If $A$ is a matrix of size $n\times n$, and $A^2+A+2I=0$

Question

If $A$ is a matrix of size $n\times n$, and $A^2+A+2I=0$, check whether $A$ is singular or not and find its inverse if it exists.

I can find the inverse by simply multiplying the given equation with $A^{-1}$.

$$A+I+2A^{-1}=0$$

$$A^{-1}=-\frac{1}{2}(A+I)$$

But how can I check if its singular or not?

The exam is tomorrow @N.S.JOHN My future will be decided tomorrow. — Aditya Dev, May 21 '16 at 15:22
@N.S.JOHN hopefully. I am just an average student. I have BITS and NIT with me. — Aditya Dev, May 21 '16 at 15:24
Rewrite as $\left(-\frac{1}{2}A-\frac{1}{2}I\right)A=I$. That shows $A$ is invertible (non-singular) by exhibiting the inverse. — André Nicolas, May 21 '16 at 15:25
An invertible matrix is non-singular, there's no point to calculate the determinant. — Bernard, May 21 '16 at 15:56
It is worth pointing out that the nonzero determinant should be viewed as a consequence, instead of the definition of invertibility/nonsingularity. Based on my experience, there are many people in this forum confused this. — Zhanxiong, May 21 '16 at 18:08

Dietrich Burde · Accepted Answer · 2016-05-21T17:36:52.287

5

Write $A(A+I)=-2I$ and take the determinant on both sides. This gives $$\det(A)\det(A+I)=(-2)^n\neq 0,$$ hence $\det(A)\neq 0$.

edited May 21 '16 at 17:36

answered May 21 '16 at 15:20

Dietrich Burde

score 4 · Answer 2 · answered May 21 '16 at 15:21

4

If $A$ is singular, then there exists a vector $v\neq 0$ such that $Av=0$. In that case $$ 0=0v=(A^2+A+2I)v=A(Av)+Av+2v=2v, $$ a contradiction.

answered May 21 '16 at 15:21

guestDiego

2 Answers2