&6\\7&8\end{pmatrix}}\end{pmatrix}}={\begin{pmatrix}5&6&10&12\\7&8&14&16\\15&18&20&24\\21&24&28&32\end{pmatrix}}}
(
1
3
2
1
0
0
1
2
2
)
⊗
(
0
5
5
0
1
1
)
=
(
1
⋅
(
0
5
5
0
1
1
)
3
⋅
(
0
5
5
0
1
1
)
2
⋅
(
0
5
5
0
1
1
)
1
⋅
(
0
5
5
0
1
1
)
0
⋅
(
0
5
5
0
1
1
)
0
⋅
(
0
5
5
0
1
1
)
1
⋅
(
0
5
5
0
1
1
)
2
⋅
(
0
5
5
0
1
1
)
2
⋅
(
0
5
5
0
1
1
)
)
=
(
0
5
0
15
0
10
5
0
15
0
10
0
1
1
3
3
2
2
0
5
0
0
0
0
5
0
0
0
0
0
1
1
0
0
0
0
0
5
0
10
0
10
5
0
10
0
10
0
1
1
2
2
2
2
)
{\displaystyle {\begin{pmatrix}1&3&2\\1&0&0\\1&2&2\end{pmatrix}}\otimes {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}={\begin{pmatrix}1\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}&3\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}&2\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}\\\\1\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}&0\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}&0\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}\\\\1\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}&2\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}&2\cdot {\begin{pmatrix}0&5\\5&0\\1&1\end{pmatrix}}\end{pmatrix}}={\begin{pmatrix}0&5&0&15&0&10\\5&0&15&0&10&0\\1&1&3&3&2&2\\0&5&0&0&0&0\\5&0&0&0&0&0\\1&1&0&0&0&0\\0&5&0&10&0&10\\5&0&10&0&10&0\\1&1&2&2&2&2\end{pmatrix}}}
A Kronecker-szorzás nem kommutatív , ami azt jelenti, hogy általában
A
⊗
B
≠
B
⊗
A
{\displaystyle A\otimes B\neq B\otimes A}
Azonban mindig vannak
P
,
Q
{\displaystyle P,Q}
permutációmátrixok , hogy
A
⊗
B
=
P
(
B
⊗
A
)
Q
{\displaystyle A\otimes B=P(B\otimes A)Q}
Hogyha
A
{\displaystyle A}
B
{\displaystyle B}
P
=
Q
T
{\displaystyle P=Q^{T}}
A Kronecker-szorzás bilineáris , vagyis
A
⊗
(
B
+
C
)
=
A
⊗
B
+
A
⊗
C
{\displaystyle A\otimes (B+C)=A\otimes B+A\otimes C}
(
B
+
C
)
⊗
A
=
B
⊗
A
+
C
⊗
A
{\displaystyle (B+C)\otimes A=B\otimes A+C\otimes A}
λ
(
A
⊗
B
)
=
(
λ
A
)
⊗
B
=
A
⊗
(
λ
B
)
{\displaystyle \lambda (A\otimes B)=(\lambda A)\otimes B=A\otimes (\lambda B)}
A Kronecker-szorzás asszociatív :
A
⊗
(
B
⊗
C
)
=
(
A
⊗
B
)
⊗
C
{\displaystyle A\otimes (B\otimes C)=(A\otimes B)\otimes C}
A transzponáltakra teljesül, hogy:
(
A
⊗
B
)
T
=
A
T
⊗
B
T
{\displaystyle (A\otimes B)^{T}=A^{T}\otimes B^{T}}
A komplex konjugált mátrixra:
A
⊗
B
¯
=
A
¯
⊗
B
¯
{\displaystyle {\overline {A\otimes B}}={\overline {A}}\otimes {\overline {B}}}
Az adjungált mátrixra teljesül, hogy:
(
A
⊗
B
)
∗
=
A
∗
⊗
B
∗
{\displaystyle (A\otimes B)^{*}=A^{*}\otimes B^{*}}
A Kronecker-szorzat rangja:
R
a
n
g
(
A
⊗
B
)
=
R
a
n
g
(
A
)
⋅
R
a
n
g
(
B
)
{\displaystyle \mathrm {Rang} (A\otimes B)=\mathrm {Rang} (A)\cdot \mathrm {Rang} (B)}
Ha
A
{\displaystyle A}
n
×
n
{\displaystyle n\times n}
B
{\displaystyle B}
m
×
m
{\displaystyle m\times m}
determinánsa
det
(
A
⊗
B
)
=
det
m
(
A
)
det
n
(
B
)
{\displaystyle \det(A\otimes B)={\det }^{m}(A)\,{\det }^{n}(B)}
Ha
(
λ
i
)
i
=
1..
