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?
Best Answer
The fact that none of the rows in the first matrix is a multiple of another does not mean that the rank of the matrix is $3$. For example, consider the matrix whose rows are $(3,5)$, $(1,2)$, $(2,3)$. No row is a multiple of another, yet the sum of the second and third is the first. And indeed this matrix cannot possibly have rank $3$ since it has only two columns. As SRX said, the way to determine the rank is to row-reduce it.