- #1
jdstokes
- 523
- 1
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.