Given that x = 3 is the solution of the equation K (X-2) / 2-k + 3x / 6 = 4 / 3k, the value of K is obtained

Given that x = 3 is the solution of the equation K (X-2) / 2-k + 3x / 6 = 4 / 3k, the value of K is obtained

Substituting x = 3 into the equation K (X-2) / 2-k + 3x / 6 = 4 / 3k, we get that:
k(3-2)/2-k+3*3/6=4/3k
k/2-k+3/2=4k/3
(4/3+1/2)k=3/2
k=9/11

If the equation (3x + 2 / X-2) + (3 + MX / 2-x) = 1 about X has no solution, find the value of M
I already know M = 5 / 2. If there is any possibility of M = 2, please write down the process,
Is waiting equal to 2

There is no possibility of M = 2
If M = 2, then there is:
(3x+2/x-2)+(3+mx/2-x)
=(3x+2/x-2)+(3+2x/2-x)=1
（3x+2-2x-3)/(x-2)=1
(x-1)/(x-2)=1
x-1=x-2
1 = 2, so it's impossible

When m is the value, the equation 3 / X-2 + 2 + 3x + m / X (X-2) = 0 has no solution

Is it [3 / (X-2)] + 2 + 3x + {M / [x (X-2)]} = 0 or what?

Let the sum of all elements of n-order matrix a be zero, and the rank of a be n-1, then the general solution of linear equation AX = 0 is______ ．

The sum of elements in each row of matrix A of order n is zero , 1) t (column vector of N ones) is a solution of AX = 0. Because the rank of a is: n-1, the dimension of basic solution system is: N-R (a), so the dimension of basic solution system of a is 1, because (1, 1 T is a solution of the equation, not 0, so the general solution of AX = 0 is: K (1,1 ，1）T．

Let the sum of all elements of n-order matrix a be zero, and the rank of a be n-1, then the general solution of linear equation AX = 0 is______ ．

The sum of elements in each row of matrix A of order n is zero , 1) t (column vector of N ones) is a solution of AX = 0. Because the rank of a is: n-1, the dimension of basic solution system is: N-R (a), so the dimension of basic solution system of a is 1, because (1, 1 T is a solution of the equation, not 0, so the general solution of AX = 0 is: K (1,1 ，1）T．

Let the sum of elements in each row of n-order matrix a be 0, and the rank of a be n-1, then the general solution of homogeneous linear equations? Search on the Internet, but I still don't understand why all elements in each row are 0, so 11111 is its general solution, not other numbers
It seems that I have a little understanding. I mainly don't understand how general solution 1111.1 is calculated

The rank of a is n-1, which indicates that the fundamental solution system of AX = 0 contains N-R (a) = 1 solution vector
The sum of elements in each row of a is 0, which means that a (1,1,..., 1) ^ t = (0,0,...,) ^ t = 0
That is, (1,1,..., 1) ^ t is the nonzero solution of AX = 0, so it is the basic solution system of AX = 0
So the general solution is K (1,1,..., 1) ^ t
Note: in fact, any other non-zero number is OK, but the first feeling of "the sum of elements in each line of a" is to add up directly, that is, to multiply all by 1
Suppose a=
1 -1 0
2 1 -3
-5 3 2
You multiply (1,1,1) ^ t by this matrix to see if it is equal to 0

Let the sum of all elements of n-order matrix a be zero, and the rank of a be n-1, then the general solution of linear equation AX = 0 is______ ．

The sum of elements in each row of matrix A of order n is zero , 1) t (column vector of N ones) is a solution of AX = 0. Because the rank of a is: n-1, the dimension of basic solution system is: N-R (a), so the dimension of basic solution system of a is 1, because (1, 1 T is a solution of the equation, not 0, so the general solution of AX = 0 is: K (1,1 ，1）T．

Let a be an mxn matrix and the rank of a be r（

r