This might be one of the easiest problems of linear algebra, but just to make sure I am not doing something completely stupid (as usual).
The exercise I have asks us to determine the rank of 2 matrices. The rank of 2 matrices is the number of linear independent rows or column. The rank of the columns and rows is equal.
The first matrix is:
$\left(\begin{matrix} 2 & -1 & 5 & -2 \\ 3 & 6 & -9 & 2 \\ -4 & 3 & 7 & 11\end{matrix}\right)$
Looking just for the rows, I can immediately see that none of the rows is a multiple of the other, so the 3 rows are linear independent. Therefore the rank of the matrix is 3.
The second matrix is:
$\left(\begin{matrix} 3 & 2 & 1 & 5 & 4 \\ 5 & 4 & 3 & 7 & 6 \\ 6 & 4 & 2 & 10 & 8 \\ -10 & -8 & -6 & -14 & -12 \\ 4 & 5 & -2 & -7 & 13 \\ \end{matrix}\right)$
We can see that the 3rd row is the double of the first one. So we remove 1 from the possible maximum rank 5. We can also observe that the 4th row is the negative double of row 2, so we exclude it.
Then the rank of this matrix would be also 3.
Am I missing something?
Edit: Let's obtain the RREF of the first matrix first.
$$\left(\begin{matrix} 2 & -1 & 5 & -2 \\ 3 & 6 & -9 & 2 \\ -4 & 3 & 7 & 11\end{matrix}\right)$$
Dividing the first row by $a_{11} = 2$ we have
$$\left(\begin{matrix} 1 & -0.5 & 2.5 & -1 \\ 3 & 6 & -9 & 2 \\ -4 & 3 & 7 & 11\end{matrix}\right)$$
Subtracting 3 times and -4 times the first row from the second row and the third row respectively we have
$$\left(\begin{matrix} 1 & -\frac{1}{2} & \frac{5}{2} & -1 \\ 0 & \frac{15}{2} & -\frac{33}{2} & 5 \\ 0 & 1 & 17 & 7\end{matrix}\right)$$
Dividing the second row by $a_{22} = \frac{15}{2}$ yields
$$\left(\begin{matrix} 1 & -\frac{1}{2} & \frac{5}{2} & -1 \\ 0 & 1 & -\frac{11}{5} & \frac{2}{3} \\ 0 & 1 & 17 & 7\end{matrix}\right)$$
Subtracting $\frac{-1}{2}$ times and 1 times the second row from the first row and the third row respectively we have
$$\left(\begin{matrix} 1 & 0 & \frac{7}{5} & -\frac{2}{3} \\ 0 & 1 & -\frac{11}{5} & \frac{2}{3} \\ 0 & 0 & \frac{96}{5} & \frac{19}{3}\end{matrix}\right)$$
Finally, dividing the third row by $a_{33} = \frac{96}{5}$ yields
$$\left(\begin{matrix} 1 & 0 & \frac{7}{5} & -\frac{2}{3} \\ 0 & 1 & -\frac{11}{5} & \frac{2}{3} \\ 0 & 0 & 1 & \frac{95}{288}\end{matrix}\right)$$
and subtracting $\frac{7}{5}$ times and $-\frac{11}{5}$ times the third row from the first row and the second row respectively we have
$$\left(\begin{matrix} 1 & 0 & 0 & -\frac{325}{288} \\ 0 & 1 & 0 & \frac{401}{288} \\ 0 & 0 & 1 & \frac{95}{288}\end{matrix}\right) = U$$
which is the RREF of A. Now what we have done so far is we have found a Gaussian elimination process $M = E_{k}E_{k-1} \ldots E_2E_1$ such that $MA = U$. It is well known that elementary matrices are nonsingular and so is M. Let's see why the above statement
is true in this particular example. Note that the fourth column $\vec{u_4}$ of $U$ can be written into a linear combination of the other three columns of $U$. $$\vec{u_4} = -\frac{325}{288}\vec{u_1} + \frac{401}{288}\vec{u_2} + \frac{95}{288}\vec{u_3}$$ Multiplying $M^{-1}$ on both sides yields $$\vec{a_4} = -\frac{325}{288}\vec{a_1} + \frac{401}{288}\vec{a_2} + \frac{95}{288}\vec{a_3}$$ which means $a_4$ can be written into a linear combination of the other three columns of $A$. Therefore the spanning set $\{\vec{a_1},\vec{a_2},\vec{a_3},\vec{a_4}\}$ for $R(A)$ can be pared down to $\{\vec{a_1},\vec{a_2},\vec{a_3}\}$. Also, $\vec{x} \in N(A)$ if and only if $x_1 = \frac{325}{288}x_4, x_2 = -\frac{401}{288}x_4, x_3 = -\frac{95}{288}x_4$ if and only if $\vec{x} \in \text{span}\{(325,-401,-95,288)^T\}$. Therefore $\text{nullity}(A) = 1$ and by rank-nullity theorem $\text{rank}(A) = n - \text{nullity}(A) = 4 - 1 = 3$. We also find a basis $\{\vec{a_1},\vec{a_2},\vec{a_3}\}$ for $R(A)$.