Let n-order square matrix a satisfy a ^ 3 + 2a-3e = 0, prove that matrix A is invertible, and write out the expression of inverse matrix of A Help me, ha ha

Let n-order square matrix a satisfy a ^ 3 + 2a-3e = 0, prove that matrix A is invertible, and write out the expression of inverse matrix of A Help me, ha ha

Because a ^ 3 + 2a-3e = 0
Deformation a ^ 3 + 2A = 3e
That is, a [1 / 3 (a ^ 2 + 2e)] = E
That is, there exists B = 1 / 3 (a ^ 2 + 2e) such that ab = Ba = E
A is reversible by definition
And the inverse matrix A ^ (- 1) = 1 / 3 (a ^ 2 + 2e)

Let n-order square matrix a satisfy (a + e) 3 = 0, prove that matrix A is invertible, and write out the expression of a inverse matrix

The direct solution of the inverse matrix shows that it is reversible
A^3+3A^2+3A+E=0
A(-A^2-3A-3E)=E
So the inverse matrix of a is - A ^ 2-3a-3e

The inverse matrix of a similar square matrix is also similar

Syndrome: B, there is a = P ^ (- 1) * b * P
Then p ^ (- 1) * B ^ (- 1) * p * a = P ^ (- 1) * B ^ (- 1) * p * P ^ (- 1) * b * P = E
Then p ^ (- 1) * B ^ (- 1) * P = a ^ (- 1)
Then a ^ (- 1) ～ B ^ (- 1)
So the inverse matrix of a similar square matrix is also similar

If the lengths of the two sides of an isosceles triangle are the two roots of the quadratic equation x ^ 2-7x + 12 = 0, then the circumference of the triangle is?

If the two sides of the equation are X1 = 3 and X2 = 4, the three sides of the triangle are 3,3,4 or 3,4,4 respectively, so the perimeter is 10 or 11

The greatest common factor of a and B is 12 and the least common multiple is 144

The product of a and B is: 12 × 144 = 1728
The second number is: 1728 △ 36 = 48
If you don't understand, please ask

X2 / A2 + Y2 / B2 = 1 (a > b > 0), m, n are two points on the ellipse symmetrical about the origin, P is any point on the ellipse, the slopes of PM, PN are K1, K2, if | k1k2 | = 1/

The question is not complete, but still found the answer, I don't know if it is what you want
Let's set P (x0, Y0, y-0), m (x1, Y1) and n (X2, Y2) (1) (1) x0 ^ 2 / A ^ 2 / 2 / 2 / b ^ 2 = 1, (2) X1 ^ 2 / A ^ 2 / A ^ 2 + 0 ^ 2 / 2 / b ^ 2 = 1, (2) X1 ^ 2 / A ^ 2 / 2 / b ^ 2 = 1, (2) (2) let (1) x0 ^ 2 / a (2) (1) x0 ^ 2 / A ^ 2 / A ^ 2 / 2 (2) or (2) x0 ^ 2 / A ^ 2 / a (2) let (1) let (x1) get (x1 (x1 ^ 2-2-2-x0-x0-x0-x0 ^ 2) / PM) as the slope of PM is (K1 = (y0-y1-y1-y1-y1-y1-y1) / (y0-y1-y1) / (x0-y1-y1-y1x2) = B ^ 2 (x0 + x2) / A ^ 2 (Y0 + Y2), K1 * K2 = [b ^ 4 (x0 + x1) (x0 + x2)] / [a ^ 4 (Y0 + Y2) (Y0 + Y1)] = | 1 / 4 |, M. N is the symmetric two points on the ellipse about the origin, X1 = - X2, Y1 = - Y2
b^2/a^2=3/4,(a^2-c^2)/a^2=3/4,e=√3/2.

Junior high school English book unit 4 vocabulary (don't review) and unit 4 to recite the article. I forgot to bring urgent!
Junior high school English textbook unit 4 vocabulary (do not review) and unit 4 to recite the article. To copy words I forgot to bring. Urgent!

It is known that the two intersections of parabola y = AX2 + BX + C (a > 0) and X axis are a (- 1,0) and B (3,0), respectively. The intersection of parabola y = AX2 + BX + C (a > 0) and Y axis is D, and the vertex is C
Make a straight line CD intersect the x-axis with E and ask if there is a point F on the y-axis, so that the triangle CEF is an isosceles right triangle

Make a straight line CD intersect the x-axis at the point E. question: is there a point F on the y-axis such that △ CEF is an isosceles right triangle? If so, ask for the value of A. if not, explain the reason
existence
∵y=a（x+1）（x-3）=ax^2-2ax-3a
∴C（1,-4a）D（0,-3a）
The analytic expression of CD is y = - ax-3a
And because y = 0, x = - 3
∴E（-3,0）
Let f (0, y)
Let ch be perpendicular to the y-axis
Isosceles right angle
∴△EFO≌△FCH
∴OF=CH
∴ y=1
EO=FH
3=y+4a
∴a=1/2
If you don't understand this question, you can ask,

Square of a + bsquare = 8A + 4b-20, find the - 2004 power of 2 * (a-b) + (- 8A cube bsquare) / (2Ab) square

Square of a + square of B = 8A + 4b-20
a²-8a+b²-4b+20=0
(a-4)²+(b-2)²=0
Then: A-4 = 0; B-2 = 0
The solution is: a = 4; b = 2
2 to the power of - 2004 * (a-b) to the power of 2004 + (- 8A cubic B squared) / (2Ab squared)
=2^(-2004)*2^2004-2a
=1-8
=-7

Polar coordinate equation of arbitrary circle
If the polar coordinate system is established by taking the origin of the original plane rectangular coordinate system as the X axis, and the positive half axis of the X axis as the polar axis, then the equation of the circle in the plane rectangular coordinate system is: (x-a) / ; (x-a); (x-a); \35; + (y-b); \\\\\178;;;; (y-b) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\cosθ, y = ρ sin θ, substituting into the above formula, we get ρ & #178; -2acosθ-2bsinθ+a²+b²-R²=0
What I want to ask is where does ρ go in the derived equation

It's a mistake. It's missing