1. **Problem 1:** Given matrix $$A = \begin{pmatrix} x - 1 & j \\ j2y - jx & j \end{pmatrix}$$ where $$j = \sqrt{-1}$$, find real values of $$x$$ and $$y$$ such that $$A$$ is singular. 2. **Recall:** A matrix is singular if its determinant is zero. 3. **Calculate determinant:** $$\det(A) = (x - 1) \cdot j - j \cdot (j2y - jx)$$ 4. Simplify the second term: $$j \cdot (j2y - jx) = j^2 2y - j^2 x = (-1) 2y - (-1) x = -2y + x$$ 5. Substitute back: $$\det(A) = j(x - 1) - (-2y + x) = j(x - 1) + 2y - x$$ 6. For $$\det(A) = 0$$, real and imaginary parts must be zero separately: - Real part: $$2y - x = 0$$ - Imaginary part: $$x - 1 = 0$$ 7. Solve: - From imaginary part: $$x = 1$$ - Substitute into real part: $$2y - 1 = 0 \Rightarrow y = \frac{1}{2}$$ --- 8. **Problem 2:** Given impedance formula for parallel circuit: $$\frac{1}{Z} = \frac{1}{R + j\omega L + \frac{1}{j\omega C}}$$ with $$R=1$$, $$L=2$$, $$C=0.5$$, find $$\omega > 0$$ for resonance (when $$Z$$ is real). 9. **Rewrite denominator:** $$R + j\omega L + \frac{1}{j\omega C} = R + j\omega L - \frac{j}{\omega C} = R + j\left(\omega L - \frac{1}{\omega C}\right)$$ 10. For $$Z$$ to be real, denominator must be purely real or imaginary part zero: $$\omega L - \frac{1}{\omega C} = 0$$ 11. Solve for $$\omega$$: $$\omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC}$$ 12. Substitute values: $$\omega^2 = \frac{1}{2 \times 0.5} = \frac{1}{1} = 1$$ 13. Therefore: $$\omega = 1$$ (since $$\omega > 0$$) **Final answers:** - For singular matrix: $$x = 1$$, $$y = \frac{1}{2}$$ - For resonance frequency: $$\omega = 1$$