İki matrisin çarpılacağını varsayalım.
A i ve B j ile terimlerin (kesikli çizgiler) toplanması aritmetik işlemi son matrisdeki ij Eğer A n × m B m × p
A
=
(
A
11
A
12
⋯
A
1
m
A
21
A
22
⋯
A
2
m
⋮
⋮
⋱
⋮
A
n
1
A
n
2
⋯
A
n
m
)
,
B
=
(
B
11
B
12
⋯
B
1
p
B
21
B
22
⋯
B
2
p
⋮
⋮
⋱
⋮
B
m
1
B
m
2
⋯
B
m
p
)
{\displaystyle \mathbf {A} ={\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1m}\\A_{21}&A_{22}&\cdots &A_{2m}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nm}\\\end{pmatrix}},\quad \mathbf {B} ={\begin{pmatrix}B_{11}&B_{12}&\cdots &B_{1p}\\B_{21}&B_{22}&\cdots &B_{2p}\\\vdots &\vdots &\ddots &\vdots \\B_{m1}&B_{m2}&\cdots &B_{mp}\\\end{pmatrix}}}
AB matris çarpma (çarpım işaretsiz veya noktasız ifade edilir), n × p
A
B
=
(
(
A
B
)
11
(
A
B
)
12
⋯
(
A
B
)
1
p
(
A
B
)
21
(
A
B
)
22
⋯
(
A
B
)
2
p
⋮
⋮
⋱
⋮
(
A
B
)
n
1
(
A
B
)
n
2
⋯
(
A
B
)
n
p
)
{\displaystyle \mathbf {A} \mathbf {B} ={\begin{pmatrix}\left(\mathbf {AB} \right)_{11}&\left(\mathbf {AB} \right)_{12}&\cdots &\left(\mathbf {AB} \right)_{1p}\\\left(\mathbf {AB} \right)_{21}&\left(\mathbf {AB} \right)_{22}&\cdots &\left(\mathbf {AB} \right)_{2p}\\\vdots &\vdots &\ddots &\vdots \\\left(\mathbf {AB} \right)_{n1}&\left(\mathbf {AB} \right)_{n2}&\cdots &\left(\mathbf {AB} \right)_{np}\\\end{pmatrix}}}
Burada her bir i, j Aik A i Bkj B j k = 1, 2, ..., m k
(
A
B
)
i
j
=
∑
k
=
1
m
A
i
k
B
k
j
.
{\displaystyle (\mathbf {A} \mathbf {B} )_{ij}=\sum _{k=1}^{m}A_{ik}B_{kj}\,.}
Girişler genellikle sayı veya ifadelerle belirtilir. Fakat matrislerin kendisi de bir giriş olabilir. (Blok matrise bakınız).
Sağdaki şekil, A B X
[
a
11
a
12
⋅
⋅
a
31
a
32
⋅
⋅
]
4
×
2
matris
[
⋅
b
12
b
13
⋅
b
22
b
23
]
2
×
3
matris
=
[
⋅
x
12
x
13
⋅
⋅
⋅
⋅
x
32
x
33
⋅
⋅
⋅
]
4
×
3
matris
{\displaystyle {\overset {4\times 2{\text{ matris}}}{\begin{bmatrix}{\color {Brown}{a_{11}}}&{\color {Brown}{a_{12}}}\\\cdot &\cdot \\{\color {Orange}{a_{31}}}&{\color {Orange}{a_{32}}}\\\cdot &\cdot \\\end{bmatrix}}}{\overset {2\times 3{\text{ matris}}}{\begin{bmatrix}\cdot &{\color {Plum}{b_{12}}}&{\color {Violet}{b_{13}}}\\\cdot &{\color {Plum}{b_{22}}}&{\color {Violet}{b_{23}}}\\\end{bmatrix}}}={\overset {4\times 3{\text{ matris}}}{\begin{bmatrix}\cdot &x_{12}&x_{13}\\\cdot &\cdot &\cdot \\\cdot &x_{32}&x_{33}\\\cdot &\cdot &\cdot \\\end{bmatrix}}}}
