Gauss Elimination from kirchoff's Law Circuit

Question

In this system there are three magnitudes of current flowing 1, 2, and 3 in the appropriate direction on picture.

a. Set up a system of linear equations that can represent the problem of electrical circuits with the help of Kirchoff's law.

b. Determine the magnitude of the electric current 1, 2, and 3 by solving the system of equations obtained in problem (a) using Gauss/Gauss-Jordan elimination.

Can anyone please check my answer:

I deleted my answer which consist the full solution since it might cause this post getting deleted.

Answer 1

From your answer you have uploaded, you have made a slight mistake in the third row.

Omitting the units, the KVL equation you should be getting is: $$ \begin{align} 90 & = 15i_3+10i_2+10i_3 \\ & =10i_2+25i_3 \\ \end{align} $$ This can be further reduced to: $$ \begin{align} 2i_2+5i_3 & = 18 \\ \end{align} $$ Hence, the corresponding KCL and KVL equations you have to form is: $$ \begin{bmatrix} 1 & -1 & 1\\ 2 & 1 & 0\\ 0 & 2 & 5\\ \end{bmatrix} \begin{bmatrix} i_1\\ i_2\\ i_3\\ \end{bmatrix} = \begin{bmatrix} 0\\ 8\\ 18\\ \end{bmatrix} $$ I leave you to solve for the currents. You should be getting $i_1 = 2A, i_2 = 4A$ and $i_3 = 2A$. Hope it helps.

Answer 2

Hint:

By KCL at junction Q (or P):

$$i_2 = i_1 +i_3$$

The potential difference across QP in the three branches are equal and by using Ohm's law:

$$-20i_1 +80V = +10i_2 = -10i_3 + 90V + 15i_3$$

You have 3 independent equations in 3 variables, $i_1, i_2, i_3$ that you can solve using Gaussian elimination.