Systems of linear equations

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  • Thread starter Carbon273
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  • #1
Carbon273
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TL;DR Summary
What conditions will there be no solutions, a unique solution and a p-parameter of solutions? If possible may you express in particular and homogenous vectors. I'm a bit rusty so bear with me, trying to reestablish my understanding here.
1569891740199.png
 

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  • #2
StoneTemplePython
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I'd strongly suggest multiplying by a permutation matrix ##\mathbf P## so that the LHS matrix is

##\mathbf {11}^T + (a-1)\mathbf I##
and the RHS has the natural progression of ##1, a, a^2 ##

assuming this is in reals, you should be able to confirm that any ##a \neq 1, -2## implies an invertible matrix. And ##a=1## still has (at least one) solution because.... and as for ##a=-2## well...
 
  • #3
Carbon273
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Hmm why is that you used that expression for the LHS. I am curious, where does the 11^T comes from?
 
  • #4
StoneTemplePython
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Hmm why is that you used that expression for the LHS. I am curious, where does the 11^T comes from?
because ##\mathbf {11}^T## is the ones matrix and it is easy to work with. The (scaled) Identity matrix is even easier to work with.

and the RHS is most natural as ##1, a, a^2## -- this is called the moment curve.
 
  • #5
mathman
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No solution when the determinant of the matrix is 0.
 
  • #6
StoneTemplePython
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No solution when the determinant of the matrix is 0.
This is not correct.

The matrix is rank one when ##a=1## and yet there are ##\gt 0## solutions. e.g. any of the standard basis vectors will do for x here.

I was trying to get OP to come to this conclusion when I said

And ##a=1## still has (at least one) solution because....
 

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