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 $2×2$ 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 $1×1$ 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
Best Answer
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$$-d\begin{vmatrix}b&c\\h&i\end{vmatrix}+e\begin{vmatrix}a&c\\g&i\end{vmatrix}-f\begin{vmatrix}a&b\\g&h\end{vmatrix}.$$