If the determinant \( \left|\begin{matrix} x^2-x+1 & x-1 \\ x+1 & x+1 \end{matrix}\right| = 16 \), then the number of distinct real solutions for \( x \) is:

Determinant \( = (x^2-x+1)(x+1) - (x-1)(x+1) = (x^3+1) - (x^2-1) = x^3 - x^2 + 2 \). Setting \( x^3 - x^2 + 2 = 16 \implies x^3 - x^2 - 14 = 0 \). Derivative \( 3x^2-2x > 0 \) outside \( [0, 2/3] \), with \( f(2) = -10 < 0 \) and \( f(3) = 4 > 0 \), there is exactly 1 real root.