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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.