# Systems of linear equations

• I
• Carbon273
In summary, the conversation discusses using a permutation matrix to manipulate a matrix equation and the significance of using a ones matrix and a scaled identity matrix. The RHS is described as a natural progression and the LHS is related to the moment curve. It is also mentioned that there is a solution even when the matrix is rank one and when the determinant is zero.

#### Carbon273

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.

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...

Hmm why is that you used that expression for the LHS. I am curious, where does the 11^T comes from?

Carbon273 said:
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.

No solution when the determinant of the matrix is 0.

mathman said:
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

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