n
{\displaystyle (\lambda _{i})_{i=1..n}\,}
A
{\displaystyle A}
(
μ
j
)
j
=
1..
m
{\displaystyle (\mu _{j})_{j=1..m}\,}
B
{\displaystyle B}
sajátértékei , akkor
(
λ
i
μ
j
)
i
=
1..
n
j
=
1..
m
{\displaystyle (\lambda _{i}\,\mu _{j})_{i=1..n \atop j=1..m}}
A
⊗
B
{\displaystyle A\otimes B}
Ha
A
,
B
{\displaystyle A,B}
invertálható , akkor
(
A
⊗
B
)
−
1
=
A
−
1
⊗
B
−
1
{\displaystyle (A\otimes B)^{-1}=A^{-1}\otimes B^{-1}}
Legyenek
A
,
B
,
C
{\displaystyle A,B,C}
D
{\displaystyle D}
A
:
m
×
n
{\displaystyle A:m\times n}
B
:
p
×
q
{\displaystyle B:p\times q}
C
:
n
×
r
{\displaystyle C:n\times r}
D
:
q
×
s
{\displaystyle D:q\times s}
dimenziókkal; ekkor léteznek az
A
C
{\displaystyle AC}
B
D
{\displaystyle BD}
[1]
A
C
⊗
B
D
=
(
A
⊗
B
)
(
C
⊗
D
)
{\displaystyle AC\otimes BD=(A\otimes B)(C\otimes D)}
A pszeudoinverzekre
(
A
⊗
B
)
+
=
A
+
⊗
B
+
{\displaystyle (A\otimes B)^{+}=A^{+}\otimes B^{+}}
Általában, ha
A
−
{\displaystyle A^{-}}
B
−
{\displaystyle B^{-}}
A
{\displaystyle A}
B
{\displaystyle B}
A
−
⊗
B
−
{\displaystyle A^{-}\otimes B^{-}}
A
⊗
B
{\displaystyle A\otimes B}
Adva legyenek az
A
∈
M
a
t
(
k
×
ℓ
)
,
B
∈
M
a
t
(
m
×
n
)
,
C
∈
M
a
t
(
k
×
n
)
{\displaystyle A\in \mathrm {Mat} (k\times \ell ),\,B\in \mathrm {Mat} (m\times n),\,C\in \mathrm {Mat} (k\times n)}
X
∈
M
a
t
(
ℓ
×
m
)
{\displaystyle X\in \mathrm {Mat} (\ell \times m)}
A
X
B
=
C
{\displaystyle AXB=C\,}
A
X
B
=
C
⟺
(
B
T
⊗
A
)
vec
(
X
)
=
vec
(
C
)
{\displaystyle AXB=C\iff (B^{T}\otimes A)\,\operatorname {vec} (X)=\operatorname {vec} (C)}
ahol
vec
{\displaystyle \operatorname {vec} }
X
∈
M
a
t
(
ℓ
×
m
)
{\displaystyle X\in \mathrm {Mat} (\ell \times m)}
x
→
1
,
.
.
.
,
x
→
m
{\displaystyle {\vec {x}}_{1},...,{\vec {x}}_{m}}
vec
(
X
)
=
(
x
→
1
⋮
x
→
m
)
{\displaystyle \operatorname {vec} (X)={\begin{pmatrix}{\vec {x}}_{1}\\\vdots \\{\vec {x}}_{m}\end{pmatrix}}}
ℓ
⋅
m
{\displaystyle \ell \cdot m}
vec
(
C
)
{\displaystyle \operatorname {vec} (C)}
m
⋅
n
{\displaystyle m\cdot n}
A vektorizáltból visszaszámítható a mátrix, így ha megvan
X
∈
M
a
t
(
ℓ
×
m
)
{\displaystyle X\in \mathrm {Mat} (\ell \times m)}
X
{\displaystyle X}
Az ekvivalencia bizonyítása
szerkesztés
Teljesül
A
X
B
=
C
⟺
A
X
(
b
→
1
,
.
