Let $$A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{bmatrix}$$ be a singular matrix. Let $$f(x) = \displaystyle\int_0^x (t^2 + 2t + 3)\,dt$$, $$x \in [1, \alpha]$$. If M and m are respectively the maximum and the minimum values of $$f$$ in $$[1, \alpha]$$, then $$3(M - m)$$ is equal to :

$$A = \begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1 \end{bmatrix}$$

$$\det(A) = 1 \cdot \begin{vmatrix} 1 & \alpha \\ 1 & -1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & -1 \\ 1 & -1 \end{vmatrix} + 0 = 0$$

$$1 \cdot (-1 - \alpha) - 2 \cdot (-3 - (-1)) = 0$$

$$-1 - \alpha + 4 = 0 \implies 3 - \alpha = 0 \implies \alpha = 3$$

Thus, the domain interval for the function is $$x \in [1, 3]$$.

$$f(x) = \int_{0}^{x} (t^2 + 2t + 3) \, dt$$

$$f(x) = \left[ \frac{t^3}{3} + t^2 + 3t \right]_{0}^{x}$$

$$f(x) = \frac{x^3}{3} + x^2 + 3x$$

$$f'(x) = x^2 + 2x + 3$$

$$m = f(1) = \frac{1^3}{3} + 1^2 + 3(1) = \frac{1}{3} + 1 + 3 = \frac{1}{3} + 4 = \frac{13}{3}$$

$$M = f(3) = \frac{3^3}{3} + 3^2 + 3(3) = 9 + 9 + 9 = 27$$

$$3(M - m) = 81 - 13 = 68$$