# 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

##### Topic

Linear Algebra

##### Difficulty Level

Medium

##### Suggested Book

Schaum’s Outline of Linear Algebra

# Start with hints

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**eigenvector**Â Â orÂ**characteristic vector**Â 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**v**Â is a vector inÂ VÂ that is not theÂ zero vector, thenÂ**v**Â 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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