Let A be a square matrix with A^2 = 0. Show that I + A is invertible and
(I + A)^−1 = I − A. Use this result to find B^−1, where
B =
(1 2 3
0 1 0
0 0 1)

Question

Let A be a square matrix with A^2 = 0. Show that I + A is invertible and
(I + A)^−1 = I − A. Use this result to find B^−1, where
B =
(1 2 3
0 1 0
0 0 1)

turingtest · Answer

$$(I-A)(I+A)=?$$

anonymous · Answer

I don't know what the dimensions of A would be. If I knew what A is I could find its inverse.

turingtest · Answer

you do not need the dimensions of A
just do the multiplication and what do you get?

anonymous · Answer

I^2 - A^2?

turingtest · Answer

yeah, and what are they telling us that A^2 equals?

anonymous · Answer

0...

turingtest · Answer

so we can write$$(I-A)(I+A)=I$$this implies that for the matrix $B=I+A$ there exists another matrix $B^{-1}$ such that$$B^{-1}B=I$$which is the definition of an invertible matrix
hence $B=I+A$ is invertible, and it's inverse is $$B^{-1}=(I+A)^{-1}=I-A$$

turingtest · Answer

is this making any sense at all to you?

anonymous · Answer

Yes. I know that B^-1(B) will give you an identity matrix.. Now I have to solve for B^-1.

turingtest · Answer

well, notice that $B^2\neq0$, but if we split B up we have $B=A+I$ where $A^2=0$
now we can just apply the theorem we just proved