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# Application of eigenvalue in degree 3 polynomial: ISI MMA 2018 Question 14

This is a cute and interesting problem based on application of eigen values in 3 degree polynomial .Here we are finding the determinant value .

# Understand the problem

Let A be a 3 x 3 real matrix with all the diagoanl entries equal to 0 . If 1 + i is an eigen value of A , the determinant of A equal ?
##### Source of the problem
Sample Questions (MMA ) :2019
Linear Algebra
Medium
##### Suggested Book
Schaum’s Outline of Linear Algebra

Do you really need a hint? Try it first!

Let A be a 3 x 3 real matrix with trace 0 Now( 1+i) be an eigen value . (Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristicÂ values, properÂ values, or latent roots & Eigen vectors areÂ InÂ linear algebra,) (AnÂ eigenvectorÂ Â orÂ characteristic vectorÂ of aÂ linear transformationÂ is a non-zeroÂ vectorÂ that changes by only a scalar factor when that linear transformation is applied to it. More formally, ifÂ TÂ is a linear transformation from aÂ vector spaceÂ VÂ over aÂ fieldÂ FÂ into itself andÂ vÂ is a vector inÂ VÂ that is not theÂ zero vector, thenÂ vÂ is an eigenvector ofÂ TÂ ifÂ T(v)Â is a scalar multiple ofÂ v. This condition can be written as the equationÂ )
So ,( 1 + i) be the roots of the characteristic poly of AÂ  Now A is a real matrix, char poly of AÂ  $\epsilon \mathbb{R}[x]$ [Right!] Therefore ( 1 – i) is also a root of char poly of A
deg( char poly of A ) =3 So , it has two imaginary roots & one real root Let real root be r Note tr(A) = 0 => r + 1 +i +1 -i = 0 , => r = -2 Can you play with the determinants now ?
We know the determinant of A is the product of all eigen valuesÂ  (-2)(1+i)(1-i) = detA => det(A) = -2[1 +1] = -4

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