1. **Problem Statement:** Solve the determinant equation for part (a): $$\begin{vmatrix} 1 & 4 & 4 \\ 1 & -2 & 1 \\ 1 & 2x & x^2 \end{vmatrix} = 0$$ 2. **Formula and Rules:** The determinant of a 3x3 matrix $$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)$$ We will apply this formula to expand the determinant. 3. **Calculate the determinant:** Let the matrix elements be: $a=1$, $b=4$, $c=4$ $d=1$, $e=-2$, $f=1$ $g=1$, $h=2x$, $i=x^2$ Calculate each minor: $ei - fh = (-2)(x^2) - (1)(2x) = -2x^2 - 2x$ $di - fg = (1)(x^2) - (1)(1) = x^2 - 1$ $dh - eg = (1)(2x) - (-2)(1) = 2x + 2$ Now substitute: $$\det = 1(-2x^2 - 2x) - 4(x^2 - 1) + 4(2x + 2)$$ Simplify: $$= -2x^2 - 2x - 4x^2 + 4 + 8x + 8$$ $$= (-2x^2 - 4x^2) + (-2x + 8x) + (4 + 8)$$ $$= -6x^2 + 6x + 12$$ 4. **Set determinant equal to zero and solve:** $$-6x^2 + 6x + 12 = 0$$ Divide both sides by -6: $$x^2 - x - 2 = 0$$ Factor: $$(x - 2)(x + 1) = 0$$ So, $$x = 2 \quad \text{or} \quad x = -1$$ **Final answer:** $x = 2$ or $x = -1$