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Clarence

  • one year ago

If A is an invertible matrix then the equation Ax = b may have two distinct non zero solutions for a non zero vector b. True or False?

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  1. clarence
    • one year ago
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    I personally think that it's false, but confirmation is always nice to have.

  2. ganeshie8
    • one year ago
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    why do you think it is false

  3. IrishBoy123
    • one year ago
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    .

  4. clarence
    • one year ago
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    Well...

  5. clarence
    • one year ago
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    If there exist two different solutions, for example Ax1 = Ax2 = b, then A(x1−x2) = 0 with x1 − x2 ≠ 0. Wouldn't that follow up to be saying that there are infinite solutions? If A were invertible, you could write 0 = A^(−1) (0) = A^(−1) (A(x1−x2)) = x1−x2, but that's impossible because x1≠x2, hence A cannot be invertible?

  6. clarence
    • one year ago
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    If that made any sense ~

  7. ganeshie8
    • one year ago
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    That looks a lot better than what I had : \(A\) is invertible, so \(Ax=b\implies x=A^{-1}b\) since the inverse of a matrix is unique (when it exists), the solution is unique.

  8. clarence
    • one year ago
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    Thank you!!

  9. ganeshie8
    • one year ago
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    np

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