Şekilde, çemberle işaretlenen hücrelerin değerleri şunlardır:
x
12
=
a
11
b
12
+
a
12
b
22
x
13
=
a
11
b
13
+
a
12
b
23
x
32
=
a
31
b
12
+
a
32
b
22
x
33
=
a
31
b
13
+
a
32
b
23
{\displaystyle {\begin{aligned}x_{12}&={\color {Brown}{a_{11}}}{\color {Plum}{b_{12}}}+{\color {Brown}{a_{12}}}{\color {Plum}{b_{22}}}\\x_{13}&={\color {Brown}{a_{11}}}{\color {Violet}{b_{13}}}+{\color {Brown}{a_{12}}}{\color {Violet}{b_{23}}}\\x_{32}&={\color {Orange}{a_{31}}}{\color {Plum}{b_{12}}}+{\color {Orange}{a_{32}}}{\color {Plum}{b_{22}}}\\x_{33}&={\color {Orange}{a_{31}}}{\color {Violet}{b_{13}}}+{\color {Orange}{a_{32}}}{\color {Violet}{b_{23}}}\end{aligned}}}
Yukarıdakiler, X
Satır vektör ve sütun vektör Aşağıdaki gibi iki matris verilsin;
A
=
(
a
b
c
)
,
B
=
(
x
y
z
)
,
{\displaystyle \mathbf {A} ={\begin{pmatrix}a&b&c\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}x\\y\\z\end{pmatrix}}\,,}
Burada matris çarpma işlemi şöyle:
A
B
=
(
a
b
c
)
(
x
y
z
)
=
a
x
+
b
y
+
c
z
,
{\displaystyle \mathbf {AB} ={\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}=ax+by+cz\,,}
Benzer şekilde;
B
A
=
(
x
y
z
)
(
a
b
c
)
=
(
x
a
x
b
x
c
y
a
y
b
y
c
z
a
z
b
z
c
)
.
{\displaystyle \mathbf {BA} ={\begin{pmatrix}x\\y\\z\end{pmatrix}}{\begin{pmatrix}a&b&c\end{pmatrix}}={\begin{pmatrix}xa&xb&xc\\ya&yb&yc\\za&zb&zc\end{pmatrix}}\,.}
AB BA 1 × 1 boyutlu matris iken, ikincisi 3 × 3 boyutlu matristir.
Kare matris ve sütun vektörüAşağıdaki gibi iki matris verilsin;
A
=
(
a
b
c
p
q
r
u
v
w
)
,
B
=
(
x
y
z
)
,
{\displaystyle \mathbf {A} ={\begin{pmatrix}a&b&c\\p&q&r\\u&v&w\end{pmatrix}},\quad \mathbf {B} ={\begin{pmatrix}x\\y\\z\end{pmatrix}}\,,}
Burada matris çarpma işlemi şöyle:
A
B
=
(
a
b
c
p
q
r
u
v
w
)
(
x
y
z
)
=
(
a
x
+
b
y
+
c
z
p
x
+
q
y
+
r
z
u
x
+
v
y
+
w
z
)
,
{\displaystyle \mathbf {AB} ={\begin{pmatrix}a&b&c\\p&q&r\\u&v&w\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}ax+by+cz\\px+qy+rz\\ux+vy+wz\end{pmatrix}}\,,}
Bu örnekte BA
Bir kare matrisi, sütun matrisi ile çarpma, doğrusal denklemleri çözme ve doğrusal dönüşümleri ifade etmek için sıkça kullanılır.
