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To solve this circuit problem using the **node-voltage method**, we will first label the nodes and then derive the matrix equation using Kirchhoff's Current Law (KCL).

Here are the steps to follow:

1. **Node labeling**:
   - There are three nodes in this circuit (excluding the reference ground), which we will call \( V_1 \), \( V_2 \), and \( V_3 \).
   - The reference ground is usually connected to the negative terminal of the circuit.

2. **KCL Equations**:
   At each node, we apply KCL: the sum of currents leaving the node is equal to the sum of currents entering the node.

### Node 1 (\( V_1 \)):

For node \( V_1 \):
- There are resistances 1Ω (to the 3A source), 5Ω (to node \( V_2 \)), and 10Ω (to node \( V_3 \)).
- The 5A current source directly enters node \( V_1 \).

Applying KCL at \( V_1 \):

\[
\frac{V_1 - 0}{1} + \frac{V_1 - V_2}{5} + \frac{V_1 - V_3}{10} = 5
\]

### Node 2 (\( V_2 \)):

For node \( V_2 \):
- There are resistances 5Ω (to node \( V_1 \)), 8Ω (to node \( V_3 \)), and 4Ω (to ground).
- The 1A current source directly enters node \( V_2 \).

Applying KCL at \( V_2 \):

\[
\frac{V_2 - V_1}{5} + \frac{V_2 - V_3}{8} + \frac{V_2 - 0}{4} = 1
\]

### Node 3 (\( V_3 \)):

For node \( V_3 \):
- There are resistances 10Ω (to node \( V_1 \)), 8Ω (to node \( V_2 \)), and 2Ω (to ground).
- The 4A current source directly enters node \( V_3 \).

Applying KCL at \( V_3 \):

\[
\frac{V_3 - V_1}{10} + \frac{V_3 - V_2}{8} + \frac{V_3 - 0}{2} = 4
\]

### Matrix Form:

These equations can now be expressed in matrix form as follows:

\[
\begin{pmatrix}
\frac{1}{1} + \frac{1}{5} + \frac{1}{10} & -\frac{1}{5} & -\frac{1}{10} \\
-\frac{1}{5} & \frac{1}{5} + \frac{1}{8} + \frac{1}{4} & -\frac{1}{8} \\
-\frac{1}{10} & -\frac{1}{8} & \frac{1}{10} + \frac{1}{8} + \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
V_1 \\
V_2 \\
V_3
\end{pmatrix}
=
\begin{pmatrix}
5 \\
1 \\
4
\end{pmatrix}
\]

Now we can simplify the values for each element in the matrix:

\[
\begin{pmatrix}
1.3 & -0.2 & -0.1 \\
-0.2 & 0.575 & -0.125 \\
-0.1 & -0.125 & 0.725
\end{pmatrix}
\begin{pmatrix}
V_1 \\
V_2 \\
V_3
\end{pmatrix}
=
\begin{pmatrix}
5 \\
1 \\
4
\end{pmatrix}
\]

This is the **node-voltage matrix equation** for the given circuit.

Would you like further details on solving this matrix or any other part of the process?

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Here are 5 questions to expand the discussion:
1. What is Kirchhoff's Current Law, and how does it apply to circuit analysis?
2. How do you determine which nodes are essential for node-voltage analysis?
3. Why do we use the node-voltage method instead of mesh analysis for certain circuits?
4. How can you solve the matrix equation obtained in node-voltage analysis?
5. What are the implications of dependent and independent sources in node-voltage analysis?

**Tip**: When writing KCL equations, always ensure that you're consistent with the direction of currents (whether entering or leaving the node) to avoid sign errors.

Write down the node-voltage matrix equation for the given circuit.

This detailed solution shows how to apply Kirchhoff's Current Law and the node-voltage method to write the matrix equation for a circuit containing resistors and current sources. Learn step-by-step how to derive the equations and construct the matrix for a complex circuit.

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