 Gaussian elimination with a row swap

Consider this original linear system and its augmented matrix:The first diagonal entry is zero, so it cannot serve as a pivot element. Interchange $R_1$ and $R_2$ to obtain a pivot element of one:This chooses a convenient nonzero entry for exact arithmetic; partial pivoting would instead interchange $R_1$ and $R_3$, selecting the largest absolute value, two. For our chosen ordering, the first-column elimination multipliers are $m_2=0$ and $m_3=2$. The elementary row operation $R_3\leftarrow R_3-2R_1$ givesThe next pivot element is two and the next elimination multiplier is $7/2$. Applying $R_3\leftarrow R_3-\frac{7}{2}{R}_2$ gives the upper triangular matrix systemEvery operation also changed the right-hand side. Back substitution now givesChecking the original equations, rather than only the reduced ones, verifies the answer by matrix multiplication:Thus a zero first diagonal entry was a row-order obstruction, not a singular matrix. Indeed, the diagonal product of the final upper triangular matrix is $7$; the single row interchange makes $\det A=-7$, since the two row additions preserve the determinant.