Electronic – AC circuit problem

I need help with the following problem:

Given the circuit of sinusoidal current (attachment 1) with given data:
$$\underline{E}=100V,\underline{E_1}=40V,\underline{Z}=(10+j10)\Omega,\omega=10^5rad/s,L=1mH,$$
$$C=0.1uF.$$
Find $$\underline{I_L},\underline{U_{16}}$$, active and reactive power in the branch 2-5.

Using the loop current analysis we can find four loops (attachment 2) that correspond to linear system of four complex equations:

$$C_1: (2\underline{Z}+jX_L)\underline{I_{C1}}-\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C3}}+\underline{Z}\underline{I_{C4}}=\underline{E_1}-\underline{E}$$

$$C_2: 2\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C1}}+\underline{Z}\underline{I_{C3}}+\underline{Z}\underline{I_{C4}}=\underline{E_1}+\underline{E}$$

$$C_3: 2\underline{Z}\underline{I_{C3}}-\underline{Z}\underline{I_{C1}}+\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C4}}=\underline{E}$$

$$C_4: (2\underline{Z}-jX_C)\underline{I_{C4}}+2\underline{Z}\underline{I_{C1}}+\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C3}}=\underline{E_1}-\underline{E}$$

This gives:
$$(20+j120)\underline{I_{C1}}-(10+j10)\underline{I_{C2}}-(10+j10)\underline{I_{C3}}+(20+j20)\underline{I_{C4}}=-60$$

$$(-10-j10)\underline{I_{C1}}+(20+j20)\underline{I_{C2}}+(10+j10)\underline{I_{C3}}+(10+j10)\underline{I_{C4}}=140$$

$$(-10-j10)\underline{I_{C1}}+(10+j10)\underline{I_{C2}}+(20+j20)\underline{I_{C3}}+(-10-j10)\underline{I_{C4}}=100$$

$$(20+j20)\underline{I_{C1}}+(10+j10)\underline{I_{C2}}-(10+j10)\underline{I_{C3}}+(20-j80)\underline{I_{C4}}=-60$$

After reducing to 3×3 system:

$$(30+j230)\underline{I_{C1}}+(-10-j10)\underline{I_{C3}}+(50+j50)\underline{I_{C4}}=20$$

$$(10+j110)\underline{I_{C1}}+(10+j10)\underline{I_{C3}}+(10+j10)\underline{I_{C4}}=20$$

$$(40+j140)\underline{I_{C1}}+(-20-j20)\underline{I_{C3}}+(40-j60)\underline{I_{C4}}=-120$$

After reducing to 2×2 system:

$$(40+j340)\underline{I_{C1}}+(60+j60)\underline{I_{C4}}=60$$

$$(-20-j320)\underline{I_{C1}}+(-60-j160)\underline{I_{C4}}=-160$$

$$
\begin{bmatrix}
40+j340 & 60+j60 \\
-20-j320 & -60-j160 \\
\end{bmatrix} \begin{bmatrix}
\underline{I_{C1}} \\
\underline{I_{C4}} \\
\end{bmatrix}=\begin{bmatrix}
60 \\
-160 \\
\end{bmatrix}\Rightarrow
$$

$$\begin{bmatrix}
40+j340 & 60+j60 & 60+j0 \\
-20-j320 & -60-j160 & -160+j0 \\
\end{bmatrix}=$$

$$\begin{bmatrix}
40 & -340 & 60 & -60 & 60 & 0 \\
340 & 40 & 60 & 60 & 0 & 60 \\
-20 & 320 & -60 & 160 & -160 & 0 \\
-320 & -20 & -160 & -60 & 0 & -160 \\
\end{bmatrix}
$$

Reduced row echelon form of this matrix is:
$$\begin{bmatrix}
1 & 0 & 0 & 0 & 1275/7481 & -240/7481 \\
0 & 1 & 0 & 0 & 240/7481 & 1275/7481 \\
0 & 0 & 1 & 0 & 303/7481 & 7688/7481\\
0 & 0 & 0 & 1 & -7688/7481 & 303/7481 \\
\end{bmatrix}$$

Now:

$$\underline{I_{C1}}=\frac{1275}{7481}+j\frac{240}{7481},\underline{I_{C4}}=\frac{303}{7481}-j\frac{7688}{7481}\Rightarrow \underline{I_{C3}}=\frac{8209}{7481}-j\frac{15089}{7481},$$$$\underline{I_{C2}}=\frac{22565}{7481}-j\frac{14675}{7481}$$

$$\underline{I_L}=\underline{I_{C1}},\underline{U_{16}}=-jX_C \underline{I_{16}},\underline{I_{16}}=\underline{I_{C2}}\Rightarrow \underline{U_{16}}=-\frac{1467500}{7481}-j\frac{2256500}{7481}$$

Active and reactive power in the branch 2-5 can be found by complex apparent power, $$\underline{S_{25}}=\underline{U_{25}}\underline{{I_{52}}^{*}}$$

$$\underline{I_{52}}=\underline{I_{C1}}+\underline{I_{C2}}+\underline{I_{C3}}=\frac{32049}{7481}-j\frac{29524}{7481}$$
$$\underline{U_{25}}=\underline{E_1}-\underline{I_{52}}\underline{Z}=-\frac{316490}{7481}-j\frac{25250}{7481}\Rightarrow \underline{S_{25}}=-\frac{9397707010}{55965361}-j\frac{10153288010}{55965361}$$

$$\Rightarrow P=-\frac{9397707010}{55965361} W,Q=-\frac{10153288010}{55965361} var$$

Question: Could someone check if the results are correct?

UPDATE:

Question: What type of simulation in OrCAD Capture CIS Lite 16.6 can be used for checking these results?