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At least in my current Linear Algebra course, exercises concerning the determinant of a matrix of dimension $n\geq 5$ more often than not require one to notice some property of the matrix so that its determinant becomes obvious and one needs not compute it in the usual way. I'm trying to find a simple way to realize that the determinant of $$\begin{pmatrix} 0 & -1 & -1 & -1 & -1 \\ 1 & 0 & -1 & -1 & -1 \\ 1 & 1 & a & -1 & -1 \\ 1 & 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 1 & 0\end{pmatrix}$$

is equal to $a$, and I would appreciate anyone who points me in the right direction.


Also, since these are the types of questions that might go to my (as well as other students') final, I'm open to users giving other examples of matrices which determinants are obvious without the need to compute them, here are a few:

$$\det \begin{pmatrix} -4 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -4 & 1 \\ 1 & 1 & -4 & 1 & 1 \\ 1 & -4 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & -4\end{pmatrix} = \det\begin{pmatrix} -4 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -4 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & -4 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & -4 \end{pmatrix}=0 $$

by adding rows $1,2,4,5$ to row $3$.

$$\det \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 6 \\ 3 & 4 & 5 & 6 & 7 \\ 4 & 5 & 6 & 7 & 8 \\ 5 & 6 & 7 & 8 & 9\end{pmatrix} =\det \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 2 & 2 \\ 4 & 5 & 6 & 7 & 8 \\ 5 & 6 & 7 & 8 & 9\end{pmatrix}=0$$

by subtracting row $1$ from row $2$ & $3$.

$$\det \begin{pmatrix} 0 & -a & -b & -d & -g \\ a & 0 & -c & -e & -h \\ b & c & 0 & -f & -i \\ d & e & f & 0 & -j \\ g & h & i & j & 0\end{pmatrix}=0$$

since $(-1)^5\det (A)=\det(A^T)=\det(A)$. All of these can be generalized to other dimensions, with the exception of the last one for which the dimension of the matrix must be odd.

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    $\begingroup$ You can compute determinants using minor expansions along any row or column, or even along any other "path" as long as it hits every column. (See Laplace's expansion in en.wikipedia.org/wiki/Determinant.) This should help with at least some of yours. $\endgroup$ – Elliot G Jan 18 at 21:40
  • $\begingroup$ I've seen that it can be done along any row or column, but any path? Are you sure? Couldn't we then compute $I_3$ along the path $a_{12}, a_{32}, a_{13}$? This would give us $0$. Maybe I'm just misunderstanding what a path is. $\endgroup$ – Leo Jan 18 at 22:06
  • $\begingroup$ You’re right; I’m confusing that fact with the definition that sums over every possible path. $\endgroup$ – Elliot G Jan 18 at 22:08
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Using cofactor expansion, the determinant is equal to $aD_4+b$ where $D_4$ and $b$ are fixed, and $a$ is the unknown entry.

If $a=0$, then the matrix is skew-symmetric of odd-dimension, which means its determinant is zero. That implies that $b = 0$. So the determinant is $aD_4$.

$D_4$ is the determinant of an even-dimensional skew-symmetric matrix with $-1$s above the diagonal. This will always be $1$ (by an admittedly nontrivial induction argument). That means the determinant is $a$.

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Denote your determinant by $f(a)$ so we get a function $f : \Bbb{R} \to M_5(\Bbb{R})$. Notice that $f(0)$ is a skew-symmetric $5 \times 5$ matrix and hence has determinant $0$ so $f(0)=0$.

The derivative of a determinant of a matrix function can be calculated by differentiating every column of the matrix and summing them up: $$f'(a) = \begin{vmatrix} 0 & -1 & 0 & -1 & -1\\1 & 0 & 0 & -1 & -1\\1 & 1 & 1 & -1 & -1\\1 & 1 & 0 & 0 & -1\\1 & 1 & 0 & 1 & 0\end{vmatrix} = \begin{vmatrix} 0 & -1 & -1 & -1\\1 & 0 & -1 & -1\\1 & 1 & 0 & -1\\1 & 1 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 0 & -1 & -1 & -1\\1 & 0 & -1 & -1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0 \end{vmatrix}.$$ by adding the second and subtracting the first row from the third and fourth. The latter matrix is block upper-triangular so $$f'(a) = \begin{vmatrix} 0 & -1 & -1 & -1\\1 & 0 & -1 & -1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 0 & -1 \\ 1 & 0\end{vmatrix}\cdot \begin{vmatrix} 0 & -1 \\ 1 & 0\end{vmatrix} = (-1)\cdot(-1)=1.$$ Therefore $f(a)= a+C$ for some constant $C \in \Bbb{R}$. From $f(0)=0$ we get $C=0$ and hence $f(a)=a$.

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  • $\begingroup$ @DavidK Thanks, it is in fact $f'(a)=1$. $\endgroup$ – mechanodroid Jan 18 at 22:50

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