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 .
E`TIBORINGIZ UCHUN RAHMAT!
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