Kare matrisler Aşağıdaki gibi iki matris verilsin;
A
=
(
a
b
c
p
q
r
u
v
w
)
,
B
=
(
α
β
γ
λ
μ
ν
ρ
σ
τ
)
,
{\displaystyle \mathbf {A} ={\begin{pmatrix}a&b&c\\p&q&r\\u&v&w\end{pmatrix}},\quad \mathbf {B} ={\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}\,,}
Burada matris çarpma işlemi şöyledir:
A
B
=
(
a
b
c
p
q
r
u
v
w
)
(
α
β
γ
λ
μ
ν
ρ
σ
τ
)
=
(
a
α
+
b
λ
+
c
ρ
a
β
+
b
μ
+
c
σ
a
γ
+
b
ν
+
c
τ
p
α
+
q
λ
+
r
ρ
p
β
+
q
μ
+
r
σ
p
γ
+
q
ν
+
r
τ
u
α
+
v
λ
+
w
ρ
u
β
+
v
μ
+
w
σ
u
γ
+
v
ν
+
w
τ
)
,
{\displaystyle \mathbf {AB} ={\begin{pmatrix}a&b&c\\p&q&r\\u&v&w\end{pmatrix}}{\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}={\begin{pmatrix}a\alpha +b\lambda +c\rho &a\beta +b\mu +c\sigma &a\gamma +b\nu +c\tau \\p\alpha +q\lambda +r\rho &p\beta +q\mu +r\sigma &p\gamma +q\nu +r\tau \\u\alpha +v\lambda +w\rho &u\beta +v\mu +w\sigma &u\gamma +v\nu +w\tau \end{pmatrix}}\,,}
Benzer şekilde;
B
A
=
(
α
β
γ
λ
μ
ν
ρ
σ
τ
)
(
a
b
c
p
q
r
u
v
w
)
=
(
α
a
+
β
p
+
γ
u
α
b
+
β
q
+
γ
v
α
c
+
β
r
+
γ
w
λ
a
+
μ
p
+
ν
u
λ
b
+
μ
q
+
ν
v
λ
c
+
μ
r
+
ν
w
ρ
a
+
σ
p
+
τ
u
ρ
b
+
σ
q
+
τ
v
ρ
c
+
σ
r
+
τ
w
)
.
{\displaystyle \mathbf {BA} ={\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}{\begin{pmatrix}a&b&c\\p&q&r\\u&v&w\end{pmatrix}}={\begin{pmatrix}\alpha a+\beta p+\gamma u&\alpha b+\beta q+\gamma v&\alpha c+\beta r+\gamma w\\\lambda a+\mu p+\nu u&\lambda b+\mu q+\nu v&\lambda c+\mu r+\nu w\\\rho a+\sigma p+\tau u&\rho b+\sigma q+\tau v&\rho c+\sigma r+\tau w\end{pmatrix}}\,.}
Bu durumda hem AB BA AB BA
Satır vektör, kare matris ve sütun vektör Aşağıdaki gibi üç matris verilsin;
A
=
(
a
b
c
)
,
B
=
(
α
β
γ
λ
μ
ν
ρ
σ
τ
)
,
C
=
(
x
y
z
)
,
{\displaystyle \mathbf {A} ={\begin{pmatrix}a&b&c\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}\,,\quad \mathbf {C} ={\begin{pmatrix}x\\y\\z\end{pmatrix}}\,,}
Burada matris çarpma işlemi şöyledir:
A
B
C
=
(
a
b
c
)
[
(
α
β
γ
λ
μ
ν
ρ
σ
τ
)
(
x
y
z
)
]
=
[
(
a
b
c
)
(
α
β
γ
λ
μ
ν
ρ
σ
τ
)
]
(
x
y
z
)
=
(
a
b
c
)
(
α
x
+
β
y
+
γ
z
λ
x
+
μ
y
+
ν
z
ρ
x
+
σ
y
+
τ
z
)
=
(
a
α
+
b
λ
+
c
ρ
a
β
+
b
μ
+
c
σ
a
γ
+
b
ν
+
c
τ
)
(
x
y
z
)
=
a
α
x
+
b
λ
x
+
c
ρ
x
+
a
β
y
+
b
μ
y
+
c
σ
y
+
a
γ
z
+
b
ν
z
+
c
τ
z
,
