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A matrix problem! 1 4 2 A= 0 -3 -2 0.43 to the nth power of A
It is found that the square of a is exactly equal to E
Therefore, when n is a positive even number, the nth power of a is equal to e; when n is a positive odd number, the nth power of a is equal to a
A problem about matrix
Let B be mxn matrix and C be nxm matrix
|λ e (order m) - BC | = λ ^ (m-n) | λ e (order n) - CB|
Using | Xe_ m-AB|=|E_ n,0\\0,xE-AB|=|E,B\\0,xE-AB|=|E,B\\A,xE_ M | = | E-X ^ {- 1} Ba, 0 \ \ 0, Xe | = λ ^ (m-n) | λ e (n) - BA|
demonstrable
A matrix problem,
Given a 25 * 25 matrix A, a ^ 4 = 0 (0 matrix), find whether (I-A) has inverse matrix
Of course
Method 1. Because the eigenvalues of matrix a satisfying the condition can only be 0, so the eigenvalues of I-A are all 1 and none of them are zero, so I-A is invertible
Method 2: from the known condition a ^ 4 = 0, so (I-A) (I + A + A ^ 2 + A ^ 3) = I-A ^ 4 = I, so I-A is reversible, and its inverse is I + A + A ^ 2 + A ^ 3
Find the zeros (1) f (x) = x ^ 4-1 (2) f (x) = x ^ 3-4x of the following functions
The zero point of F (x) is the solution of the equation f (x) = 0
（1）f(x)=x^4-1=0
x^4=1
X = 1 or - 1
Two zeros X1 = 1, X2 = - 1
(2)f(x)=x^3-4x
x^3-4x =0
x（x²-4）=0
x（x+2）（x-2）=0
x1=0,x2=-2,x3=2
So there are three zeros: X1 = 0, X2 = - 2, X3 = 2
How to prove that three points are collinear with vectors?
Know the coordinates of three points, and prove that they are collinear with vectors?
It's a plane vector
For example, if you know ABC three-point coordinates, you can express Ba vector and CB vector
Then, if Ba vector is equal to a constant multiple of CB vector, its three points are collinear
In fact, it's simpler for you to directly calculate the slope of Ba line and BC line. The essence of the two is the same
If the slope is the same, the three points are collinear
As shown in the figure, in △ ABC, CA = CB, BD ⊥ AC at point E, BD, AE at point O. (1) verification: CD = CE; (2) verification: OC ⊥ ab
It is proved that: (1) BD ⊥ AC, AE ⊥ BC,
∴∠CDB=∠CEA=90°,
∵∠C=∠C,CA=CB,
∴ΔCDB≌ΔCEA(AAS),
∴CD=CE；
(2) connect OC,
∵OC=OC,AD=AE,
∴RTΔCOD≌RTΔCOE(HL),
∴∠OCA=∠OCB,
And Ca = CB,
⊥ Co ⊥ AB (isosceles triangle with three lines in one)
It is known that (m to the power of 2-9) times x to the power of 2 - (M-3) x + 6 = 0 is a univariate linear equation with X as the unknown. If | a | is less than or equal to | m | then the value of | a + m | + | A-M | is
For linear equation of one variable, then x & # 178; coefficient is 0 and X coefficient ≠ 0
So M & # 178; - 9 = 0
m²=9
m=±3
-(m-3)≠0
m≠3
So m = - 3
|a|≤|-3|=3
So - 3 ≤ a ≤ 3
So m ≤ a ≤ - M
So A-M ≥ 0, | A-M | = a-m
a+m≤0,|a+m|=-a-m
So the original formula = - A-M + A-M = - 2m = 6
Given the points a (2,3), B (5,4), C (7,10), if the vector AP = vector AB + λ vector AC, (λ∈ R), let's ask:
What is the value of λ when the point P is on the bisector of one and three quadrants?
Let: P (x, y)
Then:
AP=(x－2,y－3)、AB=(3,1)、AC=(5,7)
The results are as follows
(x－2,y－3)=(3,1)＋λ(5,7)=(5λ＋3,7λ＋1)
Namely:
x－2=5λ＋3、y－3=7λ＋1
The results are as follows
x=5λ＋5、y=7λ＋4
Since point P is on the bisector of one or three quadrants, then:
x=y
5λ＋5=7λ＋4
The result is: λ = 1 / 2
In the triangle ABC, D E is on the edge of BC, ad = AE, BD = EC, ab = AC
AD=AE
Therefore, ADE = AED
∠ADB=∠AEC
Ad = AE
BD=EC
So △ abd is equal to △ AEC
So AB = AC
If the real number XY satisfies (square of X + square of Y + 2) times (square of X + square of Y) = 0 factorization
(x²＋y²＋2)(x²＋y²)=0
Where: let X & # 178; + Y & # 178; + 2 = 0; X & # 178; + Y & # 178; = 0; obviously (X & # 178; + Y & # 178; + 2) is larger than 0, so only X & # 178; + Y & # 178; = 0, because XY is a real number, so x = y = 0
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