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 :)
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