If matrix A is invertible, then - a must be invertible? How to prove that | - a | is not equal to 0?

If matrix A is invertible, then - a must be invertible? How to prove that | - a | is not equal to 0?

It must be like this, because a is reversible, then the determinant of a must not be equal to 0, and det (CA) = C ^ ndet (a), so the determinant of - A is equal to N times of - 1 times deta, so the determinant of - A is not equal to 0!

If the solution of the equation 9x-17 = KX about X is a positive integer, then the value of integer k is ()
A. 8B. 2C. 6，-10D. ±8

If (9-k) x = 17, the solution is x = 179 − K, ∵ x is a positive integer, ∵ 9-k = 1 or 17, the solution is k = 8 or - 8, so choose D

If the equation KX = x + 6 (x is not equal to 1) about X has a positive integer solution, find the value of K
Just this

The solution is KX = x + 6
We get (k-1) x = 6
So when x = 1, k-1 = 6, that is, k = 7
When x = 2, k-1 = 3, that is, k = 4
When x = 3, k-1 = 2, that is, k = 3
When x = 6, k-1 = 1, that is, k = 2
In conclusion, k = 2,3,4,7

A solution of the equation 2x minus KX plus 1 equals 0 is the same as the largest positive integer solution of the inequality 2 / 2 x plus 7 greater than or equal to 4x
(1) Finding K value (2) finding another solution of equation 2x minus KX plus 1 equal to 0

(1) the solution of the inequality 2 / 2 x plus 7 is greater than or equal to 4x: X ≤ 1, x = 1,
Substituting into the quadratic equation of one variable, we get 2-k + 1 = 0, k = 3
(2) the quadratic equation of one variable is 2x ^ 2-3x + 1 = 0,
(2X-1)(X-1)=0,
X1=1/2,X==1,
The other one is 1 / 2 (1 / 2)

Circle C (x + √ 3) ^ 2 + y ^ 2 = 16 a point a (√ 3,0) in the interior of circle C (x + √ 3) ^ 2 + y ^ 2 = 16 is connected with the moving point Q on the circle. The intersection of the central vertical line of AQ CQ and P is used to solve the trajectory equation of point P
Process!!!

By drawing, we can find that CP + AP = CP + PQ = R, so the sum of the distances from P point to two fixed points c and a is constant (i.e. the radius of the circle is 4), so the trajectory equation of P point can be obtained according to the definition of ellipse

Given the circle C: (x + 1) 2 + y2 = 25 and point a (1, 0), q is a point on the circle, and the vertical bisector of AQ intersects CQ at m, then the trajectory equation of point m is______ ．

From the equation of the circle, we can see that the center of the circle is C (- 1,0), and the radius is equal to 5. Let the coordinates of the point m be (x, y), and the vertical bisector of ∵ AQ intersects CQ at m, then | MQ | + | MC | = radius 5, and | MC | + | Ma | = 5 | AC |. According to the definition of ellipse, the locus of the point m is an ellipse with a and C as the focus

Given the circle C: (x + 1) 2 + y2 = 25 and point a (1, 0), q is a point on the circle, and the vertical bisector of AQ intersects CQ at m, then the trajectory equation of point m is______ ．

From the equation of the circle, we can see that the center of the circle is C (- 1,0), the radius is equal to 5, let the coordinates of the point m be (x, y), ∵ AQ's vertical bisector intersects CQ at m, | Ma | = | MQ |, and | MQ | + | MC | = radius 5, | MC | + | Ma | = 5 ＞| AC |. According to the definition of ellipse, the locus of point m is an ellipse with a and C as the focus, and 2A = 5, C = 1, | B = 212, so the elliptic equation is x2254 + & nbsp; y2214 = 1, that is 4x225 + 4y221 = 1, so the answer is 4x225 + 4y221 = 1

Make the secant of circle x + y + 2x-4y = 0 through the origin, intersect the circle at two points a and B, and find the trajectory equation of chord a and midpoint MD,
Help. Thank you

1. When the slope exists, let the straight line passing through the origin be y = KX, and bring the straight line into the circular equation, we get X & sup2; + (KX) & sup2; + 2x-4kx = 0, that is, (K & sup2; + 1) x & sup2; + (2-4k) x = 0. Obviously, both the circle and the straight line pass through the origin, that is, one of their intersection points is (0,0). According to Weida's theorem, X1 + x2 = - B / a = (4k-2) / (K & sup2; + 1), and the abscissa of the middle point is (2k-1) / (K & sup2; + 1), That is, x = (2k-1) / (K & sup2; + 1), because y = KX, then k = Y / x, bring in x = (2 (Y / x) - 1) / ((Y / x) & sup2; + 1), simplify X & sup2; + Y & sup2; + x-2y = 0
When the slope does not exist, the midpoint is (0,2), which obviously satisfies the equation x & sup2; + Y & sup2; + x-2y = 0
When the circle and the line are tangent, the tangent point is the origin, and the line and the circle only intersect at one point, and there is no midpoint
To sum up, the trajectory equation of the midpoint is a circle X & sup2; + Y & sup2; + x-2y = 0, but the origin (0,0) should be removed

Make any secant of the circle x ^ 2 + y ^ 2-2x-4y + 4 = 0 through the origin o, intersect the circle at two points a and B, and find the locus of the midpoint m of the line ab

The equation of circle: X & sup2; + Y & sup2; - 2x-4y + 4 = 0
(x-1)²+(y-2)²=1
Center (1,2) radius = 1
Let the midpoint of AB be m (x, y)
The distance from the center of a circle to m, the distance from m to the origin, and the distance from the center of a circle to the origin form a right triangle
According to Pythagorean theorem
(x-1)²+(y-2)²+x²+y²=(1-0)²+(2-0)²
2x²+2y²-2x-4y+5=5
x²+y²-x-2y=0
(x-1/2)²+(y-1)²=5/4
reference resources

⊙ o: X & sup2; + Y & sup2; = 1, ⊙ C: (x-4) & sup2; + Y & sup2; = 4, the moving circle P and ⊙ o sum are circumscribed, and the trajectory equation of the moving circle center P is

Let the radius of moving circle p be r, then | Po | = R + 1, | PC | = R + 2
So | PC | - | Po | = 1
The locus of the moving circle center P is the left branch of the hyperbola focusing on O and C
The equation is
(x-2)^2/0.25-y^2/3.75=1(x