{\displaystyle {\begin{aligned}\mathbf {ABC} &={\begin{pmatrix}a&b&c\end{pmatrix}}\left[{\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}\right]=\left[{\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}\right]{\begin{pmatrix}x\\y\\z\end{pmatrix}}\\&={\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}\alpha x+\beta y+\gamma z\\\lambda x+\mu y+\nu z\\\rho x+\sigma y+\tau z\\\end{pmatrix}}={\begin{pmatrix}a\alpha +b\lambda +c\rho &a\beta +b\mu +c\sigma &a\gamma +b\nu +c\tau \end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}\\&=a\alpha x+b\lambda x+c\rho x+a\beta y+b\mu y+c\sigma y+a\gamma z+b\nu z+c\tau z\,,\end{aligned}}}
Bu durumda CBA A (BC ) = (AB )C
Dikdörtgen matris Aşağıdaki gibi iki matris verilsin;
A
=
(
a
b
c
x
y
z
)
,
B
=
(
α
ρ
β
σ
γ
τ
)
,
{\displaystyle \mathbf {A} ={\begin{pmatrix}a&b&c\\x&y&z\end{pmatrix}}\,,\quad \mathbf {B} ={\begin{pmatrix}\alpha &\rho \\\beta &\sigma \\\gamma &\tau \\\end{pmatrix}}\,,}
Burada matris çarpma işlemi şöyledir:
A
B
=
(
a
b
c
x
y
z
)
(
α
ρ
β
σ
γ
τ
)
=
(
a
α
+
b
β
+
c
γ
a
ρ
+
b
σ
+
c
τ
x
α
+
y
β
+
z
γ
x
ρ
+
y
σ
+
z
τ
)
,
{\displaystyle \mathbf {A} \mathbf {B} ={\begin{pmatrix}a&b&c\\x&y&z\end{pmatrix}}{\begin{pmatrix}\alpha &\rho \\\beta &\sigma \\\gamma &\tau \\\end{pmatrix}}={\begin{pmatrix}a\alpha +b\beta +c\gamma &a\rho +b\sigma +c\tau \\x\alpha +y\beta +z\gamma &x\rho +y\sigma +z\tau \\\end{pmatrix}}\,,}
Benzer şekilde;
B
A
=
(
α
ρ
β
σ
γ
τ
)
(
a
b
c
x
y
z
)
=
(
α
a
+
ρ
x
α
b
+
ρ
y
α
c
+
ρ
z
β
a
+
σ
x
β
b
+
σ
y
β
c
+
σ
z
γ
a
+
τ
x
γ
b
+
τ
y
γ
c
+
τ
z
)
.
{\displaystyle \mathbf {B} \mathbf {A} ={\begin{pmatrix}\alpha &\rho \\\beta &\sigma \\\gamma &\tau \\\end{pmatrix}}{\begin{pmatrix}a&b&c\\x&y&z\end{pmatrix}}={\begin{pmatrix}\alpha a+\rho x&\alpha b+\rho y&\alpha c+\rho z\\\beta a+\sigma x&\beta b+\sigma y&\beta c+\sigma z\\\gamma a+\tau x&\gamma b+\tau y&\gamma c+\tau z\end{pmatrix}}\,.}
1. Değişme özelliği yoktur:
Genellikle:
A
B
≠
B
A
{\displaystyle \mathbf {A} \mathbf {B} \neq \mathbf {B} \mathbf {A} }
Çünkü AB BA çarpılmasına terstir. Matris çarpımını büyüklüğünü kelimelerle ifade etmek için; A B BA A C AC n > 1n × n birim matris (veya herhangi bir skaler çarpımı)dır.
