Let A be a square matrix of order 2 such that |A| = 2 and the sum of its diagonal elements is −3. If the points (x, y) satisfying $$A^2 + xA + yI = O$$ lie on a hyperbola whose length of semi major axis is x and semi minor axis is y, eccentricity is e and the length of the latus rectum is l, then $$81(e^4 + l^2)$$ is equal to ______.

We have a 2×2 matrix $$A$$ with $$|A| = 2$$ and the sum of its diagonal elements equal to $$-3$$.

Since for a 2×2 matrix the characteristic equation can be written as $$\lambda^2 - (\text{trace})\,\lambda + \det(A) = 0$$, substituting the given trace $$-3$$ and determinant $$2$$ yields $$\lambda^2 + 3\lambda + 2 = 0$$ which factors as $$(\lambda + 1)(\lambda + 2) = 0$$.

This gives the eigenvalues $$\lambda_1 = -1$$ and $$\lambda_2 = -2$$.

By the Cayley-Hamilton theorem, $$A$$ satisfies its characteristic equation, so $$A^2 + 3A + 2I = O$$.

Comparing this with the general form $$A^2 + xA + yI = O$$ shows that $$x = 3$$ and $$y = 2$$.

Interpreting these values as the semi-major and semi-minor axes gives $$a = x = 3$$ and $$b = y = 2$$, so the equation of the hyperbola becomes $$\frac{X^2}{9} - \frac{Y^2}{4} = 1$$.

Next, since $$c^2 = a^2 + b^2 = 9 + 4 = 13$$, it follows that $$e = \frac{c}{a} = \frac{\sqrt{13}}{3}$$, hence $$e^2 = \frac{13}{9}$$ and $$e^4 = \frac{169}{81}$$.

The length of the latus rectum is given by $$l = \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$$, so $$l^2 = \frac{64}{9}$$.

Finally, evaluating $$81(e^4 + l^2)$$ yields $$81\left(\frac{169}{81} + \frac{64}{9}\right) = 81 \times \frac{169}{81} + 81 \times \frac{64}{9},$$ which simplifies to $$= 169 + 9 \times 64 = 169 + 576 = \mathbf{745}$$.

The answer is $$\mathbf{745}$$.