SOLUTION: If A is a square matrix with the property (A with an exponet of 2) = 0, show that (I - A)inverse = A + I. Thank you

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Question 87107This question is from textbook
: If A is a square matrix with the property (A with an exponet of 2) = 0, show that (I - A)inverse = A + I.

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Found 2 solutions by longjonsilver, Edwin McCravy:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
Any matrix multiplied by its inverse is equal to the identity matrix, I by definition. So:

+%28I-A%29%5E%28-1%29%28I-A%29+=+I+

Post multiply both sides by (I+A):
+%28I-A%29%5E%28-1%29%28I-A%29%28I%2BA%29+=+I%28I%2BA%29+

now, +%28I-A%29%28I%2BA%29+=+I%5E2+%2B+IA+-+AI+-+A%5E2+
+%28I-A%29%28I%2BA%29+=+I%5E2+%2B+IA+-+IA+-+A%5E2+
+%28I-A%29%28I%2BA%29+=+I+%2B+IA+-+IA+-+A%5E2+
+%28I-A%29%28I%2BA%29+=+I+-+A%5E2+
+%28I-A%29%28I%2BA%29+=+I+

So, +%28I-A%29%5E%28-1%29%28I-A%29%28I%2BA%29+=+I%28I%2BA%29+ becomes
+%28I-A%29%5E%28-1%29+=+I%28I%2BA%29+

and hence +%28I-A%29%5E%28-1%29+=+%28I%2BA%29+

cheers
Jon

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
If A is a square matrix with the property (A with an exponet of 2) = 0, show
that (I - A)inverse = A + I. 
The best way to get insight into a problem like this is first (ON SCRATCH PAPER)
to assume it is true and work backwards until you run into something that is
true.  Then write the steps in reverse and discard the scratch paper.

On scratch paper (NOT TO TURN IN), assume the proposition is already known to
be true (which of course it isn't, which is why in the end you must discard the
scratch work).

SCRATCH WORK:

%28I+-+A%29%5E%28-1%29+=+A+%2B+I

Right multiply both sides by %28I-A%29

%28I-A%29%5E%28-1%29%28I-A%29+=+%28A+%2B+I%29%28I-A%29

The left side is the identity I.
And since matrix multiplication is distributive,
we may use "FOIL" to multiply out the right side:

I+=+AI+-+A%5E2+%2B+I%5E2+-+IA

Now we see on the right that AI+=+IA+=+A, making
the AI and the -IA cancel.  

I+=+-A%5E2+%2B+I%5E2

Furthermore I%5E2+=+I, so we have

I+=+-A%5E2+%2B+I

and since we are given that A%5E2+=+0

I+=+-0+%2B+I

I+=+0+%2B+I

and we are left with

I+=+I

Now this is equal, so let's turn everything backwards:

I+=+I                     The identity matrix equals itself
I+=+0+%2B+I                 Adding the 0 matrix to the right side 
I+=+-0+%2B+I                Replacing the 0 matrix by -0 
I+=+-A%5E2+%2B+I               Replacing 0 by A%5E2 (they are given equal)
I+=+-A%5E2+%2B+I%5E2              Replacing I by I%5E2 
I+=+A+-+A%5E2+%2B+I%5E2+-+A       Adding and subtracting A on the right side
I+=+AI+-+A%5E2+%2B+I%5E2+-+IA     Replacing the first A by AI and the second by IA
I+=+AI+-+AA+%2B+II+-+IA     Writing A%5E2 as AA and I%5E2 as II
I+=+A%28I+-+A%29+%2B+I%28I+-+A%29   Matrix multiplication is distributive over
                          matrix addition.
I+=+%28A+%2B+I%29%28I+-+A%29        Matrix multiplication is distributive over
                          matrix addition.
%28I-A%29%5E%28-1%29%28I-A%29+=+%28A+%2B+I%29%28I-A%29 Replacing the I on the left by the inverse of a matrix times the matrix.

[%28I-A%29%5E%28-1%29%28I-A%29]%28I-A%29%5E%28-1%29 = [%28A+%2B+I%29%28I-A%29]%28I-A%29%5E%28-1%29  Right multiply both sides by %28I-A%29%5E%28-1%29

%28I-A%29%5E%28-1%29[%28I-A%29%28I-A%29%5E%28-1%29] = %28A+%2B+I%29[%28I-A%29%28I-A%29%5E%28-1%29] Matrix multiplication is associative.

%28I-A%29%5E%28-1%29I = %28A%2BI%29I  A matrix times its inverse is I

%28I-A%29%5E%28-1%29 = A%2BI  Property of the identity matrix I

Be sure to discard the scratch work.

Edwin