.
.
,
b
→
n
)
=
(
c
→
1
,
.
.
.
,
c
→
n
)
⟺
A
X
b
i
→
=
c
i
→
⟺
(
A
X
b
→
1
⋮
A
X
b
→
n
)
=
vec
(
C
)
{\displaystyle AXB=C\iff AX\left({\vec {b}}_{1},...,{\vec {b}}_{n}\right)=\left({\vec {c}}_{1},...,{\vec {c}}_{n}\right)\iff AX{\vec {b_{i}}}={\vec {c_{i}}}\iff {\begin{pmatrix}AX{\vec {b}}_{1}\\\vdots \\AX{\vec {b}}_{n}\end{pmatrix}}=\operatorname {vec} (C)}
ahol
(
A
(
x
→
1
,
.
.
.
,
x
→
m
)
b
→
1
⋮
A
(
x
→
1
,
.
.
.
,
x
→
m
)
b
→
n
)
=
(
A
(
b
11
x
→
1
+
.
.
.
+
b
m
1
x
→
m
)
⋮
A
(
b
1
n
x
→
1
+
.
.
.
+
b
m
n
x
→
m
)
)
=
(
A
b
11
⋯
A
b
m
1
⋮
⋱
⋮
A
b
1
n
⋯
A
b
m
n
)
(
x
→
1
⋮
x
→
m
)
=
(
B
T
⊗
A
)
vec
(
X
)
{\displaystyle {\begin{pmatrix}A({\vec {x}}_{1},...,{\vec {x}}_{m}){\vec {b}}_{1}\\\vdots \\A({\vec {x}}_{1},...,{\vec {x}}_{m}){\vec {b}}_{n}\end{pmatrix}}={\begin{pmatrix}A(b_{11}{\vec {x}}_{1}+...+b_{m1}{\vec {x}}_{m})\\\vdots \\A(b_{1n}{\vec {x}}_{1}+...+b_{mn}{\vec {x}}_{m})\end{pmatrix}}={\begin{pmatrix}A\,b_{11}&\cdots &A\,b_{m1}\\\vdots &\ddots &\vdots \\A\,b_{1n}&\cdots &A\,b_{mn}\end{pmatrix}}{\begin{pmatrix}{\vec {x}}_{1}\\\vdots \\{\vec {x}}_{m}\end{pmatrix}}=(B^{T}\otimes A)\,\operatorname {vec} (X)}
Mátrix együtthatós egyenletek
szerkesztés
Az
i
=
1
,
.
.
.
,
r
{\displaystyle i=1,...,r\,}
j
=
1
,
.
.
.
,
s
{\displaystyle j=1,...,s\,}
A
i
j
∈
M
a
t
(
k
×
ℓ
)
,
B
i
j
∈
M
a
t
(
m
×
n
)
,
C
i
∈
M
a
t
(
k
×
n
)
{\displaystyle A_{ij}\in \mathrm {Mat} (k\times \ell ),\,B_{ij}\in \mathrm {Mat} (m\times n),\,C_{i}\in \mathrm {Mat} (k\times n)}
X
i
∈
M
a
t
(
ℓ
×
m
)
{\displaystyle X_{i}\in \mathrm {Mat} (\ell \times m)}
[
A
11
X
1
B
11
+
.
.
.
+
A
1
s
X
s
B
1
s
=
C
1
⋮
A
r
1
X
1
B
r
1
+
.
.
.
