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## Main Question or Discussion Point

Hi all,

I learned this stuff years ago and wasn't brilliant at it even then so I think a refresher is in order.

Suppose I have n distinct homogeneous equations in n unknowns. I want to find the solution so I write down the matrix of coefficients multiplying my vector of variables as follows

[itex]A \mathbf{x} =\mathbf{0}[/itex].

Now, we don't want [itex]\deta A \neq 0[/itex] to happen otherwise the columns of A are linearly independent so the only solution to [itex]A \mathbf{x} = \mathbf{C}_1 x_1 + \cdots \mathbf{C}_n x_n = \mathbf{0}[/itex] is [itex]\mathbf{0}[/itex].

Now how do we actually solve this for [itex]\mathbf{x}[/itex], do we just do Gaussian elimination followed by back-substitution? Is the solution unique in this case?

Now suppose the system is inhomogeneous

[itex]A\mathbf{x} = \mathbf{b}[/itex] where [itex]\mathbf{b}\neq 0[/itex]. In this case we actually want [itex]\det A \neq 0[/itex] because then we can instantly write down the unique solution

[itex]\mathbf{x} = A^{-1}\mathbf{b}[/itex].

Have I gotten the solution to square systems about right? If yes, I'll try to figure out the non-square case.

I learned this stuff years ago and wasn't brilliant at it even then so I think a refresher is in order.

Suppose I have n distinct homogeneous equations in n unknowns. I want to find the solution so I write down the matrix of coefficients multiplying my vector of variables as follows

[itex]A \mathbf{x} =\mathbf{0}[/itex].

Now, we don't want [itex]\deta A \neq 0[/itex] to happen otherwise the columns of A are linearly independent so the only solution to [itex]A \mathbf{x} = \mathbf{C}_1 x_1 + \cdots \mathbf{C}_n x_n = \mathbf{0}[/itex] is [itex]\mathbf{0}[/itex].

Now how do we actually solve this for [itex]\mathbf{x}[/itex], do we just do Gaussian elimination followed by back-substitution? Is the solution unique in this case?

Now suppose the system is inhomogeneous

[itex]A\mathbf{x} = \mathbf{b}[/itex] where [itex]\mathbf{b}\neq 0[/itex]. In this case we actually want [itex]\det A \neq 0[/itex] because then we can instantly write down the unique solution

[itex]\mathbf{x} = A^{-1}\mathbf{b}[/itex].

Have I gotten the solution to square systems about right? If yes, I'll try to figure out the non-square case.