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xkehaulanix

  • 4 years ago

True or False (and explain)? If the equation Ax = b (assuming A is a square matrix of size nxn) has at least one solution for each b in Rn, then the solution is unique for each b.

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  1. xkehaulanix
    • 4 years ago
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    Ah, I think I figured it out. If Ax = b has one solution for every b in Rn, then the matrix A must be invertible by the Invertible Matrix Theorem. Then since A is invertible, the equation Ax = b has the unique solution \[x = A ^{-1}b\]

  2. Zarkon
    • 4 years ago
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    it says' at least one solution' not one solution suppose that Ax=b has more than one solution then \(Ax_1=b\) and \(Ax_2=b\) with \(x_1\ne x_2\) then \[A(x_1-x_2)=0\] let \(x_3=x_1-x_2\) so \[Ax_3=0\] thus the nullspace in nontrivial...thus the matrix dones not have full rank...hence the columns can't span all of \(\mathbb{R}^n\) thus for some \(c\in \mathbb{R}^n\) \[Ax=c\] will not have a solution ... a contradiction...thus the solutions are unique.

  3. xkehaulanix
    • 4 years ago
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    Ah, that was a typo on my part. The theorem applies when Ax = b has at least one solution. But thank you for the proof!

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