To solve the system of equations:
[tex]\[
\begin{aligned}
12I_1 + 12I_2 - 6I_3 &= -18 \quad \quad \quad \quad(1) \\
-6I_1 + 12I_2 + 12I_3 &= 0 \quad \quad \quad \quad(2) \\
12I_1 - 6I_2 + 12I_3 &= 72 \quad \quad \quad \quad(3)
\end{aligned}
\][/tex]
we can use matrix methods to find the values for [tex]\( I_1 \)[/tex], [tex]\( I_2 \)[/tex], and [tex]\( I_3 \)[/tex].
1. Formulate the system as a matrix equation [tex]\( AX = B \)[/tex]:
First, we write the coefficient matrix [tex]\( A \)[/tex]:
[tex]\[
A = \begin{pmatrix}
12 & 12 & -6 \\
-6 & 12 & 12 \\
12 & -6 & 12
\end{pmatrix}
\][/tex]
Next, we write the constant matrix [tex]\( B \)[/tex]:
[tex]\[
B = \begin{pmatrix}
-18 \\
0 \\
72
\end{pmatrix}
\][/tex]
Then, the unknowns matrix [tex]\( X \)[/tex]:
[tex]\[
X = \begin{pmatrix}
I_1 \\
I_2 \\
I_3
\end{pmatrix}
\][/tex]
The system of equations can be rewritten as:
[tex]\[
AX = B
\][/tex]
2. Solve the matrix equation [tex]\( AX = B \)[/tex] for [tex]\(X\)[/tex]:
We use matrix operations to find [tex]\( X \)[/tex]. The solution matrix [tex]\( X \)[/tex] can be found using the inverse of matrix [tex]\( A \)[/tex] if it exists, such that:
[tex]\[
X = A^{-1}B
\][/tex]
However, for convenience, we use known methods or computational tools that provide the solution directly.
By solving:
[tex]\[
\begin{aligned}
X &= \begin{pmatrix}
I_1 \\
I_2 \\
I_3
\end{pmatrix} \\
&= \begin{pmatrix}
2.0 \\
-2.0 \\
3.0
\end{pmatrix}
\end{aligned}
\][/tex]
Thus, the solutions to the system of equations are:
[tex]\[
I_1 = 2.0, \quad I_2 = -2.0, \quad I_3 = 3.0
\][/tex]
