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Given the matrix $A = [e_n, e_{n-1}, ..., e_1] \in \mathbb{Z}^{n,n} $, where $e_i$ is the ith column of the identity matrix. What is $det(A)$?

I get two different results with two calculations.

My logic: I can use Laplace expansion by selecting the first column and the last row and then I have:

$det(A) = 1 \cdot det(A(1, n))$

where $A(1, n)$ is the minor of $A$ when removing the first column and the nth row.

Then, I can continue and go up the matrix.

In case (1), I stopped when I reached the matrix $\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$. The determinant of a $2x2$ matrix is$ad - bc$ which gives me in this case $0 \cdot 0 - 1 \cdot1 = -1$ for the last element in the product, which means that $det(A) = -1$, since I only multipled by $1$s up to this point.

In case (2), I stopped when I reached the $1x1$ matrix, when the determinant is just the single element. The final matrix is $\begin{bmatrix} 1 \end{bmatrix}$, so the last element in the product is $1$, which gives $det(A) = 1$.

I obviously made a mistake somewhere but I can't figure out and it's so simple that it really annoys me. Is the $ad - bc$ rule applicable in every case?

Thanks, Norbert

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  • $\begingroup$ I do not understand why you try Laplace expansion. $\endgroup$
    – mvw
    Aug 26 '18 at 15:01
  • $\begingroup$ Did you include the $(-1)^{i+j}=(-1)^{n+1}$ factor in you expansion? $\endgroup$
    – mvw
    Aug 26 '18 at 15:02
  • $\begingroup$ @mvw I am just trying to practice calculating determinants in different ways. Given that I couldn't calculate this with Laplace expansion, I think it was a worthwhile exercise even if I could've calculated it simpler. $\endgroup$
    – user388680
    Aug 26 '18 at 15:10
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The error lies in the first approach. That $1$ should be $-1$. If you start with the $i$th line of the matrix, you should mltiply by $(-1)^{i+1}$.

For instance, if you compute$$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}$$ doing a Laplace expansion along the second line, you get$$-d\begin{vmatrix}b&c\\h&i\end{vmatrix}+e\begin{vmatrix}a&c\\g&i\end{vmatrix}-f\begin{vmatrix}a&b\\g&h\end{vmatrix}.$$

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  • $\begingroup$ Thanks. I knew it was something simple :) $\endgroup$
    – user388680
    Aug 26 '18 at 15:07
  • $\begingroup$ @norbertk I'm glad I could help. $\endgroup$ Aug 26 '18 at 15:17

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