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Let a be a square matrix of order n, satisfying AA ^ t = e (E is the identity matrix of order n), | a|
AA ^ t = e, | a ×| a ^ t | = | a | ^ 2 = 1, | a | = 1 or - 1. | a | < 0, so | a | = - 1
A+E=A+AA^T=A(E+A^T)
|A + e | = | a | ×| e + A ^ t | = | a | ×| a + e | = - | a + e |, so | a + e | = 0
Linear algebra! Let a be an n-dimensional column vector and a ^ TA = 1, let a = e-AA ^ t, where e is the identity matrix of order n,
If R (a) = n-1, then the general solution of AX = 0 is?
R (a) = n-1. First of all, we can determine that the number of solution vectors contained in the basic solution system of a is n - (n-1) = 1. Then it's very simple to find a vector and substitute AX = 0 to make it hold. Using the hint of the topic, this vector may be a. let's try to substitute AX = 0 (e-AA ^ t) x = 0 (e-AA ^ t) a = 0A to get (E
Lagrange mean value theorem
Let y = f (x) have a second order continuous derivative in (- 1,1), and F "(x) ≠ 0. It is proved that for any x ≠ 0 in (- 1,1), there exists a unique θ (x) ∈ (0,1), so that f (x) = f (0) + X * f '[x θ (x)] holds
The explanation in the book is as follows:
For any nonzero x ∈ (- 1,1), f (x) = = f (0) + X * f '[x θ (x)], [0] are obtained from Lagrange mean value theorem
What is explained by Lagrange's mean value theorem is the existence of θ (x) (at least one)
And f '(x) in (- 1,1) single increase (or decrease) for any x, f' (x) and X are one-to-one mapping!
The corresponding x is unique, so the coefficient θ (x) is unique
With two sides are 3 / 1 decimeter square cardboard put together into a rectangle, the circumference of the rectangle is () decimeter
Can you explain how you got it
Girth
=2 × (1 / 3 × 2 + 1 / 3)
=2 × (2 / 3 + 1 / 3)
=2×1
=2 decimeters
X tends to be negative infinity to find the limit of (x + arctan e ^ - x) / radical x ^ 2-xsinx + 1
lim(x→-∞){x+arctan[e^(-x)]}/√(x²-xsinx+1)
=LIM (Y → + ∞) [arctan (e ^ y) - y] / √ (Y & # 178; - ysiny + 1) (let y = - x)
= lim(y→+∞){[arctan(e^y)]/y-1}/√(1-siny/y+1/y²)
= -1.
The focus of the ellipse X212 + Y23 = 1 is F1, and the point P is on the ellipse. If the midpoint m of the line segment Pf1 is on the Y axis, then the ordinate of the point m is ()
A. ±34B. ±32C. ±22D. ±34
Let the coordinate of point p be (m, n). According to the meaning of the title, we can know that the F1 coordinate is (3, 0) ∧ m + 3 = 0 ∧ M = - 3. Substituting it into the elliptic equation, we can get the ordinate of n = ± 32 ∧ m is ± 34, so we choose a
The difference from 1 is prime. The quotient divided by 2 is prime. The remainder divided by 9 is the largest two digit number of 5?
Let this number be in the form of 9N + 5. Since Division 2 is also prime, n is also odd. Try one by one and you'll see that 14~
It is proved that the quadratic function f (x) = ax ^ 2 + BX + Ca is less than 0 and is an increasing function in the interval (negative infinity, - B / 2a)
Why?
If X1 and X2 belong to (- ∞, - B / 2a), then ax1 + AX2 + B is less than 0
Wrong, if X1 and X2 belong to (- ∞, - B / 2a), then ax1 + AX2 + b > 0
Using the method of matching
ax²+bx+c=a(x+b/2a)²-(b²-4ac)/4a
When x ∈ (- ∞, - B / 2a), the larger x is, the smaller (x + B / 2a) & # 178; is, the larger a (x + B / 2a) & # 178; is, the larger f (x) is, so it is an increasing function
If x1 ∈ (- ∞, - B / 2a), then x1-b (because A0)
There are also 2ax2 + b > 0
The sum of the two formulas is 2ax1 + 2ax2 + 2B > 0
So ax1 + AX2 + b > 0
How many of the mathematical functions 2 (x + 1) &# 178; + (x + 1-A) &# 178; in the first year of senior high school
3x^2+(6-2a)x+a^2-2a+3
In 2,5,6,8, there are two numbers that can be coprime____ yes.
3
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