1.14. Using elementary transformations, find the rank of the matrix

$$
A=\left(\begin{array}{ccccc}
5 & 7 & 12 & 48 & -14 \\
9 & 16 & 24 & 98 & -31 \\
14 & 24 & 25 & 146 & -45 \\
11 & 12 & 24 & 94 & -25
\end{array}\right)
$$
Solution. Let's denote the $i$-th row of matrix $A$ by the symbol $\alpha_{i}$. In the first stage, we perform elementary transformations $\alpha_{2}^{\prime}=\alpha_{2}-\alpha_{3}+\alpha_{1}, \alpha_{3}^{\prime}=\alpha_{3}-\alpha_{2}-\alpha_{1} ; \alpha_{4}^{\prime}=\alpha_{4}-\alpha_{3}+\alpha_{1}$.

In the second stage, we perform the transformations

$$
\alpha_{3}^{n}=\alpha_{3}^{\prime}+\alpha_{2}^{\prime}, \alpha_{4}^{\prime \prime}=\alpha_{4}^{\prime}-\alpha_{2}^{\prime}
$$

As a result, we get

$$
\begin{aligned}
& \left(\begin{array}{ccccc}
5 & 7 & 12 & 48 & -14 \\
9 & 16 & 24 & 98 & -31 \\
14 & 24 & 25 & 146 & -45 \\
11 & 12 & 24 & 94 & -25
\end{array}\right) \xrightarrow{1}\left(\begin{array}{ccccc}
5 & 7 & 12 & 48 & -14 \\
0 & -1 & 11 & 0 & 0 \\
0 & 1 & -11 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array}\right) \xrightarrow{2} \\
& \xrightarrow{2}\left(\begin{array}{ccccc}
5 & 7 & 12 & 48 & -14 \\
0 & -1 & 11 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -4 & 6
\end{array}\right) \xrightarrow{3}\left(\begin{array}{ccccc}
5 & 7 & 12 & 48 & -14 \\
0 & -1 & 11 & 0 & 0 \\
0 & 0 & 0 & -4 & 6 \\
0 & 0 & 0 & 0 & 0
\end{array}\right) \xrightarrow{4} \\
& \xrightarrow{4}\left(\begin{array}{ccccc}
5 & 7 & 12 & 12 & -7 \\
0 & -1 & 0 & 2 & 0 \\
0 & 0 & -1 & 0 & 3 \\
0 & 0 & 0 & 0 & 0
\end{array}\right)
\end{aligned}
$$

In the third stage, we swapped the fourth row with the third row, and the third row with the fourth row. In the fourth stage, we divided the elements of the fourth and fifth columns by 4 and 2, respectively, and swapped the third and fourth columns. From the form of the matrix obtained after the fourth stage of transformation, it follows that
rank $A=3$. We could continue transforming matrix $A$, aiming to zero out the remaining elements of the matrix with different indices, but this is probably not advisable when finding the rank of the matrix. It is also worth noting that the resulting zero row from the elementary transformations could have been omitted in further transformations of the matrix, simply by deleting it, which, obviously, would not affect the rank of the original matrix.