Proof of linear algebra Let B = (E + a) - 1 (e-A) It is proved that: (E + b) (E + a) = 2E

Proof of linear algebra Let B = (E + a) - 1 (e-A) It is proved that: (E + b) (E + a) = 2E

(E + a) - 1 are you the inverse of (E + a)? If so, then
B = (E + a) - 1 (e-A) both sides multiply left (E + a)
Available
(E + a) B = e-A with (E + a) on both sides
(E+A)B+(E+A)=(E-A)+(E+A)
We get (E + a) (E + b) = 2E
Here e + A, (E + b) / 2 are inverses of each other
Thus: (E + b) (E + a) = 2E

On the method of equivalent substitution in the proof of linear algebra
The necessary condition to prove that vector groups α 1 + α 2, α 2 + α 3 and α 3 + α 1 are linearly independent is that when vector groups α 1, α 2 and α 3 are linearly independent:
Because α 1 + α 2, α 2 + α 3, α 3 + α 1 are linearly independent,
So K1 (α 1 + α 2) + K2 (α 2 + α 3) + K3 (α 3 + α 1) = 0 (K1 ∈ R, K2 ∈ R, K3 ∈ R) only holds when K1 = K2 = K3 = 0
That is, (K1 + K3) α 1 + (K1 + K2) α 2 + (K2 + K3) α 3 = 0 (K1 ∈ R, K2 ∈ R, K3 ∈ R) only holds when K1 + K3 = K1 + K2 = K2 + K3 = 0
So α 1, α 2 and α 3 are linear independent
Is it right to prove that this can be replaced by equivalent?
I think it's absolutely right, but my teacher insists it's not OK
The key point is why it can't be like this. I also know other methods. I just want to find the right and wrong basis of this practice, not the result

Hehe, this is not so good! I'll give you a way to deal with this kind of problem. Prove: because (α 1 + α 2, α 2 + α 3, α 3 + α 1) = (α 1, α 2, α 3) P, where p = 10110011, because | P | = 2 ≠ 0, P is invertible. So the rank of two vector groups is equal

There are the following four mathematical propositions. Correct propositions are proved, and wrong propositions are given counter examples,
1. If a and B are not equal irrational numbers, then AB + A-B is irrational;
2. If a and B are unequal irrational numbers, then (a-b) / (a + b) is irrational;
3. If a and B are not equal irrational numbers, then the radical a + B are irrational numbers;
4. If both a and B are positive rational numbers and both radical a and B are irrational numbers, then radical a + radical B are irrational numbers

1. Error, counterexample: a = radical 2 + 1, B = radical 2-1
2. Error, counter example: a = 2 * radical 2, B = radical 2
3. Error, counter example: a = (2 + radical 3) ^ 2, B = (2-radical 3) ^ 2
4. Error, counter example: a = (2 + radical 3) ^ 2, B = (2-radical 3) ^ 2

3-x / X-2 △ x + 2 - (5 / X-2)) simplification
X-2 / 3-x divided by (x + 2 - X-2 / 5)

3-x/(x-2)÷(xx-4-5)/(x-2)=3-x/(x-2)÷(x+3)(x-3)/(x-2)=3-x/(x+3)(x-3)=(3xx-27-x)/(x+3)(x-3)

Vector a = (x, y + 2), B = (x, Y-2), and the module of a plus the module of B = 8, find the point (x, y) trajectory equation

The trajectory is an ellipse
The problem can be transformed into: finding the locus of a point whose distance sum of points (0,2) and (0, - 2) is equal to 8. Therefore, we can see that the locus is an ellipse with Y-axis as its major axis, where 2A = 8 and C = 2
A = 4, and a ^ 2-B ^ 2 = C ^ 2, so B ^ 2 = 12
The equation is: y ^ 2 / 16 + x ^ 2 / 12 = 1

In the triangle ABC, ∠ B is equal to 60 ° and AD and AE are bisectors of ∠ BAC and ∠ BCA, respectively. AD and CE intersect at point F
Write the relationship between Fe and FD

Analysis:
EF=DF,
prove:
FM ⊥ AB over F,
Through F as FN ⊥ AC in N,
C as cm '⊥ AB in M',
Let a be an '⊥ BC be n',
It is better to set ∠ BAC > BCA,
It is easy to get the angle bisector from ∠ B = 60 ° and ad, CE
∠DFN
=∠DAN'
=(1/2)∠BAC-(90°-∠B)
=(1/2)∠BAC-30°,
∠EFM
=∠ECM'
=(90°-∠B)-(1/2)∠BCA
=30°-(1/2)∠BCA,
BF is also the bisector of B, FM = FN,
∵∠DFN-∠EFM
=(1/2)∠BAC-30°-[30°-(1/2)∠BCA]
=(1/2)(∠BAC+∠BCA)-60°
=(1/2)*(180°-60°)-60°
=0,
∴∠EFM=∠DFN,
∴FE
=FM/cos∠EFM
=FN/cos∠DFN
=FD
That is, EF = DF
That's it!

Decomposition factor. ① the square of a + 2Ab + B + 2A + 2B; ② the square of 49 + 14x + X - the square of Y

(1)a^2+2ab+b^2+2a+2b
=(a+b)^2+2(a+b)
=(a+b)(a+b+2)
(2)x^2+14x+49-y^2
=(x+7)^2-y^2
=(x+y+7)(x-y+7)

In the polar coordinate system, the distance between the circle C: P = 2 √ 2Sin (θ + π / 4) and the line L: PCOS θ = 2 is 1
number

The general equation of circle C: P = 2cos θ + 2Sin θ is x ^ 2 + y ^ 2-2x-2y = 0, ①
The general equation of line L is x = 2,
The abscissa of the point whose distance from the line L is 1 is 1 or 3,
Substituting x = 1 into (1) yields y ^ 2-2y-1 = 0, y = 1, soil √ 2;
Substituting x = 3 into (1) yields y ^ 2-3y + 3 = 0 with no real root
The coordinate of the point is (1,1 √ 2)

As shown in the figure, in △ ABC, D is the point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB

Prove: in △ ACD, because ad = AC and AE ⊥ CD, according to the intersection point of the vertical line and the bottom of the isosceles triangle, we can prove that e is the midpoint of CD, and because f is the midpoint of CB, so EF ∥ BD, and EF is the median line of △ BCD, so EF = 12bd, that is BD = 2ef

How much water is a liter equal to

In mathematics, it's 2 jin