# Find the determinant of $A$ satisfying $A^{-1}=I-2A.$

I am stuck with the following problem:

Let $A$ be a $3\times 3$ matrix over real numbers satisfying $A^{-1}=I-2A.$
Then find the value of det$(A).$

I do not know how to proceed. Can someone point me in the right direction? Thanks in advance for your time.

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I'm not too well-versed in matrices, but I have a feeling that you can left-multiply both sides by $A$ and rewrite it as a 2nd degree polynomial. The roots would then be the possible determinants. –  Ryan Dec 17 '12 at 5:41
Well. The least I can say is that $\det(A) \neq 0$. Honestly I must say I'm clueless for the moment. –  Patrick Da Silva Dec 17 '12 at 6:01
Since $$2A^2 - A + I = 0,$$ the minimal polynomial of $A$ divides $2\lambda ^2 - \lambda + 1$, so the possible eigenvalues are $$\frac{1 \pm \sqrt 7 i}{4}.$$ Don't know if that helps. –  Patrick Da Silva Dec 17 '12 at 6:03
@user52976 Could you confirm what you have written is in fact the right problem? –  user17762 Dec 17 '12 at 6:10
@PatrickDaSilva thank you sir for your effort and concern.I can understand it. –  user52976 Dec 17 '12 at 6:42
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No such $A$ exists. Hence we cannot speak of its determinant.

Suppose $A$ is real and $A^{-1}=I-2A$. Then $A^2-\frac12A+\frac12I=0$. Hence the minimal polynomial $m_A$ of $A$ must divide $x^2-\frac12x+\frac12$, which has no real root. Therefore $m_A(x)=x^2-\frac12x+\frac12$. But the minimal polynomial and characteristic polynomial $p_A$ of $A$ must have identical irreducible factors, and this cannot happen because $p_A$ has degree 3 and $m_A$ is an irreducible polynomial of degree 2.

Edit: The OP says that the question appears on an extrance exam paper, and four answers are given: (a) $1/2$, (b) $−1/2$, (c) $1$, (d) $2$. It seems that there's a typo in the exam question and $A$ is probably 2x2. If this is really the case, then the above argument shows that the characteristic polynomial of $A$ is $x^2-\frac12x+\frac12$. Hence $\det A = 1/2$.

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You have shown that no real matrix $A$ exists. A complex matrix $A$ probably exists. –  Patrick Da Silva Dec 17 '12 at 6:10
@PatrickDaSilva Sure, but the OP says that $A$ is real. –  user1551 Dec 17 '12 at 6:11
Or if we allowed $A$ to be $4$ by $4$. –  JSchlather Dec 17 '12 at 6:13
@JacobSchlather Yep. Maybe there's a typo in the OP's textbook/assignment. –  user1551 Dec 17 '12 at 6:19
@PatrickDaSilva Nevermind. I misread questions 53.487561% of the time. –  user1551 Dec 17 '12 at 6:36
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Let $\lambda$ be a real eigenvalue with eigenvector $x$ (there is a real root to the characteristic equation). Since $A$ is invertible, $\lambda\neq 0$, so $Ax = \lambda x$ and $A^{-1}x=\lambda^{-1}x.$ Putting these into $A^{-1}x=x-2Ax$ gives $2\lambda^2-\lambda+1=0,$ contradicting that $\lambda$ is real. Hence no such $A$ exists.

It is now abundantly clear there was a typo in the question. I showed above no such real matrix exists. Even if we allow complex entries, the characteristic polynomial has the form $$(x^2-x/2+1/2)(x-z)$$ for some $z\in \mathbb{C}\setminus{\mathbb{R}}$ and the determinant is not real, so not one of the options.

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a real matrix does not necessarily have real eigenvalues –  Jonathan Dec 17 '12 at 6:37
@Jonathan A real 3 by 3 matrix has at least one real eigenvalue. –  Ragib Zaman Dec 17 '12 at 6:38
ah, true. characterlimit –  Jonathan Dec 17 '12 at 6:40
There is a formula for the characteristic polynomial for $3 \times 3$ matrices which says that $$c_A(\lambda) = \det(A - \lambda I) = -\lambda^3 + c_2 \lambda^2 - c_1 \lambda + \det(A).$$ (There is an analog for $2 \times 2$ matrices but I don't think there is a general formula for every $n$.) Since $$2A^2 - A + I = 0,$$ we know that the minimal polynomial $m_A(\lambda)$ satisfies $$m_A(\lambda) \, | \, 2\lambda^2 - \lambda + 1 = 2 \left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right) \left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right).$$ Therefore, the characteristic polynomial is among the following : $$-\left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right)^3 \\ -\left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right)^2 \left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right) \\ -\left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right) \left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right)^2 \\ -\left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right)^3.$$ This gives you only four options for the determinant (I didn't compute them, it probably gives ugly complex numbers). I don't know how to distinguish the four possibilities though. Maybe $\det(A) \neq 0$ can help. Maybe.
As pointed out in other answers, no such real matrix exist. For complex matrix $A$, note that $$I=2AA^{-1}=A(I-2A)=A-2A^2$$ Now let us assume $A$ is diagonal with all the diagonal entries equal to $a$, then each diagonal entry of $A-2A^2$ equals $a-2a^2$. Now from the equation $a-2a^2=1$, we have $a=(1\pm\sqrt{7}i)/4$. So for complex $A$ possible values of $\det (A)$ are $a^3.$