2
$\begingroup$

We have an $n{\times}n$ matrix, with two's everywhere, expect the diagonal, where there are one's. We're asked to calculate the determinant, and I'm having trouble understading step of the solution:

\begin{vmatrix} 1 & 2 & 2 & \ldots & 2 \\ 2 & 1 & 2 & \ldots & 2 \\ 2 & 2 & 1 & \ldots & 2 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 2 & 2 & 2 & \ldots & 1 \notag \end{vmatrix}

\begin{align}= \end{align} \begin{align}(2n-1) \begin{vmatrix} 1 & 1 & 1 & \ldots & 1 \\ 2 & 1 & 2 & \ldots & 2 \\ 2 & 2 & 1 & \ldots & 2 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 2 & 2 & 2 & \ldots & 1 \notag \end{vmatrix} \end{align}

I know there's a property that if you multiply a row by a constant, then the determinant is multiplied by that constant. But here we are multiplying each element by a different constant.

And also wanted to ask if there some algorithm or pattern to solve this type of problems? I'm doing problems on determinants and every one has a very unique solution, that there's no way I could come up with myself. So can I approach them?

Edit: this is the rest of the solution, in case someone comes across this question and needs the whole solution:

You then add the first row multiplied by (-2) to all the others and get this:

\begin{align}= \end{align} \begin{align}(2n-1) \begin{vmatrix} 1 & 1 & 1 & \ldots & 1 \\ 0 & -1 & 0 & \ldots & 0 \\ 0 & 0 & -1 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & -1 \notag \end{vmatrix} \end{align}

So the answer is 1 times the det of a diagonal matrix will all (-1)'s.

$$=(2n-1)(-1)^{n-1} $$

$\endgroup$
3
  • 1
    $\begingroup$ Other than brute force calculation, there is no general approach that works for all problems. Sometimes problems of this type (where the dimension is general $n\times n$) can be solved by induction, e.g. if you develop the determinant with respect to the first row, the first cofactor will be exactly the same determinat but of dimension $(n-1)\times (n-1)$ plus/minus other cofactors. Then you notice that other cofactors differ from each other only by how the rows are arranged. Sometimes it is worthwhile to plug in a specific $n$ to see what's going on. $\endgroup$ Dec 6 '19 at 0:07
  • 1
    $\begingroup$ You can take look at the answer to a question about a slightly more general determinant. $\endgroup$
    – Bernard
    Dec 6 '19 at 0:17
  • 1
    $\begingroup$ There’s an IMO simpler way to compute this determinant if you know that it’s the product of the eigenvalues. $\endgroup$
    – amd
    Dec 6 '19 at 2:57
1
$\begingroup$

Hint: Show that the determinant doesn't change from you add a row/column to another row.

Step which the solution didn't explain: Now, to the first row, add all the other rows.
Clearly each entry in the first row is $2n-1$, which we can factor out.


For the general problem, we can try various techniques similar to this, including expanding along cofactors, using induction, etc.

$\endgroup$
4
  • $\begingroup$ Sorry, I forgot to mention, this is only a part of the solution, then they do what you say, add the first row (*-2) to the others... but I don't understand this step in particular. Where does the (2n-1) come from? $\endgroup$
    – Belen
    Dec 5 '19 at 23:59
  • $\begingroup$ @SofiaB.Lopez in each column the entry $2$ appears $2(n-1)=2n-2$ times and $1$ appears once. So if you add that all together you get $2(n-1)+1=2n-1$. $\endgroup$ Dec 6 '19 at 0:10
  • $\begingroup$ Okay, that makes sense, but this is like a rule, that I can turn a single row into 1's and multiply the whole thing by its sum? I think there's something obvious I'm not seeing $\endgroup$
    – Belen
    Dec 6 '19 at 8:19
  • $\begingroup$ I just got it!! Thank you for helping! I hadn't understood the part where you add all the other rows to the first one. $\endgroup$
    – Belen
    Dec 6 '19 at 8:54

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.