Resistor Coefficient Voltage Matrix Doesn't have a Solution
In the following schematic, \$I_1\$, \$I_2\$, and \$I_3\$ are the interior loops, which I consider to be walked "clockwise" starting always at their lower left corner. Those are set up with KVL. There is also \$V_\text{A}\$ and \$V_\text{B}\$. Those are set up with KCL. This would make for 5 equations and 5 unknowns:







simulate this circuit &ndash; Schematic created using CircuitLab

So, it's my belief that you were being asked to solve for all three loop currents using KVL and both unknown node voltages using KCL. The equations would be:

$$\begin{align*}
0\:\text{V}+8\:\text{V}-4\:\text{k}\Omega\,\cdot\,I_1-2\:\text{k}\Omega\,\cdot\left(I_1-I_2\right)&amp;= 0\:\text{V}\\\\
0\:\text{V}-2\:\text{k}\Omega\,\cdot\left(I_2-I_1\right)-1\:\text{k}\Omega\,\cdot\,I_2+4\:\text{V}&amp;=0\:\text{V}\\\\
0\:\text{V}-4\:\text{V}-6\:\text{k}\Omega\,\cdot\,I_3-1\:\text{k}\Omega\,\cdot\,I_3&amp;=0\:\text{V}\\\\
\frac{V_\text{A}}{4\:\text{k}\Omega}+\frac{V_\text{A}}{2\:\text{k}\Omega}+\frac{V_\text{A}}{1\:\text{k}\Omega}&amp;=\frac{8\:\text{V}}{4\:\text{k}\Omega}+\frac{0\:\text{V}}{2\:\text{k}\Omega}+\frac{-4\:\text{V}}{1\:\text{k}\Omega}\\\\
\frac{V_\text{B}}{6\:\text{k}\Omega}+\frac{V_\text{B}}{1\:\text{k}\Omega}&amp;=\frac{-4\:\text{V}}{6\:\text{k}\Omega}+\frac{0\:\text{V}}{1\:\text{k}\Omega}\\\\
\end{align*}$$

I'll let you sort out the matrix from the above statements. It will have a solution.

The results will be \$I_1= 2\frac27\:\text{mA}\$, \$I_2= 2\frac67\:\text{mA}\$, \$I_3= -\frac47\:\text{mA}\$, \$V_\text{A}=-1\frac17\:\text{V}\$, and \$V_\text{B}=-\frac47\:\text{V}\$.

From that you can, of course, completely work out all necessary voltage drops across the resistors, too.