+
A
r
s
X
s
B
r
s
=
C
r
]
{\displaystyle {\begin{bmatrix}A_{11}X_{1}B_{11}+...+A_{1s}X_{s}B_{1s}&=&C_{1}\\&\vdots &\\A_{r1}X_{1}B_{r1}+...+A_{rs}X_{s}B_{rs}&=&C_{r}\\\end{bmatrix}}}
egyenleteket. Ez ekvivalens a következő egyenletrendszer megoldásával:
(
B
11
T
⊗
A
11
⋯
B
1
s
T
⊗
A
1
s
⋮
⋱
⋮
B
r
1
T
⊗
A
r
1
⋯
B
r
s
T
⊗
A
r
s
)
(
vec
X
1
⋮
vec
X
s
)
=
(
vec
C
1
⋮
vec
C
r
)
{\displaystyle {\begin{pmatrix}B_{11}^{T}\otimes A_{11}&\cdots &B_{1s}^{T}\otimes A_{1s}\\\vdots &\ddots &\vdots \\B_{r1}^{T}\otimes A_{r1}&\cdots &B_{rs}^{T}\otimes A_{rs}\\\end{pmatrix}}{\begin{pmatrix}\operatorname {vec} \,X_{1}\\\vdots \\\operatorname {vec} \,X_{s}\end{pmatrix}}={\begin{pmatrix}\operatorname {vec} \,C_{1}\\\vdots \\\operatorname {vec} \,C_{r}\end{pmatrix}}}
Kapcsolat a tenzorszorzással
szerkesztés
Adva legyenek a véges dimenziós vektorterek közötti
φ
1
:
V
1
⟶
W
1
{\displaystyle \varphi _{1}:V_{1}\longrightarrow W_{1}}
φ
2
:
V
2
⟶
W
2
{\displaystyle \varphi _{2}:V_{2}\longrightarrow W_{2}}
φ
1
⊗
φ
2
:
V
1
⊗
V
2
⟶
W
1
⊗
W
2
{\displaystyle \varphi _{1}\otimes \varphi _{2}:V_{1}\otimes V_{2}\longrightarrow W_{1}\otimes W_{2}}
a
φ
1
⊗
φ
2
(
v
1
⊗
v
2
)
=
φ
1
(
v
1
)
⊗
φ
2
(
v
2
)
{\displaystyle \varphi _{1}\otimes \varphi _{2}(v_{1}\otimes v_{2})=\varphi _{1}(v_{1})\otimes \varphi _{2}(v_{2})}
Hogyha bázist választunk az
V
1
,
W
1
,
V
2
{\displaystyle V_{1},W_{1},V_{2}}
W
2
{\displaystyle W_{2}}
φ
1
{\displaystyle \varphi _{1}}
lineáris leképezés ábrázolható egy mátrixszal. Jelölje ezt a mátrixot
A
{\displaystyle A}
φ
2
{\displaystyle \varphi _{2}}
B
{\displaystyle B}
A
⊗
B
{\displaystyle A\otimes B}
φ
1
⊗
φ
2
{\displaystyle \varphi _{1}\otimes \varphi _{2}}
(
e
1
,
e
2
,
…
,
e
n
)
{\displaystyle (e_{1},e_{2},\ldots ,e_{n})}
V
1
{\displaystyle V_{1}}
(
f
1
,
f
2
,
…
,
f
p
)
{\displaystyle (f_{1},f_{2},\ldots ,f_{p})}
V
2
{\displaystyle V_{2}}
(
e
1
⊗
f
1
,
e
1
⊗
f
2
,
…
,
e
1
⊗
f
p
,
e
2
⊗
f
1
,
…
,
e
n
⊗
f
p
−
1
,
e
n
⊗
f
p
)
{\displaystyle (e_{1}\otimes f_{1},e_{1}\otimes f_{2},\ldots ,e_{1}\otimes f_{p},e_{2}\otimes f_{1},\ldots ,e_{n}\otimes f_{p-1},e_{n}\otimes f_{p})}
A Kronecker-szorzást használják például az általánosított regressziós analízisben a korrelált hibák kovarianciamátrixának előállításához. Az eredmény egy blokkdiagonális mátrix.
A kvantummechanikában több részecskés rendszereket írnak le a segítségével, ahol minden részecske spektruma korlátos. Nem korlátos spektrum esetén csak a Kronecker-szorzat algebrai szerkezete marad meg, mivel ekkor nem nem ábrázolható mátrixokkal.
