Quick and easy operation: three and seven eighths minus one fourth plus three fourths plus one eighth, five minus one sixth plus five minus one

Quick and easy operation: three and seven eighths minus one fourth plus three fourths plus one eighth, five minus one sixth plus five minus one

Three and seven eighths minus one fourth plus three fourths plus one eighth
=3 and 7 / 8 + 1 / 8 - (1 / 4 + 3 / 4)
=4-1
=3
Five minus one sixth plus five minus one sixth
=5-1+（5/6-1/6）
=4+2/3
=4 and 2 / 3

Prove the invertibility of matrix
Now we have a matrix A to construct a matrix n, whose columns form the basis of NULA, (NULA is the zeroing space of matrix A, that is, the solution space of AX = 0) to construct a matrix R, whose rows form the basis of rowa. (row a is the row space of matrix a) it is proved that s = (RT, n) invertible RT denotes the transpose of R

First of all, the matrix here needs to be real, otherwise there are counter examples
For example, take the second order complex matrix A = [1, - I; I, 1], then s can be [1,1; - I, - I], it is easy to see that s is irreversible
Let B 'denote the transpose of B. for a real matrix, it can be proved as follows
Let a be a matrix of order n, we know that the dimension of nul A is N-R (a), so n is n × (N-R (a)) matrix
The dimension of row a is R (a), so r is R (a) × n matrix
So s = [R ', n] is a square matrix of order n
According to the selection of N, an = 0, and then RN = 0
It can be calculated that s's has block form [RR ', RN; n'r', n'n] = [RR ', RN; (RN)', n'n] = [RR ', 0; 0, n'n]
Then R (s) = R (s's) = R (RR ') + R (n'n) = R (R) + R (n) (for any real matrix B, R (BB') = R (B'b) = R (b) (*))
From R (n) = N-R (a), R (R) = R (a), R (s) = n is obtained, so s is a full rank square matrix of order n, that is, an invertible matrix of order n
Note 1: if we study inner product space, we can understand this result easily
An = 0 indicates that the column vectors of a 'and N are orthogonal to each other, that is, the transpose of row a and nul a are orthogonal subspaces
The two dimensions complement each other, so their respective bases can be combined into a set of bases in the whole space
So the column vector of S is a set of bases of the whole space, and s can be inversed with full rank
Note 2: about the conclusion (*), this is a common problem, which can be proved by the same solution of BX = 0 and b'bx = 0
The condition of real matrix can not be removed, which directly leads to the problem also needs the condition of real matrix

The least common multiple is represented by []. For example, [6,3] = 6. How can the greatest common factor be represented by sign?

（）
Put in parentheses

X + y = 17, X & # 178; + Y & # 178; = 169, how to solve

y=17-x
x²+﹙17-x﹚²=169
x²-17x+60=0
x₁=12,x₂=5
y₁=5,y₂=12

1. How many centimeters is 17 / 50 meters? 2. How many meters is 5000 / 8 meters? 3. How many minutes is 8 / 15 meters?
4. How many cubic decimeters is 39 / 250 cubic meters? How many square meters is 8 / 125 hectare?

1.34
two point six two five
three point three two
4.156,640

Factorization of (x2 minus ax) 2 minus 4 (x2 minus ax minus 1)
Factorization of (x2 minus ax) 2 minus 4 (x2 minus ax minus 1)
(x + y) 4 minus (x + y) 2 minus 45

﹙x²－ax﹚²－4﹙x²－ax－1﹚
＝﹙x²－ax﹚²－4﹙x²－ax﹚＋4
＝[﹙x²－ax﹚－2]²
＝﹙x²－ax－2﹚²

5 out of 12 times 7 out of 8 plus 1 out of 8 times 5 out of 12 minus 5 out of 12
It should be easy to calculate

5 out of 12 times 7 out of 8 plus 1 out of 8 times 5 out of 12 minus 5 out of 12
=5/12x(7/8+1/8-1)
=5/12x0
=0

Length: 0.5m width: 0.3m height: 0.2m surface area and volume of a cuboid

Surface area 0.5 * 0.3 * 2 + 0.5 * 0.2 * 2 + 0.3 * 0.2 * 2 = 0.3 + 0.2 + 0.12 = 0.62m2
0.5*0.2*0.3=0.03M3

The quotient 26 divided by 39 = () 49 divided by 14 = () 24 divided by 18 = () 18 divided by 15 = ()

26 divided by 39 = (2 / 3)
49 divided by 14 = (7 / 2)
24 divided by 18 = (4 / 3)
18 divided by 15 = (6 / 5)

In the 100 natural numbers of 1-100, what is the sum of all the numbers that are both multiples of 2 and multiples of 3

If the least common multiple of 2 and 3 is 6, then 6, 12, 18 and so on are all multiples of 2 and multiples of 3, and the last bit is 96
Add all the numbers to get 6 + 6 * 2 + 6 * 3.. + 6 * 16 = 6 * (1 + 2 + 3 +. + 16)
=6*[（1+16）/2*16]=6*8*17=48*17=816