I have two $3\times3$ matrices $A$ and $B$ where $$A = [c_1 : c_2 : c_3]$$ $$B = [c_1 : c_1 + c_2 : c_1+c_2+c_3]$$ where $c_i$ is the $i^{th}$ column of $A$. Given that $|A| = 1$, I am to find the rank and determinant of $B$. Now since $A$ has a non-zero determinant, it will have a full rank, so $rank(A) = 3$ and all its columns will be linearly independent. It is obvious that $B$'s columns will also be linearly independent, so its rank will also be 3. Also, since the determinant is invariant under column operations and $B$ can be obtained by applying column operations to $A$, $|B| = |A| = 1$.
However, I was trying to follow an alternate line of reasoning to arrive at $|B|$...
Since the vectors $c_1, c_2$, and $c_3$ are linearly independent and have a size of 3, they are a basis for $\mathbb{R^3}$. (I am not too sure about this part but I cannot think of how 3 linearly independent vectors of size 3 would not span $\mathbb{R^3}$). Off course, $\mathbb{R^3}$ is also spanned by the basis $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ which can be represented by the matrix
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$
Which means we can obtain $A$ and $B$ by applying column operations to $I$ because they all have the same column space. Hence, $|A| = |B| = |I| = 1$. However, my reasoning would also imply that every $3\times3$ matrix with a non-zero determinant (which will give it full rank) must have a determinant equaling 1.
Did I do something wrong?
Edit: Okay, so this will not be true for real numbers but will this be true for matrices with positive non-zero integers?
$B= \begin{bmatrix} 1& 0 & 0 \\ 1 & 1 & 0 \\ 1& 1 & 1 \end{bmatrix} A$.
Since $\det (CD) = \det C \det D$ you can see that $\det B = \det A$. This gives you the rank of $B$ in this case.
Note: In this example, you can reduce $B$ to $A$ by elementary row operations all of which have determinant 1. This is not true in general, if you multiply by a row scaling operation then the determinant changes.