Dizi gösterimi:
∑
k
A
i
k
B
k
j
≠
∑
k
B
i
k
A
k
j
{\displaystyle \sum _{k}A_{ik}B_{kj}\neq \sum _{k}B_{ik}A_{kj}}
2. Matrisin toplama üzerine dağılma özelliği vardır:
A
(
B
+
C
)
=
A
B
+
A
C
{\displaystyle \mathbf {A} (\mathbf {B} +\mathbf {C} )=\mathbf {AB} +\mathbf {AC} }
Sağ dağılım:
(
A
+
B
)
C
=
A
C
+
B
C
{\displaystyle (\mathbf {A} +\mathbf {B} )\mathbf {C} =\mathbf {AC} +\mathbf {BC} }
Dizi gösteriminde sırasıyla bunlar:
∑
k
A
i
k
(
B
k
j
+
C
k
j
)
=
∑
k
A
i
k
B
k
j
+
∑
k
A
i
k
C
k
j
{\displaystyle \sum _{k}A_{ik}(B_{kj}+C_{kj})=\sum _{k}A_{ik}B_{kj}+\sum _{k}A_{ik}C_{kj}}
∑
k
(
A
i
k
+
B
i
k
)
C
k
j
=
∑
k
A
i
k
C
k
j
+
∑
k
B
i
k
C
k
j
{\displaystyle \sum _{k}(A_{ik}+B_{ik})C_{kj}=\sum _{k}A_{ik}C_{kj}+\sum _{k}B_{ik}C_{kj}}
3. Skaler çarpma , matris çarpımı ile uyumludur:
λ
(
A
B
)
=
(
λ
A
)
B
{\displaystyle \lambda (\mathbf {AB} )=(\lambda \mathbf {A} )\mathbf {B} }
(
A
B
)
λ
=
A
(
B
λ
)
{\displaystyle (\mathbf {A} \mathbf {B} )\lambda =\mathbf {A} (\mathbf {B} \lambda )}
Burada λ skalerdir . Eğer matrisin tüm girişleri reel veya karmaşık sayı ise, tüm dört miktarda eşit olur. Daha genel bir ifade ile, eğer λ X λ X = X λ
λ
∑
k
(
A
i
k
B
k
j
)
=
∑
k
(
λ
A
i
k
)
B
k
j
=
∑
k
A
i
k
(
λ
B
k
j
)
{\displaystyle \lambda \sum _{k}(A_{ik}B_{kj})=\sum _{k}(\lambda A_{ik})B_{kj}=\sum _{k}A_{ik}(\lambda B_{kj})}
∑
k
(
A
i
k
B
k
j
)
λ
=
∑
k
(
A
i
k
λ
)
B
k
j
=
∑
k
A
i
k
(
B
k
j
λ
)
{\displaystyle \sum _{k}(A_{ik}B_{kj})\lambda =\sum _{k}(A_{ik}\lambda )B_{kj}=\sum _{k}A_{ik}(B_{kj}\lambda )}
4. Transpoze :
(
A
B
)
T
=
B
T
A
T
{\displaystyle (\mathbf {AB} )^{\mathrm {T} }=\mathbf {B} ^{\mathrm {T} }\mathbf {A} ^{\mathrm {T} }}
Burada T , transpozeyi ifade eder.
Dizi gösteriminde:
[
(
A
B
)
T
]
i
j
=
(
A
B
)
j
i
=
∑
k
(
A
)
j
k
(
B
)
k
i
=
∑
k
(
A
T
)
k
j
(
B
T
)
i
k
=
∑
k
(
B
T
)
i
k
(
A
T
)
k
j
=
[
(
B
T
)
(
A
T
)
]
i
j
{\displaystyle {\begin{aligned}\left[(\mathbf {AB} )^{\mathrm {T} }\right]_{ij}&=\left(\mathbf {AB} \right)_{ji}\\&=\sum _{k}\left(\mathbf {A} \right)_{jk}\left(\mathbf {B} \right)_{ki}\\&=\sum _{k}\left(\mathbf {A} ^{\mathrm {T} }\right)_{kj}\left(\mathbf {B} ^{\mathrm {T} }\right)_{ik}\\&=\sum _{k}\left(\mathbf {B} ^{\mathrm {T} }\right)_{ik}\left(\mathbf {A} ^{\mathrm {T} }\right)_{kj}\\&=\left[\left(\mathbf {B} ^{\mathrm {T} }\right)\left(\mathbf {A} ^{\mathrm {T} }\right)\right]_{ij}\end{aligned}}}
5. Karmaşık eşlenik :A B
(
A
B
)
⋆
=
A
⋆
B
⋆
{\displaystyle (\mathbf {AB} )^{\star }=\mathbf {A} ^{\star }\mathbf {B} ^{\star }}
olur. Burada * , bir matrisin karmaşık eşleniğini ifade eder.
