# Square Matrix, is it regular ?

#### Yankel

##### Active member
A is a square matrix such as:

A^2 + A + I = 0

a. A^-1 = A
b. A^-1 = A^2
c. We can't tell if A is invertable
d.A is not invertable
e. A^-1 = A + I

I have tried to play with the equation, I tried to multiply in A^-1 and to isolate it, but it didn't get me anywhere....I need help

thanks

#### PaulRS

##### Member
Note that A^2 + A + I = 0 implies A^2 + A = - I and so - A^2 - A = I

But - A^2 - A = A ( -A - I) thus A ( - A - I ) = I . Now, clearly this means A^{-1} = -A - I because det (A) det (-A-I) = det ( A (-A - I) ) = det I = 1 thus it is invertible.

Finally A^2 + A + I = 0 implies A^2 = - A - I ... so in fact A^{-1} = A^2 .

#### CaptainBlack

##### Well-known member
A is a square matrix such as:

A^2 + A + I = 0

a. A^-1 = A
b. A^-1 = A^2
c. We can't tell if A is invertable
d.A is not invertable
e. A^-1 = A + I

I have tried to play with the equation, I tried to multiply in A^-1 and to isolate it, but it didn't get me anywhere....I need help

thanks
We can rewrite the equation that A satisfies as:

-(A^2+A)=I

hence:

-(A+I)A=I

and:

-A(A+I)=I

So -(A+I) is the inverse of A and -(A+I)=A^2

CB

---------- Post added at 11:16 AM ---------- Previous post was at 11:12 AM ----------

Note that A^2 + A + I = 0 implies A^2 + A = - I and so - A^2 - A = I

But - A^2 - A = A ( -A - I) thus A ( - A - I ) = I . Now, clearly this means A^{-1} = -A - I because det (A) det (-A-I) = det ( A (-A - I) ) = det I = 1 thus it is invertible.

Finally A^2 + A + I = 0 implies A^2 = - A - I ... so in fact A^{-1} = A^2 .
You don't need the discussion of determinants, A is invertible if there exists a matrix B such that AB=BA=I.

That is if you show that B is both a right and a left inverse of A then A is invertible and B is the inverse.

CB

Last edited:

#### Yankel

##### Active member
Thanks guys !!

I am ashamed, I actually got to the point of A(-A-I)=I and did figure out that A^-1=-A-I.....silly of me not to to go back and see that A^2 is also -A-I, it's so obvious from the question....thanks for that !