Clarence
  • Clarence
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?
Mathematics
jamiebookeater
  • jamiebookeater
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Clarence
  • Clarence
I personally think that it's false, but confirmation is always nice to have.
ganeshie8
  • ganeshie8
why do you think it is false
IrishBoy123
  • IrishBoy123
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Clarence
  • Clarence
Well...
Clarence
  • Clarence
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?
Clarence
  • Clarence
If that made any sense ~
ganeshie8
  • ganeshie8
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.
Clarence
  • Clarence
Thank you!!
ganeshie8
  • ganeshie8
np

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