We are going to use following elementary row operations:

\(\circ \text{Interchange i-th and j-th row:} R_i \leftrightarrow R_j\)

\circ \text{Multiply i-th row by scalar} \alpha R_i \leftrightarrow \alpha \cdot R_i\)

\(\text{Add} \alpha \text{times i-th row to j-th row: } R_j \rightarrow R_j+\alpha \cdot R_i\)

\(\begin{bmatrix} 2 &3 & 5\\ -\frac{2}{3} & -1 & -2 \end{bmatrix} R_2\rightarrow3 \cdot R_2 \begin{bmatrix} 2 &3 & 5\\ -2 & -3 & -6 \end{bmatrix}\)

\(R_2 \rightarrow R_2+R_1 \begin{bmatrix} 2 &3 & 5\\ 0 & 0 & -1 \end{bmatrix}\)

The second equation is 0=-1 . Contradiction , so the system doesnt have a solution.

\(\circ \text{Interchange i-th and j-th row:} R_i \leftrightarrow R_j\)

\circ \text{Multiply i-th row by scalar} \alpha R_i \leftrightarrow \alpha \cdot R_i\)

\(\text{Add} \alpha \text{times i-th row to j-th row: } R_j \rightarrow R_j+\alpha \cdot R_i\)

\(\begin{bmatrix} 2 &3 & 5\\ -\frac{2}{3} & -1 & -2 \end{bmatrix} R_2\rightarrow3 \cdot R_2 \begin{bmatrix} 2 &3 & 5\\ -2 & -3 & -6 \end{bmatrix}\)

\(R_2 \rightarrow R_2+R_1 \begin{bmatrix} 2 &3 & 5\\ 0 & 0 & -1 \end{bmatrix}\)

The second equation is 0=-1 . Contradiction , so the system doesnt have a solution.