This sentence begins with "I"

This sentence begins with "I"

This sentence starts with an "I".

The rank of multiplication of two matrices
Exercise: let AB be all nonzero matrices of order n, and ab = 0, then the rank of AB is less than n
In the process of solving the problem, it is said that because AB = 0, the rank (a) + rank (b) ≤ n, and then AB is not zero, so the rank is greater than or equal to 1,
The problem is that rank (a) + rank (b) ≤ n. rank (AB) = rank (a) + rank (b)? If it is, then AB is a zero matrix, rank is 0, and a and B are non-zero matrices, so n is not equal to 0. That should be rank (a) + rank (b) < n? I'm very confused. I hope an expert can solve it,
Without a few commas, it's troublesome to see that a and B are both non-zero matrices. How to find the rank of a × B,

Theorem: if AB = 0, then rank (a) + rank (b) ≤ n
It is proved that the column vector of matrix B is denoted as Bi. ∵ AB = 0, the ∵ ABI = 0,
Let Bi be the solution of AX = 0
The fundamental solution system of ∵ AX = 0 contains n-rank (a) linearly independent solutions,
Ψ rank (b) ≤ n-rank (a),
That is, rank (a) + rank (b) ≤ n
PS: this conclusion is often used in proving or choosing to fill in the blanks. It needs to be remembered and applied~

Divide the eight numbers 6, 4, 8, 30, 21, 35, 33 and 22 into two groups to make the product of the four numbers in each group equal,

Step 1: only 30 and 35 of the 8 numbers have a factor of 5, so these two numbers should be in two groups respectively
Only 21 and 35 of the 8 numbers have a factor of 7, so these two numbers should be in two groups respectively
Therefore, 30 and 21 were in one group and 35 in another group
Step 2: only 6, 30, 21 and 33 of the eight numbers have a factor of 3, so two of the four numbers should be in one group and the other two in one group,
Because 30 and 21 are in one group, 6 and 33 are in another group, that is, 30 and 21 are in one group, and 6, 35 and 33 are in another group
Step 3: only 33 and 22 of the 8 numbers have a factor of 11, so these two numbers should be in two groups respectively, that is, 30, 21 and 22 in one group
Group 6, 35, 33 in the other group
Step 4: 30 × 21 × 22 = (3 × 2 × 5) × (3 × 7) × (11 × 2) = (2 × 2) × (3 × 3 × 5 × 7 × 11)
6×35×33=（2×3）×（5×7）×（3×11）=2×（3×3×5×7×11）
So 4 should be in the upper group and 8 in the lower group
4 × 30 × 21 × 22 = 8 × 6 × 35 × 33

How to divide the 8 numbers 6, 4, 8, 30, 21, 35, 33, 22 into two groups so that the product of 4 numbers of each group is equal?

6=2*3
4=2*2
8=2*2*2
30=2*3*5
21=3*7
35=5*7
33=3*11
22=2*11
After combination, it can be divided into two groups
2*2*2*2*3*3*5*7*11=2*2*2*2*3*3*5*7*11
So the difference is
4, 30, 21, 22
6, 8, 35, 33

Divide the eight numbers 26, 8, 4130, 21, 35, 33 and 22 into two groups, and the product of the four numbers in each group is equal

Divide the eight numbers 26, 8, 4130, 21, 35, 33 and 22 into two groups, and the product of the four numbers in each group is equal
analysis:
26=2×13
8=2×2×2
4=2×2
130=2×5×13
21=3×7
35=5×7
33=3×11
22=2×11
After these numbers are decomposed into prime factors,
There are: 8 2, 2 3, 2 5, 2 7, 2 11, 2 13
They were divided into two groups: 4 2, 1 3, 1 5, 1 7, 1 11 and 1 13 in each group
That is to say:
The first group: 2 × 13 [26], 3 × 11 [33], 5 × 7 [35], 2 × 2 × 2 [8]
The second group: 2 × 5 × 13 [130], 2 × 11 [22], 3 × 7 [21], 2 × 2 [4]
Group 1: 26, 33, 35, 8
The second group: 130, 21, 22, 4

Divide the eight numbers 26, 8, 4, 130, 21, 35, 33 and 22 into two groups, so that the product of the four numbers in each group is equal,

8 number decomposition factor: 13 * 2,2 ^ 3,2 ^ 2,13 * 5 * 2,7 * 3,7 * 5,11 * 3,11 * 2
1. Divide factor 13 into two groups
13*2,7*5,11*3,　8
13*5*2,7*3,11*2,4
Divide the other five into the previous group and write directly on it
Write another factor of 7 to the next group and write it directly above
Write another factor of 3 to the previous group and write it directly above
Write another factor of 11 to the next group
The upper 2 factor index is 2, and the lower 2 group index is 2, so put 8 on the upper side and 4 on the lower side

Decompose the factor (1) 27 ^ 3 + 8 (2) 4x ^ 4 + 3x ^ 2-1 in the complex range
The title is wrong. It should be (1) 27x ^ 3 + 8

(1)27x^3+8=(3x)^3+2^3=(3x+2)*(9x^2-6x+4)=(3x+2)*9*[x-(1+√3i)/3][x-(1-√3i)/3]=(3x+2)*[3x-(1+√3i)][3x-(1-√3i)](2)4x^4+3x^2-1=(4x^2-1)(x^2+1)=(2x-1)(2x+1)(x-i)(x+i)

Why do you use the plural for $10 in English and don't use the plural for $10?

Dollar has a plural form, dollars, so add s
The yuan of RMB and the Yan of yen (I remember it seems to be spelled like this) are both foreign currencies, so it's more convenient to use transliterated single and multiple homographs for convenience. After all, as an American, you can't know how to count all the currencies in the world, so it's more convenient to use transliterated single and multiple homographs, For example, we Chinese think that if all the English sentences are consistent, it will be more labor-saving. Americans think that it is more convenient for Chinese grandparents to call them Grandpa, right,

How to divide the eight numbers 6, 4, 8, 30, 21, 35, 33 and 22 into two groups so that the four product of each group is equal?

21,30,22,4
35,22,6,8
The product of the two groups is 55440

4, 6, 8, 21, 22, 30 and 35 are divided into two groups to make the product of four numbers in each group equal. How to divide them
Fast
wrote it wrong. There's another 33

There are 22, there is no solution