Matrix equation system

[Dell]

given the matrix A and vectors X & B
A=
1 2 3
2 k 6
0 0 k-4
k 2k 3k

X=
x1
x2
x3

B=
2
4
0
2k

what values of K(if any) will give AX=B--
a) with no possible x1 x2 x3
b) with a single x1 x2 x3, what are they
c) with infinite options, including 1 free parameter? what is this solutions
d) with infinite options including 2 free parameters? what is theis solutions
-----------------------------------------------------------------------
i make an extended matrix Ae, with B connected to A, and then analyze

first thing i see is that i have 4 equations but 2 of them are definitely the same, (in A : R1=K*R4) so i can esssentialy ingore R4,
next thing i see is that if K is 4, R3 is all 0's, and R2=2R1, so all I am left with is R1, from here i can clearly see the answer to d) k=4, infinite options including 2 free parameters, x3=u x2=t x1=2-2t-3u

now if K is 0, then R3 &4 still fall away, and I am left with R1 R2, since R1 and R2 are exactly the same other than their second digit, i subtract one from the other and i get that x2=0 and therefore i get the answer to c) k=0, infinite options, including 1 free parameter x3=u x2=0 x1=2-3u

if K is anything else then R4 still falls away, and X3=0, so now i can subtract R2-2R1 and i will get that x2 is also 0 therefore x1= 2 so this answers b) K$$\neq$$0,4 a single x1 x2 x3

from what i see there is no answer to a) no possible x1 x2 x3 which i think is okay since i have the exact amount of equations or less equations, never more than 3, and i only have x1...x3,.

i have found alll of these answers without any real mathematical operations. how would i do this with equations or math operations,,, originally what i tried to do is take the matrix A, without R3, call it A', now i know that there will be either no solution if A' is singular, so i checked detA' and i got detA'=(K-4)(K-4), if i compare this to 0 i get k=4 which is fine, but how do i get the k=0 solution?

 

[mighty2000]

To find the solution for k=0, you can use the inverse of A' to solve for X. First, find the inverse of A' by using the determinant you found earlier. So the inverse of A' would be:

A'^-1 = 1/((K-4)(K-4)) *
[ K-6 2K-6 3K-9;
2 K-4 3;
0 0 K-4 ]

Then, multiply A'^-1 by B to get X:

X = A'^-1 * B = 1/((K-4)(K-4)) *
[ K-6 2K-6 3K-9;
2 K-4 3;
0 0 K-4 ] * [2;4;0;2K] =
[ 2-3K; 0; 0; 2K ]

So for k=0, the solution would be x1=2-3K=2, x2=0, x3=0, which matches the solution you found earlier.

To find the solutions for k=4, you can use the same method. The inverse of A' would be:

A'^-1 = 1/0 *
[ 0 0 0;
2 0 3;
0 0 0 ]

Since the determinant is 0, the inverse does not exist and there is no solution for k=4. This confirms the result you found earlier.

Overall, using the inverse of A' can be a more systematic way of finding solutions for different values of k.