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Eigen value of A:  1 , -1 and x

18 is eigen value of $A^2 +3A$==>A's 3rd Eigen value of A be either 3 or -6.

|A|=product of eigen value=1*-1*(3||6) !=0 So invertible and for |$A^2 +3A$ |=18*4*(-2) which is also non zero.

Hence Both invertible.

can  u @srestha @Shaik Masthan plz verify.

@Abhisek Tiwari 4

yes correct I think :)

Thanks.. ;)