A matrix exercise 1. Known matrix A= 1　-1　1 2　 4 x -3 -3 5 And B= 2　0　0 0　2 0 0　0　Y Where p ^ - 1AP = b Find x, y, matrix P

A matrix exercise 1. Known matrix A= 1　-1　1 2　 4 x -3 -3 5 And B= 2　0　0 0　2 0 0　0　Y Where p ^ - 1AP = b Find x, y, matrix P

Because p ^ - 1AP = B, a and B are similar
So the eigenvalues of a are 2,2, y
So trace (a) = 1 + 4 + 5 = 2 + 2 + y
Y = 6
And | a | = 2 * 2 * y = 24
And | a | = 6x + 36 = 6 (x + 6)
So x = - 2

On a problem of matrix
A 2 2
2 - 2, which seems to be such a matrix, finding the reciprocal of the sum of squares of A

A²
=(2 2
2 -2)*
(2 2
2 -2)
=(8 0
0 8)
Namely
A*(1/8 A)
=(1 0
0 1)
therefore
A^(-1)
=1/8 A
=1/8( 2 2
2 -2)
=(1/4 1/4
1/4 -1/4)

If the distance to the left focus of the ellipse x28 + y24 = 1 is equal to the distance to the fixed line x = 2, the trajectory equation of the moving point is______ ．

Ellipse x28 + y24 = 1, the left focus coordinate is (- 2, 0), defined by the parabola: the moving point trajectory whose distance to the left focus (- 2, 0) is equal to the distance to the fixed line x = 2 is a parabola with (- 2, 0) as the focus and x = 2 as the guide line, and the trajectory equation of the moving point is y2 = - 8x. So the answer is y2 = - 8x

As shown in the figure, in △ ABC, ab = AC, arbitrarily extend CA to P, and then extend AB to Q, so that AP = BQ. Prove that the outer center O of △ ABC is co circular with points a, P and Q

To connect OP, OQ, ob, OA, OQ, ob, OA, and connect OP, OQ, OQ, ob, OA, and OA, connect OP, OQ, OQ, ob, OA, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\qand Q are in the same circle The outer center O of △ ABC is concentric with points a, P and Q

Let Z be an imaginary number, w = Z + A / Z be a real number (a is a normal number)
1, if a = 1 and the real part of Z is √ 2 / 2, find Z
2. Find | Z |. Explain what graph is the set of points Z corresponding to Z

1、z+a/z=w
=>z²-wz+a=0
z=[w±i√(4a-w²]/2(4a>w²)
According to the meaning of the title, w / 2 = √ 2 / 2
Then z = √ 2 / 2 ± I √ 2 / 2
2、
z=[w±i√(4a-w²]/2(4a>w²)
|z|²=(w/2)²+(4a-w²)/4=a²
|z|=|a|
The set graph of Z is a circle with radius a (except left and right vertices)

Given the function f (x) = ln (1 + x ^ 2) - 1 / 2x ^ 2 + 1 / 2, find the extremum

F (x) = ln (1 + x ^ 2) - 1 / 2x ^ 2 + 1 / 2 domain x belongs to R
f'(x)=2x/(1+x^2)-x
Let f '(x) = 0
Then 2x / (1 + x ^ 2) - x = 0
The solution is x = 0 or x = 1
So f (x) has an extreme value on x = 0 or x = 1
f(0)=1/2,f(1)=ln2-1/2+1/2=ln2
f（1)>f（0）
So in the maximum f (1) = LN2, the minimum f (0) = 1 / 2

As shown in the figure is a right triangle, with ab side as the axis to rotate a circle, the body is a triangle______ And its volume is______ Cubic centimeter

With ab as the axis, the volume of the cone is: 13 × 3.14 × 22 × 6 = 13 × 3.14 × 4 × 6 = 25.12 (cubic centimeter); answer: with ab side as the axis, the body is a cone, its volume is 25.12 cubic centimeter. So the answer is: cone, 25.12

Calculation formula of total electric energy of parallel circuit

If we know the voltage and total current: w = uit
If you know the current and resistance: w = I1 ^ 2 * R1 * t + I2 ^ 2 * R2 * t
If we know the power of each resistor: w = P1 * t + P2 * t
Similarly, there are other cases

If a ∈ m, then 1 + A / 1-A ∈ m (a ≠ ± 1 and a ≠ 0). Given 3 ∈ m, please find out all the elements of the set M
The answer to this question is like this
Let a = 3
So, a ∈ M
Because, (1 + a) / (1-A) ∈ M
So, (1 + 3) / (1-3) = - 2 ∈ M
Therefore, [1 + (- 2)] / [[1 - (- 2)] = - 1 / 3 is also ∈ M
So, [1 + (- 1 / 3)] / [1 - (- 1 / 3)] = 1 / 2 ∈ M
So, (1 + 1 / 2) / (1-1 / 2) = 3
It starts to repeat, so m = {3, - 2, - 1 / 3, 1 / 2}
If the answer is right, why do you want a = 3? Is there any reason?

Because we know that 3 ∈ M
So a = 3 is a case
When a = 3, 1 + A / 1-A = - 2
The set of M is {3, - 2}
In the second case, when a = - 2
1+a/1-a=-1/3
The M set is {3, - 2, - 1 / 3}
In the third case, a = - 1 / 3
1+a/1-a=1/2
The set of M is {3, - 2,1 / 2, - 1 / 3}
In the third case, when a = 1 + A / 1-A
a^2+1=0
A has no solution
M is an empty set
To sum up, M set is {3, - 2, - 1 / 3,1 / 2, empty set}

As shown in the figure, point C is the midpoint of AB, ad = CE, CD = be

In △ ACD and △ CBE, ad = CECD = beac = CB, (5 points) ≌ ACD ≌ CBE (SSS). (6 points)