Question

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?

Answer 1

I personally think that it's false, but confirmation is always nice to have.

Answer 2

why do you think it is false

Answer 3

.

Answer 4

Well...

Answer 5

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?

Answer 6

If that made any sense ~

Answer 7

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.

Answer 8

Thank you!!

Answer 9

np