Given that A = [[4, 9, 2], [-8, 6, -3], [3, 1, -5]], Evaluate |A| + |A^T|. Hence or otherwise solve the equations 4x + 9y + 2z = 21, -8x + 6y − 3z = 41, 3x + y - 5z = -73.

$$|A| = 4 \begin{vmatrix} 6 & -3 \\ 1 & -5 \end{vmatrix} - 9 \begin{vmatrix} -8 & -3 \\ 3 & -5 \end{vmatrix} + 2 \begin{vmatrix} -8 & 6 \\ 3 & 1 \end{vmatrix}$$ $$|A| = 4((6)(-5) - (-3)(1)) - 9((-8)(-5) - (-3)(3)) + 2((-8)(1) - (6)(3))$$ $$|A| = 4(-30 + 3) - 9(40 + 9) + 2(-8 - 18)$$ $$|A| = 4(-27) - 9(49) + 2(-26)$$ $$|A| = -108 - 441 - 52$$ $$|A| = -601$$

Therefore, $|A| + |A^T| = |A| + |A| = 2|A|$.

$$|A| + |A^T| = 2(-601) = -1202$$

The value of $|A| + |A^T|$ is $\boxed{-1202}$.

• Solve the equations: $4x + 9y + 2z = 21$ $-8x + 6y - 3z = 41$ $3x + y - 5z = -73$

We use Cramer's Rule to solve for $x, y, z$. We already found $|A| = -601$.
Calculate the determinant of $A_x$ (matrix $A$ with the first column replaced by the constant terms):

$$A_x = \begin{bmatrix} 21 & 9 & 2 \\ 41 & 6 & -3 \\ -73 & 1 & -5 \end{bmatrix}$$ $$|A_x| = 21((6)(-5) - (-3)(1)) - 9((41)(-5) - (-3)(-73)) + 2((41)(1) - (6)(-73))$$ $$|A_x| = 21(-30 + 3) - 9(-205 - 219) + 2(41 + 438)$$ $$|A_x| = 21(-27) - 9(-424) + 2(479)$$ $$|A_x| = -567 + 3816 + 958 = 4207$$ $$x = \frac{|A_x|}{|A|} = \frac{4207}{-601} = -7$$

$$A_y = \begin{bmatrix} 4 & 21 & 2 \\ -8 & 41 & -3 \\ 3 & -73 & -5 \end{bmatrix}$$ $$|A_y| = 4((41)(-5) - (-3)(-73)) - 21((-8)(-5) - (-3)(3)) + 2((-8)(-73) - (41)(3))$$ $$|A_y| = 4(-205 - 219) - 21(40 + 9) + 2(584 - 123)$$ $$|A_y| = 4(-424) - 21(49) + 2(461)$$ $$|A_y| = -1696 - 1029 + 922 = -1803$$ $$y = \frac{|A_y|}{|A|} = \frac{-1803}{-601} = 3$$

$$A_z = \begin{bmatrix} 4 & 9 & 21 \\ -8 & 6 & 41 \\ 3 & 1 & -73 \end{bmatrix}$$ $$|A_z| = 4((6)(-73) - (41)(1)) - 9((-8)(-73) - (41)(3)) + 21((-8)(1) - (6)(3))$$ $$|A_z| = 4(-438 - 41) - 9(584 - 123) + 21(-8 - 18)$$ $$|A_z| = 4(-479) - 9(461) + 21(-26)$$ $$|A_z| = -1916 - 4149 - 546 = -6611$$ $$z = \frac{|A_z|}{|A|} = \frac{-6611}{-601} = 11$$

The solution to the system of equations is $\boxed{x = -7, y = 3, z = 11}$.

Q2.(a) Evaluate in polar form

• (i) $\frac{16 \angle 75^\circ}{2 \angle 15^\circ}$

$$\frac{16 \angle 75^\circ}{2 \angle 15^\circ} = \left(\frac{16}{2}\right) \angle (75^\circ - 15^\circ)$$ $$= 8 \angle 60^\circ$$

The result is $\boxed{8 \angle 60^\circ}$.

• (ii) $\frac{10 \angle \frac{\pi}{4} \times 12 \angle \frac{\pi}{2}}{6 \angle -\frac{\pi}{3}}$

Multiply magnitudes and add angles.

$$10 \angle \frac{\pi}{4} \times 12 \angle \frac{\pi}{2} = (10 \times 12) \angle \left(\frac{\pi}{4} + \frac{\pi}{2}\right)$$ $$= 120 \angle \left(\frac{\pi}{4} + \frac{2\pi}{4}\right)$$ $$= 120 \angle \frac{3\pi}{4}$$

Divide magnitudes and subtract angles.

$$\frac{120 \angle \frac{3\pi}{4}}{6 \angle -\frac{\pi}{3}} = \left(\frac{120}{6}\right) \angle \left(\frac{3\pi}{4} - \left(-\frac{\pi}{3}\right)\right)$$ $$= 20 \angle \left(\frac{3\pi}{4} + \frac{\pi}{3}\right)$$ $$= 20 \angle \left(\frac{9\pi}{12} + \frac{4\pi}{12}\right)$$ $$= 20 \angle \frac{13\pi}{12}$$

The result is $\boxed{20 \angle \frac{13\pi}{12}}$.

Q2.(b) If $Z = R + j\omega L + \frac{1}{j\omega C}$, express $Z$ in $(a+jb)$ form when $R=10, L=5, C=0.04$ and $\omega=4$.

$$\frac{1}{j(0.16)} = \frac{1}{j(0.16)} \times \frac{j}{j} = \frac{j}{j^2(0.16)}$$

Since $j^2 = -1$:

$$= \frac{j}{-(0.16)} = -\frac{1}{0.16}j$$

Convert the fraction to a decimal: $\frac{1}{0.16} = \frac{100}{16} = \frac{25}{4} = 6.25$.
So, $\frac{1}{j(0.16)} = -6.25j$.

$$Z = 10 + j20 - j6.25$$ $$Z = 10 + j(20 - 6.25)$$ $$Z = 10 + j13.75$$

The expression for $Z$ in $(a+jb)$ form is $\boxed{10 + j13.75}$.