# Thread: what will be the determinant of the matrix A

1. ## what will be the determinant of the matrix A

Let A be 3 by 3 matrix over real numbers satisfying A⁻¹= I -2A. Then what will be the

determinant of the matrix A?

2. ## Re: what will be the determinant of the matrix A

Originally Posted by sorv1986
Let A be 3 by 3 matrix over real numbers satisfying A⁻¹= I -2A. Then what will be the determinant of the matrix A?
Perhaps (perhaps) that is not the exact formulation of the problem. Could you review it?

3. ## Re: what will be the determinant of the matrix A

There are two possible answers. First, the determinant of a matrix is the same as the product of all of its eigenvalues. Second, if A satisfies that equation, its eigenvalues must also. You can manipulate the equation to a quadratic that has two solutions. Since A is 3 by 3, it has three eigenvalues. Since only two numbers sastisfy that equation, one of them must be a duplicate eigenvalue. The determinant depends upon which of them is the duplicate.

4. ## Re: what will be the determinant of the matrix A

Originally Posted by FernandoRevilla
Perhaps (perhaps) that is not the exact formulation of the problem. Could you review it?
i m sry .but that was the only information given about the matriX A.

5. ## Re: what will be the determinant of the matrix A

det(AB) = det(A)det(B)

Determinant is non-zero if it is invertible

det(A) = 1/(det(A^-1))

6. ## Re: what will be the determinant of the matrix A

Originally Posted by HallsofIvy
The determinant depends upon which of them is the duplicate.
But that is not possible. We have $\displaystyle A^{-1}=I-2A\Leftrightarrow 2A^2-A+I=0$ . That is , $\displaystyle p(\lambda)=2(\lambda-\lambda_1)(\lambda-\lambda_2)$ is an annihilator polynomial of $\displaystyle A$ being $\displaystyle \lambda_1=(1+\sqrt{7}i)/4$ and $\displaystyle \lambda_2=(1-\sqrt{7}i)/4$ . As $\displaystyle A\in\mathbb{R}^{3\times 3}$ , $\displaystyle A-\lambda_1I\neq 0$ and $\displaystyle A-\lambda_2I\neq 0$ . This means that the minimal polynomial of $\displaystyle A$ is just $\displaystyle (1/2)p(\lambda)$ and $\displaystyle \lambda_1,\lambda_2$ are eigenvalues of $\displaystyle A$ as you said. The third one $\displaystyle \lambda_3$ should be necessarily real and $\displaystyle A$ is diagonalizable in $\displaystyle \mathbb{C}$ . So, $\displaystyle A$ is similar (in $\displaystyle \mathbb{C}$) to $\displaystyle D=\textrm{diag}\;(\lambda_1,\lambda_2,\lambda_3)$ and as a consequence:

$\displaystyle 2D^2-D+I\Rightarrow 2\lambda_i^2-\lambda_i+1=0\;(i=1,2,3)$

Again, as you said $\displaystyle \lambda_3=\lambda_1$ or $\displaystyle \lambda_3=\lambda_2$ which implies $\displaystyle \det A=\det D=\lambda_1\lambda_2\lambda_3=(1/2)\lambda_3\not\in\mathbb{R}$ (contradiction).

7. ## Re: what will be the determinant of the matrix A

Originally Posted by CaramelCardinal
Determinant is non-zero if it is invertible det(A) = 1/(det(A^-1))
Rigorously true.

8. ## Re: what will be the determinant of the matrix A

Originally Posted by FernandoRevilla
But that is not possible. We have $\displaystyle A^{-1}=I-2A\Leftrightarrow 2A^2-A+I=0$ . That is , $\displaystyle p(\lambda)=2(\lambda-\lambda_1)(\lambda-\lambda_2)$ is an annihilator polynomial of $\displaystyle A$ being $\displaystyle \lambda_1=(1+\sqrt{7}i)/4$ and $\displaystyle \lambda_2=(1-\sqrt{7}i)/4$ . As $\displaystyle A\in\mathbb{R}^{3\times 3}$ , $\displaystyle A-\lambda_1I\neq 0$ and $\displaystyle A-\lambda_2I\neq 0$ . This means that the minimal polynomial of $\displaystyle A$ is just $\displaystyle (1/2)p(\lambda)$ and $\displaystyle \lambda_1,\lambda_2$ are eigenvalues of $\displaystyle A$ as you said. The third one $\displaystyle \lambda_3$ should be necessarily real and $\displaystyle A$ is diagonalizable in $\displaystyle \mathbb{C}$ . So, $\displaystyle A$ is similar (in $\displaystyle \mathbb{C}$) to $\displaystyle D=\textrm{diag}\;(\lambda_1,\lambda_2,\lambda_3)$ and as a consequence:

$\displaystyle 2D^2-D+I\Rightarrow 2\lambda_i^2-\lambda_i+1=0\;(i=1,2,3)$

Again, as you said $\displaystyle \lambda_3=\lambda_1$ or $\displaystyle \lambda_3=\lambda_2$ which implies $\displaystyle \det A=\det D=\lambda_1\lambda_2\lambda_3=(1/2)\lambda_3\not\in\mathbb{R}$ (contradiction).
how [COLOR=rgb(0, 0, 0)]lambda3=lambda1 or lambda3= lambda2[/COLOR] is possible? lambda_3 must be real as the matrix is 3 by 3.Is that problem is wrong or something? Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me

9. ## Re: what will be the determinant of the matrix A

Originally Posted by CaramelCardinal
det(AB) = det(A)det(B)

Determinant is non-zero if it is invertible

det(A) = 1/(det(A^-1))
how the det of A^-1 i.e. det (I-2A) would be calculated?

10. ## Re: what will be the determinant of the matrix A

Originally Posted by sorv1986
Is that problem is wrong or something?
Plainly: $\displaystyle A$ does not exist.

Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me
The minimal polynomial $\displaystyle \mu (\lambda)$ of $\displaystyle A$ divides to any annihilating polynomial of $\displaystyle A$ so $\displaystyle \mu(\lambda)=\lambda-\lambda_1$ or $\displaystyle \mu(\lambda)=\lambda-\lambda_2$ or $\displaystyle \mu(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)$ . As $\displaystyle A-\lambda_1I\neq 0$ and $\displaystyle A-\lambda_2I\neq 0$ we conclude that $\displaystyle \mu(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)$ is the minimal polynomial of $\displaystyle A$ .