A Tracy‑Singh és a Khatri–Rao-szorzatok A m × n -es mátrix mi × nj méretű A ij B p × q -s mátrix pk × ql méretű B kl i mi m , Σj nj n , Σk pk p és Σl ql q .
A Tracy–Singh-szorzat definíciója:[2] [3]
A
∘
B
=
(
A
i
j
∘
B
)
i
j
=
(
(
A
i
j
⊗
B
k
l
)
k
l
)
i
j
{\displaystyle \mathbf {A} \circ \mathbf {B} =(\mathbf {A} _{ij}\circ \mathbf {B} )_{ij}=((\mathbf {A} _{ij}\otimes \mathbf {B} _{kl})_{kl})_{ij}}
ahol a szorzat ij indexű blokkja az mi p × nj q méretű A ij B mátrix, ahol is (kl )-edik blokk az mi pk × nj ql méretű A ij B kl
Példa:
Legyenek A és B mindketten 2 × 2-es blokkmátrixok:
A
=
[
A
11
A
12
A
21
A
22
]
=
[
1
2
3
4
5
6
7
8
9
]
,
B
=
[
B
11
B
12
B
21
B
22
]
=
[
1
4
7
2
5
8
3
6
9
]
,
{\displaystyle \mathbf {A} =\left[{\begin{array}{c | c}\mathbf {A} _{11}&\mathbf {A} _{12}\\\hline \mathbf {A} _{21}&\mathbf {A} _{22}\end{array}}\right]=\left[{\begin{array}{c c | c}1&2&3\\4&5&6\\\hline 7&8&9\end{array}}\right],\quad \mathbf {B} =\left[{\begin{array}{c | c}\mathbf {B} _{11}&\mathbf {B} _{12}\\\hline \mathbf {B} _{21}&\mathbf {B} _{22}\end{array}}\right]=\left[{\begin{array}{c | c c}1&4&7\\\hline 2&5&8\\3&6&9\end{array}}\right],}
kapjuk, hogy:
A
∘
B
=
[
A
11
∘
B
A
12
∘
B
A
21
∘
B
A
22
∘
B
]
=
[
A
11
⊗
B
11
A
11
⊗
B
12
A
12
⊗
B
11
A
12
⊗
B
12
A
11
⊗
B
21
A
11
⊗
B
22
A
12
⊗
B
21
A
12
⊗
B
22
A
21
⊗
B
11
A
21
⊗
B
12
A
22
⊗
B
11
A
22
⊗
B
12
A
21
⊗
B
21
A
21
⊗
B
22
A
22
⊗
B
21
A
22
⊗
B
22
]
{\displaystyle \mathbf {A} \circ \mathbf {B} =\left[{\begin{array}{c | c}\mathbf {A} _{11}\circ \mathbf {B} &\mathbf {A} _{12}\circ \mathbf {B} \\\hline \mathbf {A} _{21}\circ \mathbf {B} &\mathbf {A} _{22}\circ \mathbf {B} \end{array}}\right]=\left[{\begin{array}{c | c | c | c }\mathbf {A} _{11}\otimes \mathbf {B} _{11}&\mathbf {A} _{11}\otimes \mathbf {B} _{12}&\mathbf {A} _{12}\otimes \mathbf {B} _{11}&\mathbf {A} _{12}\otimes \mathbf {B} _{12}\\\hline \mathbf {A} _{11}\otimes \mathbf {B} _{21}&\mathbf {A} _{11}\otimes \mathbf {B} _{22}&\mathbf {A} _{12}\otimes \mathbf {B} _{21}&\mathbf {A} _{12}\otimes \mathbf {B} _{22}\\\hline \mathbf {A} _{21}\otimes \mathbf {B} _{11}&\mathbf {A} _{21}\otimes \mathbf {B} _{12}&\mathbf {A} _{22}\otimes \mathbf {B} _{11}&\mathbf {A} _{22}\otimes \mathbf {B} _{12}\\\hline \mathbf {A} _{21}\otimes \mathbf {B} _{21}&\mathbf {A} _{21}\otimes \mathbf {B} _{22}&\mathbf {A} _{22}\otimes \mathbf {B} _{21}&\mathbf {A} _{22}\otimes \mathbf {B} _{22}\end{array}}\right]}
=
[
1
2
4
7
8
14
3
12
21
4
5
16
28
20
35
6
24
42
2
4
5
8
10
16
6
15
24
3
6
6
9
12
18
9
18
27
8
10
20
32
25
40
12
30
48
12
15
24
36
30
45
18
36
54
7
8
28
49
32
56
9
36
63
14
16
35
56
40
64
18
45
72
21
24
42
63
48
72
27
54
81
]
.
