Let L, M, N be non-singular 3 X 3 matrices where L2 = L-1, M = L8, and N = L2. What is the determinant of (M-N)?

The correct option is 0.

To understand why the determinant of $M-N$ is equal to 0, let us analyze the given matrix equations step-by-step.

We are given that $L$ is a non-singular $3\times 3$ matrix. By definition, a non-singular matrix has a non-zero determinant ($det\left(L\right)\ne 0$), which also means its inverse $L^{-1}$ exists.

We are given the relation:
$L^2=L^{-1}$

Let us multiply both sides of this equation by $L$ from either the left or the right:
$L·L^2=L·L^{-1}$
Since $L·L^{-1}=I$ (where $I$ is the $3\times 3$ identity matrix), we get:
$L^3=I$

Now, let us look at the expressions for the matrices $M$ and $N$ in terms of $L$:
$M=L^8$
$N=L^2$

We can simplify $M=L^8$ using the relation $L^3=I$:
$M=L^8=L^{6+2}=L^6·L^2={\left(L^3\right)}^2·L^2$
Substituting $L^3=I$:
$M=I^2·L^2=I·L^2=L^2$

We now have:
$M=L^2$
And we were given:
$N=L^2$

Let us find the difference matrix $M-N$:
$M-N=L^2-L^2=O$
where $O$ represents the zero matrix.

Finally, we calculate the determinant of $M-N$:
$det(M-N)=det\left(O\right)=0$

Thus, the determinant of $M-N$ is indeed 0.