Suppose a real matrix A satisfies (A^3)=A,A is not equal to I[identity matrix] or 0[null matrix].

If rank((A))=r and trace((A))=t ,then –

a)r>=t and r+t is odd

b)r>=t and r+t is even

c)r<t and r+t is odd

d)r<t and r+t is even

Let us focus on the given information: A^3=A and A is not equal to I(The Identity matrix) and also A is not equal to O(The Null matrix)

Now A^3=A

i.e., A^3-A=O

i.e., A(A^2-I)=O

i.e., either A=O or A^2-I=O but it is already given to us that A is not equal to O , Hence we have A^2-I=O, i.e., A^2=I which implies that A=+I,-I.

Again we have that A is not equal to +I, therefore we take A=-I which implies

rank(A)=r=n(If A and I is of order n) and trace(A)=t=-n (adding -1, n times). Hence we arrive at r+t=0 which is even and r>=t essentially.

Therefore option b) is the correct one.