{\displaystyle =\left[{\begin{array}{c c | c c c c | c | c c}1&2&4&7&8&14&3&12&21\\4&5&16&28&20&35&6&24&42\\\hline 2&4&5&8&10&16&6&15&24\\3&6&6&9&12&18&9&18&27\\8&10&20&32&25&40&12&30&48\\12&15&24&36&30&45&18&36&54\\\hline 7&8&28&49&32&56&9&36&63\\\hline 14&16&35&56&40&64&18&45&72\\21&24&42&63&48&72&27&54&81\end{array}}\right].}
A Kronecker-szorzatot Leopold Kronecker után nevezték el, aki elsőként definiálta és használta. Korábban néha Zehfuss-mátrixnak nevezték, Johann Georg Zehfuss nyomán.
↑ Steeb, Willi Hans : Kronecker Product of Matrices and Applications . BI-Wiss.Verlag, 1991, ISBN 3-411-14811-X , S.16↑ Tracy, DS, Singh RP. 1972. A new matrix product and its applications in matrix differentiation. Statistica Neerlandica 26: 143–157.
↑ Liu S. 1999. Matrix results on the Khatri-Rao and Tracy-Singh products. Linear Algebra and its Applications 289: 267–277. (pdf Archiválva 2011. január 27-i dátummal a Wayback Machine -ben)
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![{\displaystyle \mathbf {A} \circ \mathbf {B} =\left[{\begin{array}{c | c}\mathbf {A} _{11}\circ \mathbf {B} &\mathbf {A} _{12}\circ \mathbf {B} \\\hline \mathbf {A} _{21}\circ \mathbf {B} &\mathbf {A} _{22}\circ \mathbf {B} \end{array}}\right]=\left[{\begin{array}{c | c | c | c }\mathbf {A} _{11}\otimes \mathbf {B} _{11}&\mathbf {A} _{11}\otimes \mathbf {B} _{12}&\mathbf {A} _{12}\otimes \mathbf {B} _{11}&\mathbf {A} _{12}\otimes \mathbf {B} _{12}\\\hline \mathbf {A} _{11}\otimes \mathbf {B} _{21}&\mathbf {A} _{11}\otimes \mathbf {B} _{22}&\mathbf {A} _{12}\otimes \mathbf {B} _{21}&\mathbf {A} _{12}\otimes \mathbf {B} _{22}\\\hline \mathbf {A} _{21}\otimes \mathbf {B} _{11}&\mathbf {A} _{21}\otimes \mathbf {B} _{12}&\mathbf {A} _{22}\otimes \mathbf {B} _{11}&\mathbf {A} _{22}\otimes \mathbf {B} _{12}\\\hline \mathbf {A} _{21}\otimes \mathbf {B} _{21}&\mathbf {A} _{21}\otimes \mathbf {B} _{22}&\mathbf {A} _{22}\otimes \mathbf {B} _{21}&\mathbf {A} _{22}\otimes \mathbf {B} _{22}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc17e704f3021d67d304e588037dde99a0c663b1)
![{\displaystyle =\left[{\begin{array}{c c | c c c c | c | c c}1&2&4&7&8&14&3&12&21\\4&5&16&28&20&35&6&24&42\\\hline 2&4&5&8&10&16&6&15&24\\3&6&6&9&12&18&9&18&27\\8&10&20&32&25&40&12&30&48\\12&15&24&36&30&45&18&36&54\\\hline 7&8&28&49&32&56&9&36&63\\\hline 14&16&35&56&40&64&18&45&72\\21&24&42&63&48&72&27&54&81\end{array}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/582820ef0780c4644d9a7eb92747b29ce0983efb)
