# 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.

• Can you tell us the text where this problem comes from? Just a wild guess: Is this somewhere near Cramer's rule? Jun 7 '16 at 12:25
• Wouldn't multiplying a 3x3 Identity matrix with the vector $x$ just give $x$? The determinant of $x$ isn't defined is it Jun 7 '16 at 12:26
• It is from a previous exam question, the exact question is here: i.imgur.com/UuuAU8N.png It has to do with determinants in my linear algebra text book by david lay, but the exact question isn't in the book as far as I could tell, but it has to come from the same sections that deal with Cramer's rule. Jun 7 '16 at 12:35
• The notation $I_2(x)$ must not mean a matrix-vector multiplication, but rather the result of replacing the second column of identity matrix $I$ with the column vector $x$. Once that is understood, the problem becomes a straightforward application of Cramer's Rule. Jun 7 '16 at 13:15

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 $$I_2$$ ?

• That is the exact answer! Thank you. I do not have enough reputation to up vote though :( Jun 7 '16 at 12:44
• I imagine the instructors sitting on swivel chair's, puffing cigars, and laughing uncontrollably at the notion of using uncommon notation in exams. Jun 7 '16 at 12:47
• I did not suggest that, but one thing is sure : having good/thorough notations is at the very basis of mathematical activity. Jun 7 '16 at 12:52
• You are a sorcerer! Jun 7 '16 at 18:25

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.

• Another way to see this is that you can obtain $I_j(x)$ from $I$ by a sequence of elementary row operations. The effects of elementary row operations on the determinant are discussed here: How row operations affect the determinant. Jun 7 '16 at 18:40

Using the Weinstein-Aronszajn determinant identity,

$$\det \left(\mathrm{I}_n - \mathrm{e}_i \mathrm{e}_i^\top + \mathrm{x} \mathrm{e}_i^\top \right) = \det \left( \mathrm{I}_n + (\mathrm{x} - \mathrm{e}_i) \, \mathrm{e}_i^\top \right) = 1 + \mathrm{e}_i^\top (\mathrm{x} - \mathrm{e}_i) = 1 + x_i - 1 = \color{blue}{x_i}$$