Use Gauss Elimination method to solve the linear systems of equations. x1 x2 + x3 = 0 -x1 + x2 - x3 = 0 10x1 + 25x3 = 90 20x1 + 10x2 = 80

$R_2 \leftarrow R_2 + R_1$
$R_3 \leftarrow R_3 - 10R_1$
$R_4 \leftarrow R_4 - 20R_1$

Swap $R_2$ with $R_3$.

Swap $R_3$ with $R_4$.

$R_2 \leftarrow \frac{1}{10}R_2$

$R_3 \leftarrow R_3 - 30R_2$

$R_3 \leftarrow -\frac{1}{65}R_3$

Simplify the fraction: $\frac{190}{65} = \frac{38}{13}$.

From the third row: $x_3 = \frac{38}{13}$.
From the second row: $x_2 + 1.5x_3 = 9$
$x_2 + \frac{3}{2}\left(\frac{38}{13}\right) = 9$
$x_2 + \frac{57}{13} = 9$
$x_2 = 9 - \frac{57}{13} = \frac{117 - 57}{13} = \frac{60}{13}$.
From the first row: $x_1 - x_2 + x_3 = 0$
$x_1 - \frac{60}{13} + \frac{38}{13} = 0$
$x_1 - \frac{22}{13} = 0$
$x_1 = \frac{22}{13}$.
The solution is $x_1 = \boxed{\frac{22}{13}}$, $x_2 = \boxed{\frac{60}{13}}$, $x_3 = \boxed{\frac{38}{13}}$.

Substitute the given values:

The characteristic equation for the homogeneous part is $0.1r^2 + 11r + 100 = 0$, or $r^2 + 110r + 1000 = 0$.
Using the quadratic formula, $r = \frac{-110 \pm \sqrt{110^2 - 4(1000)}}{2} = \frac{-110 \pm \sqrt{12100 - 4000}}{2} = \frac{-110 \pm \sqrt{8100}}{2} = \frac{-110 \pm 90}{2}$.
The roots are $r_1 = -10$ and $r_2 = -100$.
So, $Q_h(t) = A e^{-10t} + B e^{-100t}$.

For the particular solution $Q_p(t)$, we use the phasor method for steady-state current.
The impedance $Z = R + j\left(\omega L - \frac{1}{\omega C}\right)$.
$\omega L = 120\pi \times 0.1 = 12\pi \approx 37.6991\Omega$.
$\frac{1}{\omega C} = \frac{1}{120\pi \times 10^{-2}} = \frac{1}{1.2\pi} \approx 0.2653\Omega$.
$Z = 11 + j(12\pi - \frac{1}{1.2\pi}) \approx 11 + j(37.6991 - 0.2653) = 11 + j37.4338\Omega$.
The magnitude of impedance $|Z| = \sqrt{11^2 + (37.4338)^2} = \sqrt{121 + 1399.98} = \sqrt{1520.98} \approx 38.9997\Omega$.
The phase angle of impedance $\phi_Z = \arctan\left(\frac{37.4338}{11}\right) \approx 73.61^\circ \approx 1.2848$ radians.
The peak current $I_{peak} = \frac{E_{peak}}{|Z|} = \frac{110}{38.9997} \approx 2.8205$ A.
The steady-state current is $I_p(t) = I_{peak} \sin(\omega t - \phi_Z) = 2.8205 \sin(120\pi t - 73.61^\circ)$.
The steady-state charge $Q_p(t)$ leads the current by $90^\circ$ in a capacitive circuit, or lags the voltage by $90^\circ$.
$Q_p(t) = \frac{I_{peak}}{\omega} \sin(\omega t - \phi_Z - \frac{\pi}{2}) = \frac{2.8205}{120\pi} \sin(120\pi t - 73.61^\circ - 90^\circ)$
$Q_p(t) = 0.00747 \sin(120\pi t - 163.61^\circ)$.
Alternatively, $Q_p(t) = C_1 \cos(120\pi t) + C_2 \sin(120\pi t)$, where $C_1 \approx -0.002108$ and $C_2 \approx -0.007173$ (from previous thought process).

$Q(t) = A e^{-10t} + B e^{-100t} + 0.00747 \sin(120\pi t - 163.61^\circ)$.
$I(t) = \frac{dQ}{dt} = -10A e^{-10t} - 100B e^{-100t} + 2.8205 \sin(120\pi t - 73.61^\circ)$.

Using $Q(0)=0$:
$0 = A e^0 + B e^0 + 0.00747 \sin(-163.61^\circ)$
$0 = A + B + 0.00747(-0.2822)$
$0 = A + B - 0.002108$
$A + B = 0.002108 \quad (Eq. 1)$

Using $I(0)=0$:
$0 = -10A e^0 - 100B e^0 + 2.8205 \sin(-73.61^\circ)$
$0 = -10A - 100B + 2.8205(-0.9599)$
$0 = -10A - 100B - 2.707$
$-10A - 100B = 2.707 \quad (Eq. 2)$

From (Eq. 1), $A = 0.002108 - B$. Substitute into (Eq. 2):
$-10(0.002108 - B) - 100B = 2.707$
$-0.02108 + 10B - 100B = 2.707$
$-90B = 2.707 + 0.02108$
$-90B = 2.72808$
$B = -\frac{2.72808}{90} \approx -0.030312$
$A = 0.002108 - (-0.030312) = 0.002108 + 0.030312 = 0.03242$.

The current $I(t)$ is: