Kirchoff's Law equation 3 unknowns Help

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SUMMARY

The discussion focuses on solving a system of equations derived from Kirchoff's Law for an electrical circuit with three unknown currents: i1, i2, and i3. The equations provided are: i1 + 8i2 + 3i3 = -31, 3i1 - 2i2 + i3 = -5, and 2i1 - 3i2 + 2i3 = 6. Through Gaussian elimination, the intermediate steps have been shown, leading to a reduced row echelon form that simplifies the solution process. The final step requires back substitution to determine the values of i1, i2, and i3.

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JonB
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Please could somebody help me with this Kirchoff's law question with explanations/working out? I've been on it for hours and cannot see where I'm going wrong!

Q. An electrical circuit comprises three closed loops giving the following equations for the currents i1, i2 and i3.

i1 + 8i2 + 3i3 = -31
3i1 - 2i2 + i3 = -5
2i1 - 3i2 + 2i3 = 6

Solve for i1, i2 and i3

Any help appreciated!
 
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We have
$$
\left[\begin{array}{rrr|r}
1 &8 &3 &-31 \\
3 &-2 &1 &-5 \\
2 &-3 &2 &6
\end{array} \right]_{\begin{array}{rr}-3R_1+R_2 \mapsto R_2 \\ -2R_1+R_3 \mapsto R_3\end{array}} \to
\left[\begin{array}{rrr|r}
1 &8 &3 &-31 \\
0 &-26 &-8 &88 \\
0 &-19 &-4 &68
\end{array} \right]_{-\frac{19}{26}R_2 + R_3 \mapsto R_3} \to
\left[\begin{array}{rrr|r}
1 &8 &3 &-31 \\
0 &-26 &-8 &88 \\
0 &0 &\frac{24}{13} &\frac{48}{13}
\end{array} \right].
$$
Can you finish?
 

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