# Matrix equation system

1. Jan 30, 2009

### 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
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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 im 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 im 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???