Abdishukurov shaxzod


Matritsalar va ular ustida amallar



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Abdishukurov Shaxzod matematika

Matritsalar va ular ustida amallar


𝐵𝐴 ko’paytmani topamiz:

𝐵𝐴 =


−1 5

−2 −3


3 4

4 −5 8



1 3 −1

=

(−1) ⋅ 4 + 5 ⋅ 1



(−2) ⋅ 4 + (−3) ⋅ 1

3 ⋅ 4 + 4 ⋅ 1

(−1) ⋅ (−5) + 5 ⋅ 3

(−2) ⋅ (−5) + (−3) ⋅ 3

3 ⋅ (−5) + 4 ⋅ 3

(−1) ⋅ 8 + 5 ⋅ (−1)

(−2) ⋅ 8 + (−3) ⋅ (−1)

3 ⋅ 8 + 4 ⋅ (−1)

=

1 20 −13


−11 1 −13

16 −3 20


.

Shunday qilib, 𝐴𝐵 ≠ 𝐵𝐴 ekan.


Matritsalar va ular ustida amallar


Misol 2. 𝐴𝐵 va 𝐵𝐴 ko’paytmalarni toping.

𝐴 =


3 5

1 2


, 𝐵 =

1 −5


−1 2

.

Hisoblaymiz:



𝐴𝐵 =

3 5 1 −5



1 2 −1 2

=

3 ⋅ 1 + 5 ⋅ (−1)



1 ⋅ 1 + 2 ⋅ (−1)

3 ⋅ (−5) + 5 ⋅ 2

1 ⋅ (−5) + 2 ⋅ 2

=

−2 −5



−1 −1

,

𝐵𝐴 =



1 −5 3 5


−1 2 1 2

=

1 ⋅ 3 + (−5) ⋅ 1



(−1) ⋅ 3 + 2 ⋅ 1

1 ⋅ 5 + (−5) ⋅ 2

(−1) ⋅ 5 + 2 ⋅ 2

=

−2 −5



−1 −1

.

Shunday qilib, 𝐴𝐵 = 𝐵𝐴 ekan.


Matritsalar va ular ustida amallar


Misol 3. 𝐴𝐵 𝐶 va 𝐴 𝐵𝐶 ko’paytmalarni toping.

𝐴 =


1 3

−1 1


2 5

, 𝐵 =


2 −6 1

1 3 −1


, 𝐶 =

−1

2



4

.

Ko’paytmalarni hisoblaymiz:



5

3

−2

−7

𝐴𝐵 =

−1

9

−2

,

𝐴𝐵 𝐶 =

11

,

9

3

−3

−15

𝐵𝐶 =

−10


1

, 𝐴 𝐵𝐶 =


−7

11

−15



,

ya`ni


𝐴𝐵 𝐶 = 𝐴 𝐵𝐶 .

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