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Please have a look on the linear system below:
ax + by + cz + dt = m
a'x + b'y + c'z + d't = n
(where a,b,c,d,a',b',c',d',m,n are real)
Supporse the vectors (a,b,c,d) and (a',b',c',d') are not linear dependent.
1) Prove the system has got a soution.
2) Prove there exist three vectors {u0,u1,u2} in R4 as each solution can be expressed by the form: S = u0 + j(u1) + k(u2)
(j and k are scalar)
Thank you very much..
No Idea? anyone?!
Well, this isn't comprehensive, but:
Both equations represent 3-D surfaces.
Just as:
two 'non-parallel' 2-D surfaces (planes) intersect in a 1-D surface (line) which can be represented with 2 vectors
and:
two 'non-parallel' 1-D surfaces (lines) intersect in a 0-D surface (point) which can be represented with 1 vector
in general:
two 'non-parallel' n-D surfaces intersect in a (n-1)-D surface which can be represented with n vectors.
So, here we have two 3-D surfaces, not 'parallel' as the given vectors are not linear dependant, so they intersect at the 'solution' which is a 2-D plane, which has the required form.
I meant for a more algebraic proof. but thanks anyway.