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: