Ring and Group Properties
Создано: 19 декабря 2024 г.
Создано: 19 декабря 2024 г.
(A) Suppose there is a ring. If , where and is not the identity for , then .
(B) Suppose that is a ring and is not empty. If for all and for all , where denotes the inverse of under , then is also a ring.
(C) If is a field, then is also an integral domain.
(D) Suppose that and are two groups. A mapping is a group isomorphism, if for all .
(E) If is a group and , then is also a group, where is the set of integers.
The correct statements are:
(B), (C), and (E)
(A)
There are 40 different ways to store 1, 2, 3, 4, 5 in A[1:5] (each number stored in a memory location) so that A[i] ≠ i for all 1 ≤ i ≤ 5.
(B)
f(x) = x/(1 − x)2 is the ordinary generating function for 0, 1, 2, 3, 4, ….
(C)
There are 105 different integer solutions for x1 + x2 + x3 + x4 = 24, where 3 ≤ xi ≤ 8 for i = 1, 2, 3, 4.
(D)
There are (58 − 38)/2 different ways to distribute 8 different objects O1, O2, …, O8 to five different boxes B1, B2, …, B5 provided an even number of objects are distributed to B5.
(E)
Let an be the number of different quaternary (i.e., {0, 1, 2, 3}) sequences of length n that have an even number of 1's, where n ≥ 2. Then, an = 2an−1 + 4n−1.
The correct statements are:
(B) and (E)
正確的選項為:(1) 與 (4)
1
{
ax+by=2,
cx+dy=1
−
1
{
ax+by=−1,
cx+dy=−1
的增廣矩陣經過相同的列運算後,分別得到
[
1
−
1
3
0
1
2
]
,
[
1
−
1
2
0
1
−
1
]
[
1
0
−1
1
3
2
],[
1
0
−1
1
1
{
ax+by=0,
cx+dy=1
的解為
根據題意,已有兩組線性方程組,在同一組列運算(行列變換)後分別得到了階梯型的增廣矩陣。設
已知三組方程組及其解為:
(\begin{cases} a x + b y = 2 \ c x + d y = 1 \end{cases})
(\begin{cases} a x + b y = -1 \ c x + d y = -1 \end{cases})
(\begin{cases} a x + b y = 0 \ c x + d y = 1 \end{cases})
前兩組方程組在經過同樣的列運算後,得到對應的最後階梯型增廣矩陣分別為:
(1) (\begin{pmatrix} 1 & -1 & 3 \ 0 & 1 & 2 \end{pmatrix}) ,因此第一組方程組解為:
故第一組解為 。
(2) (\begin{pmatrix} 1 & -1 & 2 \ 0 & 1 & -1 \end{pmatrix}) ,因此第二組方程組解為:
故第二組解為 。
由於前兩組方程組有相同的係數矩陣 且經過相同的列運算變換至標準形,故可由這兩組已知解來求出 ,再應用於第三組方程組。
從第一組解得知:
從第二組解得知:
令 。
則:
以及
由 。
將上述方程組合併整理可求得:
再解 :
故有:
現在利用此反矩陣求第三組方程組 的解:
故第三組聯立方程組的解為 。
Γ 的半徑為 。