Dizi gösteriminde:
[
(
A
B
)
⋆
]
i
j
=
[
∑
k
(
A
)
i
k
(
B
)
k
j
]
⋆
=
∑
k
(
A
)
i
k
⋆
(
B
)
k
j
⋆
=
∑
k
(
A
⋆
)
i
k
(
B
⋆
)
k
j
=
(
A
⋆
B
⋆
)
i
j
{\displaystyle {\begin{aligned}\left[(\mathbf {AB} )^{\star }\right]_{ij}&=\left[\sum _{k}\left(\mathbf {A} \right)_{ik}\left(\mathbf {B} \right)_{kj}\right]^{\star }\\&=\sum _{k}\left(\mathbf {A} \right)_{ik}^{\star }\left(\mathbf {B} \right)_{kj}^{\star }\\&=\sum _{k}\left(\mathbf {A} ^{\star }\right)_{ik}\left(\mathbf {B} ^{\star }\right)_{kj}\\&=\left(\mathbf {A} ^{\star }\mathbf {B} ^{\star }\right)_{ij}\end{aligned}}}
6. Eşlenik transpozesi :
Eğer A B
(
A
B
)
†
=
B
†
A
†
{\displaystyle (\mathbf {AB} )^{\dagger }=\mathbf {B} ^{\dagger }\mathbf {A} ^{\dagger }}
Burada † , bir matrisin karmaşık transpozesini ifade eder.
Dizi gösteriminde:
[
(
A
B
)
†
]
i
j
=
[
(
A
B
)
⋆
]
j
i
=
∑
k
(
A
⋆
)
j
k
(
B
⋆
)
k
i
=
∑
k
(
A
†
)
k
j
(
B
†
)
i
k
=
∑
k
(
B
†
)
i
k
(
A
†
)
k
j
=
[
(
A
†
)
(
B
†
)
]
i
j
{\displaystyle {\begin{aligned}\left[(\mathbf {AB} )^{\dagger }\right]_{ij}&=\left[\left(\mathbf {AB} \right)^{\star }\right]_{ji}\\&=\sum _{k}\left(\mathbf {A} ^{\star }\right)_{jk}\left(\mathbf {B} ^{\star }\right)_{ki}\\&=\sum _{k}\left(\mathbf {A} ^{\dagger }\right)_{kj}\left(\mathbf {B} ^{\dagger }\right)_{ik}\\&=\sum _{k}\left(\mathbf {B} ^{\dagger }\right)_{ik}\left(\mathbf {A} ^{\dagger }\right)_{kj}\\&=\left[\left(\mathbf {A} ^{\dagger }\right)\left(\mathbf {B} ^{\dagger }\right)\right]_{ij}\end{aligned}}}
7. İlkköşegen toplamı :AB A B
t
r
(
A
B
)
=
t
r
(
B
A
)
{\displaystyle \mathrm {tr} (\mathbf {AB} )=\mathrm {tr} (\mathbf {BA} )}
Dizi gösteriminde:
t
r
(
A
B
)
=
∑
i
∑
k
A
i
k
B
k
i
=
∑
k
∑
i
B
k
i
A
i
k
=
t
r
(
B
A
)
{\displaystyle {\begin{aligned}\mathrm {tr} (\mathbf {AB} )&=\sum _{i}\sum _{k}A_{ik}B_{ki}\\&=\sum _{k}\sum _{i}B_{ki}A_{ik}\\&=\mathrm {tr} (\mathbf {BA} )\end{aligned}}}
1. Birim matris :
Eğer A
A
I
=
I
A
=
A
{\displaystyle \mathbf {AI} =\mathbf {IA} =\mathbf {A} }
Burada I
2. Tersinir matris :
Eğer A A −1
A
A
−
1
=
A
−
1
A
=
I
{\displaystyle \mathbf {A} \mathbf {A} ^{-1}=\mathbf {A} ^{-1}\mathbf {A} =\mathbf {I} }
Bu durumda aşağıdaki eşitlik sağlanır;
(
A
B
)
−
1
=
B
−
1
A
−
1
{\displaystyle (\mathbf {AB} )^{\mathrm {-1} }=\mathbf {B} ^{\mathrm {-1} }\mathbf {A} ^{\mathrm {-1} }}
3. Determinant :AB A B
det
(
A
B
)
=
det
(
A
)
det
(
B
)
{\displaystyle \det(\mathbf {AB} )=\det(\mathbf {A} )\det(\mathbf {B} )}
det(A ) ve det(B ) yalnızca sayıdır. Bu yüzden, AB ≠ BA det(AB ) = det(BA ) olur.
