Please look at the problem about matrix for me Let the fourth-order matrix A = (x, - R2, R3, - R4) B = (y, R2, - R3, R4), where x, y, R2, R3, R4 are all four-dimensional column vectors, and the determinant IAI = 4, IBI = 1, then the determinant ia-bi =? I I is the determinant of a matrix

Please look at the problem about matrix for me Let the fourth-order matrix A = (x, - R2, R3, - R4) B = (y, R2, - R3, R4), where x, y, R2, R3, R4 are all four-dimensional column vectors, and the determinant IAI = 4, IBI = 1, then the determinant ia-bi =? I I is the determinant of a matrix

|A-B|
=|（X,-R2,R3,-R4) -(Y,R2,-R3,R4)|
=|=|（X-Y,-2R2,2R3,-2R4)|
=|（X,-2R2,2R3,-2R4)|-|（-Y,-2R2,2R3,-2R4)|
=8×|（X,-R2,R3,-R4)|-8×|（Y,R2,-R3,R4)|
=24

Matrix topics
It is known that the real matrix A = (AIJ) 3 * 3 satisfies the following conditions: (1) AIJ = AIJ, AIJ is the algebraic cofactor of AIJ (I, j = 1,2,3) (2) a11 does not = 0. The value of determinant A is calculated

The value of a is 1
The general answer is as follows:
It can be seen from (1) that matrix A is the adjoint matrix of A
From the formula AA * = | a | e, (a * denotes the adjoint matrix), taking determinants on both sides, we can get | a | = 0 or 1
From the expansion theorem, we can know that | a | = a11 * a11 + A12 * A12 + A13 * A13 = the square of a11 + the square of A12 + the square of A13. From (2), we can know that | a | > 0, all can only take 1
The answer is 1

Can the sum of several plus signs in the formula 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 be made equal to - 7 by changing them to minus signs? If so, please give an example; if not, please
Give reasons

1-2+3-4+5-6+7+8-9-10=-7

Quadratic function and quadratic equation of one variable
It is known that the quadratic function y = x ^ + (2k-1) x + K ^ - 1 of X, and the sum of squares of the two roots of the quadratic equation x ^ + (2k-1) x + K ^ - 1 = 0 of X is equal to 9. If the image of the quadratic function and the X axis intersect at points a and B from left to right, ask whether there is a point m on the image of the symmetry axis and the right, so that the area of the acute angle △ AMB is equal to 3? If there is, Request the coordinates of point m; if not, please explain the reason

I think it's not difficult. Let me give you some ideas. First, use Weida's theorem to work out a ^ 2 + B ^ 2 = (a + b) ^ 2-2ab = 9. It should be able to work out K, then two, and then calculate the length of ab. the length of ab multiplied by the absolute value of M, and the ordinate = 6, can work out the absolute value of M ordinate. In this way, it should be able to work out four ordinates of M, and you can omit several of them. Finally, it is the coordinate of M

9 8 7 6 5 4 3 2 1 two plus signs, two minus signs, two multiplication signs, two division signs, the result is an integer, what is the maximum value?

Because there is a division sign, so 1 and 2 must be △ because there is a multiplication sign, so 9 and 8 and 7 must be × 6 5 4 3, followed by a minus sign and a plus sign, these four numbers should be + 6 + 5-4-3, the guarantee number is the largest. To sum up: 9 × 8 × 7 + 6 + 5-4-3 △ 1 △ 2 multiply and divide first, then add and subtract = 504 + 6 + 5-4-1.5, because there is 1.5, which does not conform to the meaning of the whole number, so

Write the common factor; the square of 3 (a-b) - 4 (B-A)

The square of 3 (a-b) and the common factor of - 4 (B-A) are (a-b), or (B-A)
The steps of factorization are as follows
The square of 3 (a-b) - 4 (B-A)
=Square of 3 (a-b) + 4 (a-b)
=(a-b)({3(a-b)+4}
=(a-b)(3a-3b+4)

Given the regression linear equation y = BX + A, where a = 3 and the center of sample point is (1,2), then the regression linear equation is ()
A. y=x+3B. y=-2x+3C. y=-x+3D. y=x-3

The regression linear equation is y = BX + 3, ∵ the center of sample point is (1,2), ∵ 2 = B + 3, ∵ B = - 1, ∵ the regression linear equation is y = - x + 3

Calculate a definite integral problem,
∫ (1-sin2x) radical DX
Upper limit π, lower limit 0
Root 1-sin2x)

Estimate your integrand function should be under the root sign (1-sin2x), the prompt is as follows: 1-sin2x = 1-2sinxcosx = (SiNx) ^ 2 + (cosx) ^ 2-2sinxcosx = (SiNx cosx) ^ 2, the rest of your own calculation! Don't forget to divide the integral interval into two parts, 0-pi / 4, and PI / 4-pi

Sixth grade math exercise book 39 pages

94%
126%
43kg
Girls surpass, boys don't
(if you want a formula, ask)

Given that the point P is on the ellipse y * 2 / A * 2 + X * / b * 2, F1 and F2 are the focus of the ellipse, find the value range of Pf1 * PF2

The definition of ellipse is: Pf1 + PF2 = 2bpf1 * PF2 = Pf1 (2b-pf1) = - (Pf1) ^ 2 + 2B * Pf1. The focal length of ellipse is ABS (b ^ 2-A ^ 2) under 2 * root sign. You can discuss it in two cases. The focus is on the x-axis and the focus is on the y-axis