1. Let a be a matrix of order 3, and | a | = 6. If an eigenvalue of a is 2, then a * must have an eigenvalue of order 3 1. Let a be a matrix of order 3, and | a | = 6. If an eigenvalue of a is 2, then a * must have an eigenvalue of? 2. Let a be a matrix of order 3 and | a | = 3, then | (- a) ^ - 1 | =?

1. Let a be a matrix of order 3, and | a | = 6. If an eigenvalue of a is 2, then a * must have an eigenvalue of order 3 1. Let a be a matrix of order 3, and | a | = 6. If an eigenvalue of a is 2, then a * must have an eigenvalue of? 2. Let a be a matrix of order 3 and | a | = 3, then | (- a) ^ - 1 | =?

Finding eigenvalues and eigenvectors of matrix A = (4 25 / / 6 4 - 9 / / 5 3 - 7)

|A-λE|=
4-λ 2 5
6 4-λ -9
5 3 -7-λ
=
- λ^3 + λ^2 + 50λ - 20
There's a reason why it can't be broken down. There's a problem with this topic

Given that the eigenvalues of a matrix of order 3 are 2, - 3,6, then R (2a-6e)=
It's better to have a specific process

R (2a-6e) = 3, because 2a-6e = - 2 (3e-a), and 3 is not the eigenvalue of a, 2a-6e = - 2 (3e-a), so det (2a-6e) ≠ 0, so r (2a-6e) = 3,

To find the eigenvalues and eigenvectors of the following matrices, please write down the procedure
1 2 3
2 1 3
This is the topic

The second column is multiplied by - 1 to the first column, the first row is added to the second row, and then according to a11, it is B (B + 1) (B-9) with B as the eigenvalue
So the eigenvalue is 0 - 1 9
The eigenvectors are substituted into the eigenvectors respectively
b=0
-1 -2 -3
-2 -1 -3
-3 -3 -6
a1=(-1,-1,1)T
b=-1
-2 -2 -3
-2 -2 -3
-3 -3 -7
a2=(-1,1,0)T
b=9
8 -2 -3
-2 8 -3
-3 -3 3
a3=(1,1,2)T

Let the sum of the first n terms of the sequence {an} be Sn, and Sn = (x + 1) - xan, where x is a constant not equal to - 1 and 0. ① prove that {an} is an equal ratio sequence; ② let
Let the sum of the first n terms of the sequence {an} be Sn, and Sn = (x + 1) - xan, where x is a constant not equal to - 1 and 0
① It is proved that {an} is an equal ratio sequence; ② let the common ratio Q of {an} be f (x), and the sequence {BN} satisfy B1 = 1 / 3, BN = f {B (n-1)} (n belongs to N, n > = 2), then the first n terms and TN of {1 / BN} can be obtained
Ask for detailed explanation of the second question, the first question has been solved

(1)
Sn=(x+1)-xan (1)
S(n-1)=(x+1) -xa(n-1) (2)
(1) -(2)
an = xan-xan(n-1)
an/a(n-1) = x/(x-1)
=>The {an} is an equal ratio sequence
(2)
q=f(x) =x/(x-1)
bn=f(b(n-1))
= b(n-1)/[b(n-1) -1]
1/bn = [b(n-1) -1]/b(n-1)
= 1- 1/b(n-1)
1/bn -1/2 = -(1/b(n-1) -1/2 )
= (-1)^(n-1) .( 1/b1- 1/2)
= (5/2) .(-1)^(n-1)
1/bn = 1/2 +(5/2) .(-1)^(n-1)
Tn = 1/b1+1/b2+...+1/bn
= n/2 + (5/4)( 1- (-1)^n )

A wire can be enclosed into a square with a side length of 12.56 cm. If the wire is used to enclose a circle, how many square decimeters is the area of the circle?

Let the side length of the square be perimeter / 4 = 12.56 / 4 = 3.14
2 * 3.14 * radius = 12.56
So radius = 2
So the area is 3.14 * 2 * 2 = 12.56

Half, three quarters, seven eighths, fifteen sixteenth... Write the nth number

The nth number = the nth power of 2 (nth power of 2-1)

Given that the average of data x1, X2, X3, ··· xn is x, calculate the average of the following groups of data
（1）X1+5,X2+5,X3+5,···Xn+5； （2）ax1+k,ax2+k,ax3+k,···axn+k

The average of ∵ x1, X2, X3, ··· xn is X
Ψ X1 + x2 + X3 +... + xn = nx
The average of X1 + 5, X2 + 5, X3 + 5, ··· xn + 5 is
(x1+5+x2+5+.+xn+5)/n
=(NX + 5N) / n
=X + 5
The average of ax1 + K, AX2 + K, AX3 + K, ··· axn + k
=[a(x1+x2+...+xn)+nk]/n
=(anx + NK) / n
=MAX + k

If the length of the diagonal of a square is l, find its perimeter and area
Detailed process perimeter = 4x side length

Area: L ^ 2 / 2
Perimeter: 2 √ 2L

It is better to use the mean inequality to find the minimum value of F (x) = (x ^ 2-2x + 4) / X (x is greater than 0) and take the minimum value of X

【1】 Split term: function f (x) = (X & sup2; - 2x + 4) / x = - 2 + X + (4 / x). [2] ∵ x ＞ 0, we can know from "mean inequality": x + (4 / x) ≥ 2 √ [x (4 / x)] = 4. The equal sign is obtained only when x = 2.. x + (4 / x) ≥ 4. (x ＞ 0). [3] ∵ x + (4 / x) ≥ 4... - 2 + X + (4 / x) ≥ 2. That is, f (x) ≥ 2 = f (2).. f (x) min = f (2) = 2