Let A be an n x n matrix such that the det(A)=5; Let E be an m x m matrix such that the det(E)=4; Let F be an n x m matrix.

Find the det\begin{bmatrix}0&A\\E&F\end{bmatrix}

The answer can be expressed in terms of n and m.

I'm having trouble with this question as why my answer is wrong is throwing me off. Using properties of the determinant, I can rewrite this question as ;

det(0)det(F) - det(A)det(E)

My assumption was that det(0)det(F) equates to 0, but I'm not so sure anymore considering that my answer of -20 is incorrect. Any clues on maybe how to express the F determinant (which is non-square)

  • $\begingroup$ Your $O, F$ are not square matrices and do not have determinants. $\endgroup$
    – markvs
    Commented Nov 16, 2021 at 23:54

1 Answer 1


By switching the "block-columns", we end up with $$ \det \pmatrix{0 & A\\ E & F} = (-1)^{m(m+n-1)}\det \pmatrix{A & 0\\F & E}. $$ On the other hand, using the formula for the determinant of a block upper-triangular matrix (proved here for instance) gives us $$ \det\pmatrix{A & 0\\F & E} = \det(A) \det(E) = 20. $$

Regarding the exponent of $(-1)$: to get from $\pmatrix{0 & A\\ E & F}$ to $\pmatrix{A & 0\\F & E}$, we apply a permutation of the columns of the block matrix. The value $(-1)^{m(m+n-1)}$ is the "parity" or "sign" of this permutation. The exponent of $(-1)$ here is any number of transpositions (i.e. column-switches) with which one could achieve this permutation.

So, why $m(m+n-1)$? Here is how one could achieve the desired permutation using exactly $m(m+n-1)$ transpositions. First, we consider the cyclic permutation $$ [2\ \ 3\ \ \cdots \ \ m \ \ \cdots \ \ (m + n)\ \ 1] $$ (where I have expressed the permutation in one-line notation). This permutation can be implemented in $(m+n-1)$ switches: first switch the $1$st and $2$nd elements, then the $2$nd and $3$rd, and so forth until you switch $(m + n - 1)$th and $(m + n)$th.

Now, to do what we wanted to do, which was the permutation $$ [m+1 \ \ \cdots \ \ (m + n - 1)\ \ 1 \ \ \cdots\ \ m], $$ it suffices to apply the first permutation $m$ times in a row. If we want to break it down into transpositions, we can repeat the above sequence $m$ times. Thus, we can achieve this permutation with $m(m + n - 1)$ transpositions in total.

  • $\begingroup$ That was spot on thanks man. The answer ended up being (-1)^(m(m+n-1))*20. Could you maybe re-explain how you got the (-1)^m... as I don't quite understand where these manipulations came from? $\endgroup$
    – aort01
    Commented Nov 17, 2021 at 0:21
  • 1
    $\begingroup$ Well, can't "re-explain" it if I didn't explain it at all in the first place. I'll add a note to my answer. $\endgroup$ Commented Nov 17, 2021 at 0:24
  • 1
    $\begingroup$ @Alex03 See my latest edit. $\endgroup$ Commented Nov 17, 2021 at 0:42

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