Is the Square Matrix A Invertible Given A^2 + A + I = 0?

In summary, A is a square matrix satisfying A^2 + A + I = 0. Based on the equation, it can be concluded that A is invertible and its inverse is equal to -A-I or A^2. This can be shown by multiplying A with -A-I and proving it to be the identity matrix.
  • #1
Yankel
395
0
A is a square matrix such as:

A^2 + A + I = 0

Which answer is correct ?

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 (Worried)

thanks
 
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  • #2
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 .
 
  • #3
Yankel said:
A is a square matrix such as:

A^2 + A + I = 0

Which answer is correct ?

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 (Worried)

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 ----------

PaulRS said:
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:
  • #4
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 !

:-)
 
  • #5


The answer to this question is c. We cannot determine if A is invertible based on the given information. In order to determine if a square matrix is regular or invertible, we need to know more about its properties, such as its determinant or rank. Without this information, it is not possible to determine if A is invertible.
 

FAQ: Is the Square Matrix A Invertible Given A^2 + A + I = 0?

What is a square matrix?

A square matrix is a type of matrix where the number of rows is equal to the number of columns. It is represented by the notation n x n, where n represents the number of rows and columns.

What does it mean for a square matrix to be regular?

A square matrix is considered regular if it is non-singular, meaning that it has an inverse. This means that the matrix can be multiplied by its inverse to produce the identity matrix.

How can I determine if a square matrix is regular?

To determine if a square matrix is regular, you can calculate its determinant. If the determinant is non-zero, then the matrix is regular. If the determinant is zero, then the matrix is singular and not regular.

Why is it important to know if a square matrix is regular?

Knowing if a square matrix is regular is important in many areas of mathematics, including linear algebra, differential equations, and optimization. It allows for easier computation and solving of equations involving the matrix.

Can a square matrix be regular and singular at the same time?

No, a square matrix cannot be both regular and singular at the same time. If the determinant is non-zero, then the matrix is regular. If the determinant is zero, then the matrix is singular. The two terms are mutually exclusive.

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