1. **Problem 1:** Solve for $x$ and $y$ such that matrix $$A=\begin{pmatrix} j2y - jx & j \\ j & j2y - jx \end{pmatrix}$$ is singular, where $j=\sqrt{-1}$. 2. **Recall:** A matrix is singular if its determinant is zero. For a 2x2 matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix},$$ the determinant is $$ad - bc.$$ 3. **Calculate determinant of $A$: $$\det(A) = (j2y - jx)(j2y - jx) - (j)(j) = (j2y - jx)^2 - j^2.$$** 4. **Simplify:** Since $j^2 = -1$, we have $$\det(A) = (j2y - jx)^2 - (-1) = (j2y - jx)^2 + 1.$$ 5. **Expand:** $$(j2y - jx)^2 = (j2y)^2 - 2(j2y)(jx) + (jx)^2 = j^2 4y^2 - 2j^2 2yx + j^2 x^2 = -4y^2 + 4yx - x^2.$$ 6. **Substitute back:** $$\det(A) = (-4y^2 + 4yx - x^2) + 1 = -4y^2 + 4yx - x^2 + 1.$$ 7. **Set determinant to zero for singularity:** $$-4y^2 + 4yx - x^2 + 1 = 0.$$ 8. **Rewrite:** $$-4y^2 + 4yx - x^2 = -1.$$ Multiply both sides by $-1$: $$4y^2 - 4yx + x^2 = 1.$$ 9. **Recognize quadratic form:** $$4y^2 - 4yx + x^2 = (2y - x)^2 = 1.$$ 10. **Solve:** $$2y - x = \pm 1.$$ 11. **Express $y$ in terms of $x$: $$y = \frac{x \pm 1}{2}.$$** --- 12. **Problem 2:** Find $$\omega$ for resonance in the parallel circuit with impedance $$\frac{1}{Z} = \frac{1}{R + j \omega L} + j \omega C,$$ where $R=1$, $L=2$, $C=0.5$, and $\omega > 0$. 13. **Resonance condition:** Impedance $Z$ is real, so imaginary part of $\frac{1}{Z}$ must be zero. 14. **Calculate:** $$\frac{1}{Z} = \frac{1}{1 + j2\omega} + j0.5\omega.$$ 15. **Rationalize denominator:** $$\frac{1}{1 + j2\omega} = \frac{1 - j2\omega}{(1)^2 + (2\omega)^2} = \frac{1 - j2\omega}{1 + 4\omega^2}.$$ 16. **Separate real and imaginary parts:** $$\frac{1}{Z} = \frac{1}{1 + 4\omega^2} - j \frac{2\omega}{1 + 4\omega^2} + j0.5\omega = \frac{1}{1 + 4\omega^2} + j\left(0.5\omega - \frac{2\omega}{1 + 4\omega^2}\right).$$ 17. **Set imaginary part to zero:** $$0.5\omega - \frac{2\omega}{1 + 4\omega^2} = 0.$$ 18. **Solve:** Multiply both sides by $1 + 4\omega^2$: $$0.5\omega (1 + 4\omega^2) = 2\omega.$$ 19. **Divide both sides by $\omega$ (since $\omega > 0$):** $$0.5 (1 + 4\omega^2) = 2.$$ 20. **Simplify:** $$0.5 + 2\omega^2 = 2 \Rightarrow 2\omega^2 = 1.5 \Rightarrow \omega^2 = 0.75.$$ 21. **Final:** $$\omega = \sqrt{0.75} = \frac{\sqrt{3}}{2}.$$