1. **State the problem:** We need to find the value of $x$ for which the matrix $$\begin{pmatrix}8 & x & 0 \\ 4 & 0 & 2 \\ 12 & 6 & 0\end{pmatrix}$$ becomes singular. 2. **Recall the definition:** A matrix is singular if its determinant is zero. 3. **Write the determinant formula for a 3x3 matrix:** $$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ where the matrix is $$\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}$$ 4. **Identify elements:** $$a=8, b=x, c=0, d=4, e=0, f=2, g=12, h=6, i=0$$ 5. **Calculate the determinant:** $$\det = 8(0 \cdot 0 - 2 \cdot 6) - x(4 \cdot 0 - 2 \cdot 12) + 0(4 \cdot 6 - 0 \cdot 12)$$ $$= 8(0 - 12) - x(0 - 24) + 0$$ $$= 8(-12) - x(-24)$$ $$= -96 + 24x$$ 6. **Set determinant to zero for singularity:** $$-96 + 24x = 0$$ 7. **Solve for $x$:** $$24x = 96$$ $$x = \frac{96}{24} = 4$$ **Final answer:** The matrix is singular when $x = 4$.