2
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Can you tell us the text where this problem comes from? Just a wild guess: Is this somewhere near Cramer's rule? $\endgroup$ Jun 7 '16 at 12:25
  • 3
    $\begingroup$ Wouldn't multiplying a 3x3 Identity matrix with the vector $x$ just give $x$? The determinant of $x$ isn't defined is it $\endgroup$
    – Joshua Lin
    Jun 7 '16 at 12:26
  • $\begingroup$ 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. $\endgroup$
    – Simon
    Jun 7 '16 at 12:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – hardmath
    Jun 7 '16 at 13:15
9
$\begingroup$

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 $I_2$ ?

$\endgroup$
4
  • $\begingroup$ That is the exact answer! Thank you. I do not have enough reputation to up vote though :( $\endgroup$
    – Simon
    Jun 7 '16 at 12:44
  • $\begingroup$ I imagine the instructors sitting on swivel chair's, puffing cigars, and laughing uncontrollably at the notion of using uncommon notation in exams. $\endgroup$
    – Simon
    Jun 7 '16 at 12:47
  • $\begingroup$ I did not suggest that, but one thing is sure : having good/thorough notations is at the very basis of mathematical activity. $\endgroup$
    – Jean Marie
    Jun 7 '16 at 12:52
  • $\begingroup$ You are a sorcerer! $\endgroup$ Jun 7 '16 at 18:25
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Ken Duna
    Jun 7 '16 at 18:40
0
$\begingroup$

Using the Weinstein-Aronszajn determinant identity,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.