#csa#written_test

Question

If A³ =0 , then det of A-3I = ?

Answer 1

This is Nilpotent Matrix

All the eigenvalues of Nilpotent matrix are all 0

Let the eigenvalues of A(n X n matrix) be $\lambda _1, \lambda _2, \lambda _3, .........................\lambda _n$

$\lambda _1 = 0$

$\lambda _1 = 0$

$\lambda _2 = 0$

$\lambda _3 = 0$

$.$

$.$

$.$

$\lambda _n = 0$

eigenvalues of $A - 3.I \ be \ \lambda _1 -3, \lambda _2-3, \lambda _3-3, .........................\lambda _n-3$         $\because eigenvalues \ of \ A - k.I \ is \ \lambda -k, \ k \ is \ scalar$

 

eigenvalues of $A - 3.I \ be \  -3, -3, -3, .........................-3$

$determinant \ of \ matrix = product \ of \ eigenvalues$

$(-3) * (-3) * (-3) * (-3) ...................*(-3)$

$(-1)^n.3^n$