It is proved that for any nonzero n-dimensional column vector x, x'ax > 0, x'bx > 0, X '(2a + 3b) x = 2x'ax because a and B are all n-order positive definite matrices+ What's the meaning of "one after X"

X 'is transposed, and column vectors are transposed into row vectors

If the equations am * NX = B (M
Let a = {α 1, α 2, α 3}, α 1 = (1, x, - 3) t, α 2 = (- 1, - 3) t, α 3 = (1, y, 5) t, matrix A have three linearly independent eigenvectors, and λ = 2 is the double eigenvalue of a, then x = y =?

It is proved that AX = B always has a solution for any B
It is equivalent to that any m-dimensional column vector can be expressed linearly by a group of column vectors A1, A2,..., an of A
In particular, the group of m-dimensional fundamental vectors ε 1, ε 2,..., ε M can be expressed linearly by A1, A2,..., an
And any m-dimensional sequence vector can be expressed linearly by ε 1, ε 2,..., ε M
So the vector group A1, A2,..., an is equivalent to ε 1, ε 2,..., ε M
So r (A1, A2,..., an) = R (ε 1, ε 2,..., ε m) = M
That is, R (a) = M

How to prove that a is M &; n matrix, B is m-dimensional sequence vector, for any system of non-homogeneous equations, B always has a solution equivalent to a, and the obtained sequence vector can represent any m-dimensional vector

This can be seen from the relationship between the linear representation and the solution of the system of linear equations
The vector form of AX = B is X1A1 +... + xnan = B
So AX = B has a solution if and only if B can be expressed linearly by the column vectors A1,..., an of A

Let a be a matrix of order m * n, for any m-dimensional sequence vector B, ax = B, then why is at * a invertible

"For any m-dimensional sequence vector B, ax = B has a solution"
This shows that R (a) = M
(A^TA) = r(A) = m
But a ^ TA is a square matrix of order n, and N may be greater than m
So a ^ TA is not necessarily reversible

If y = 0, then the value of K is ()
A. 4B. -4C. 2D. -2

Substituting y = 0 into the equation 2x + 5Y = - 4, we get x = - 2, so x = - 2Y = 0 is the solution of the equation system KX − 3Y = 82x + 5Y = - 4. Then substituting x = - 2, y = 0 into the equation kx-3y = 8, we get k = - 4

When k is less than - 3, the real number solutions of the equations (1) x-ky = 0 (2) 2x-y ^ 2-9 = 0 have（
A. 4 groups B.3 group C.2 group D.1 please write the specific steps,

By substituting x = KY into the second equation, we can get y ^ 2-2ky + 9 = 0 = > Δ = B ^ 2-4ac = (2k) ^ 2-4 * 9 = 4K ^ 2-36
K ^ 2 > 9 = > 4 * k ^ 2 > 36 = > Δ > 0, so the equation y ^ 2-2ky + 9 = 0 will have two different real roots, so the equation system will have two groups of real solutions

Given the solution of the system of equations {3mx + 2y-6m, 2x + y = 8, we can get the solution of the system of equations (1) and the solution of M (2)

3mx+2y=6m
2x+y=8
x+y=10
From the simultaneous solution of the latter two equations, x = - 2, y = 12
Put it in the first equation
-The solution of 6m + 24 = 6m is m = 2

If the system of equations {x + y = 4-m, XY + m (x + y) = 5 has a real solution, the value range of M is obtained

｛x+y=4-m
xy+m(x+y)=5
xy+m(4-m)=5
x(4-m-x)+m(4-m)=5
-x^2+(4-m)x-(m^2-4m)-5=0
x^2+(m-4)x+(m^2-4m+5)=0
Discriminant = (M-4) ^ 2-4 * 1 * (m ^ 2-4m + 5)
=m^2-8m+16-4m^2+16m-20
=-3m^2+8m-4
The equations {x + y = 4-m XY + m (x + y) = 5 have real solutions
Then there is the discriminant > = 0
-3m^2+8m-4>=0
3m^2-8m+4

It is known that x = - 0.5, y = 3 is a real number solution of the system of equations x + y = m, xy = n. another real number solution of the system of equations is? Please tell me the method briefly

X + y = m, xy = n, we know that XY is the solution of the equation a ^ 2-ma + n = 0
∵ x = - 0.5, y = 3 is a real solution of the system of equations x + y = m, xy = n, ∵ M = 2.5, n = - 1.5
The equation is a ^ 2-5A / 2-3 / 2 = 0 2A ^ 2-5a-3 = 0 A1 = - 1 / 2 A2 = 3
The other solution is x = 3, y = - 0.5

When m is less than - 2, the number of real number solutions of the system of equations x = my Y ^ 2-x + 1 = O with respect to XY is——

y^2-x+1=o
y^2-my+1=0
△=m^2-4
∵m4
∴△>0
That is, the equation has two different real roots

It is proved that for any nonzero n-dimensional column vector x, x'ax > 0, x'bx > 0, X '(2a + 3b) x = 2x'ax because a and B are all n-order positive definite matrices+ What's the meaning of "one after X"