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
Difficulty Level
Suggested Book
Schaum’s Outline of Linear Algebra

Start with hints

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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, characteristic values, proper values, or latent roots & Eigen vectors are In linear algebra,) (An eigenvector or characteristic vector of a linear transformation is a non-zero vector that changes by only a scalar factor when that linear transformation is applied to it. More formally, if T is a linear transformation from a vector space V over a field F into itself and v is a vector in V that is not the zero vector, then v is an eigenvector of T if T(v) is a scalar multiple of v. This condition can be written as the equation )
So ,( 1 + i) be the roots of the characteristic poly of A Now A is a real matrix, char poly of A \(\epsilon \mathbb{R}[x]\) [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 values (-2)(1+i)(1-i) = detA => det(A) = -2[1 +1] = -4

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