Show the determinant of an identity matrix multiplied by a vector is equal to an element of the vector I'm working out a few exercises for an exam, this is an interesting problem that should be simple (about 2 marks) but I can't seem to wrap my head around it. The question is:
Let $I$ be the $3\times 3$ Identity matrix. Show that: 
$$
\det(I_2(x))=x_2,\quad where \quad x=(x_1,x_2,x_3)\in\mathbb R^3
$$
(the $x$ is simply a vector in 3 dimensional space, I couldn't figure out how to do the correct notation on this site)
Now, I don't know what that subscript 2 of the Identity matrix means, but either it is the $2\times2$ Identity matrix which makes no sense or it means something else.
If it was a typing mistake where I got the question from and it was supposed to be a 3, then multiplying the $3\times3$ Identity matrix with $x$ and calculating the determinant gives me: $$\det(I_3(x))=x_1x_2x_3$$
Otherwise the secret might just be that the subscript 2 means calculating some sort of limited determinant?
Can anyone make sense of the question? Thank you in advance.
 A: I think I have understood that your (non standard !) notation $I_2$ means (by Cramer's rule) substitution of the second column of $I$ (whence number 2) by vector $x$:
$$\begin{vmatrix}1&x_1&0\\0&x_2&0\\0&x_3&1\end{vmatrix}$$
which in fact is $x_2$.
Remark : Has your lecturer advised you to use notation $I_2$ ?  
A: That notation is terrible, but let's look at the general phenomenon. Let $I$ be an $n \times n$ identity matrix, and $x = (x_1,x_2,\dots,x_n)$. Let $I_j(x)$ denote the matrix obtained by replacing column $j$ of $I$ by $x$. 
Claim: $\det(I_j(x)) = x_j$.
To compute the determinant of a matrix we can expand by minors along any row or column. Let's expand along the the $j^{th}$ row.
Notice that the only $\textit{potentially}$ nonzero entry in the $j^{th}$ row is $x_j$. Let $B$ be the matrix obtained by deleting the $j^{th}$ row and column of $I_j(x)$. Then $$\det(I_j(x)) = x_j \det(B)$$ 
But $B$ is an $(n-1)\times (n-1)$ identity matrix! So $\det(B)=1$. Thus $$\det(I_j(x)) = x_j.$$
Note that when you expand by minors there are some pesky alternating negative signs. But in $I_j(x)$, $x_j$ sits on the main diagonal and entries on the main diagonal always have a positive effect in the determinant sum.
A: Using the Weinstein-Aronszajn determinant identity,
$$\det \left(\mathrm{I}_n - \mathrm{e}_i \mathrm{e}_i^\top + \mathrm{x} \mathrm{e}_i^\top \right) = \det \left( \mathrm{I}_n + (\mathrm{x} - \mathrm{e}_i) \, \mathrm{e}_i^\top \right) = 1 + \mathrm{e}_i^\top (\mathrm{x} - \mathrm{e}_i) = 1 + x_i - 1 = \color{blue}{x_i}$$