![{\displaystyle {\begin{aligned}\mathbf {ABC} &={\begin{pmatrix}a&b&c\end{pmatrix}}\left[{\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}\right]=\left[{\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}\alpha &\beta &\gamma \\\lambda &\mu &\nu \\\rho &\sigma &\tau \\\end{pmatrix}}\right]{\begin{pmatrix}x\\y\\z\end{pmatrix}}\\&={\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}\alpha x+\beta y+\gamma z\\\lambda x+\mu y+\nu z\\\rho x+\sigma y+\tau z\\\end{pmatrix}}={\begin{pmatrix}a\alpha +b\lambda +c\rho &a\beta +b\mu +c\sigma &a\gamma +b\nu +c\tau \end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}\\&=a\alpha x+b\lambda x+c\rho x+a\beta y+b\mu y+c\sigma y+a\gamma z+b\nu z+c\tau z\,,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43c6d62200e683fe070d5e66e7e94d11ffdfad2d)
![{\displaystyle {\begin{aligned}\left[(\mathbf {AB} )^{\mathrm {T} }\right]_{ij}&=\left(\mathbf {AB} \right)_{ji}\\&=\sum _{k}\left(\mathbf {A} \right)_{jk}\left(\mathbf {B} \right)_{ki}\\&=\sum _{k}\left(\mathbf {A} ^{\mathrm {T} }\right)_{kj}\left(\mathbf {B} ^{\mathrm {T} }\right)_{ik}\\&=\sum _{k}\left(\mathbf {B} ^{\mathrm {T} }\right)_{ik}\left(\mathbf {A} ^{\mathrm {T} }\right)_{kj}\\&=\left[\left(\mathbf {B} ^{\mathrm {T} }\right)\left(\mathbf {A} ^{\mathrm {T} }\right)\right]_{ij}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bc6a19c27313397d4104044470bfbf1c7ab4476)
![{\displaystyle {\begin{aligned}\left[(\mathbf {AB} )^{\star }\right]_{ij}&=\left[\sum _{k}\left(\mathbf {A} \right)_{ik}\left(\mathbf {B} \right)_{kj}\right]^{\star }\\&=\sum _{k}\left(\mathbf {A} \right)_{ik}^{\star }\left(\mathbf {B} \right)_{kj}^{\star }\\&=\sum _{k}\left(\mathbf {A} ^{\star }\right)_{ik}\left(\mathbf {B} ^{\star }\right)_{kj}\\&=\left(\mathbf {A} ^{\star }\mathbf {B} ^{\star }\right)_{ij}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9013561a49a213ca7718b1933374923a3fa093c6)
![{\displaystyle {\begin{aligned}\left[(\mathbf {AB} )^{\dagger }\right]_{ij}&=\left[\left(\mathbf {AB} \right)^{\star }\right]_{ji}\\&=\sum _{k}\left(\mathbf {A} ^{\star }\right)_{jk}\left(\mathbf {B} ^{\star }\right)_{ki}\\&=\sum _{k}\left(\mathbf {A} ^{\dagger }\right)_{kj}\left(\mathbf {B} ^{\dagger }\right)_{ik}\\&=\sum _{k}\left(\mathbf {B} ^{\dagger }\right)_{ik}\left(\mathbf {A} ^{\dagger }\right)_{kj}\\&=\left[\left(\mathbf {A} ^{\dagger }\right)\left(\mathbf {B} ^{\dagger }\right)\right]_{ij}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ce25a89d4e1e89214866c30d89f78243a785bc9)
