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Let $\mathbf{A}$ and $\mathbf{B}$ be $k \times k$ matrices and $\mathbf{M}$ is the block matrix

$$\mathbf{M} = \begin{pmatrix}0 & \mathbf{B} \\ \mathbf{A} & 0\end{pmatrix}.$$

How to prove that $\det(\mathbf{M}) = (-1)^k \det(\mathbf{A}) \det(\mathbf{B})$?

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Here is one way among others:

$$

\left( \matrix{0&B\\A&0}\right)=\left( \matrix{0&I_k\\I_k&0}\right)\left( \matrix{A&0\\0&B}\right).

$$

I assume you are allowed to use the block diagonal case, which gives $\det A\cdot\det B$ for the matrix on the far right. Just in case, this follows for instance from Leibniz formula.

Now it boils down to

$$

\det\left( \matrix{0&I_k\\I_k&0}\right)=(-1)^k.

$$

This is the matrix of a permutation made of $k$ transpositions. So the determinant is $(-1)^k$ after $k$ elementary operations (transpositions in this case) leading to the identity matrix.

Note that by performing $k$ columns swaps you can obtain the matrix

$$\mathbf{M}^{‘} = \begin{pmatrix}\mathbf{B} & 0 \\ 0 & \mathbf{A}\end{pmatrix}.$$

Obviously $det(\mathbf{M}^{‘})=(-1)^kdet(\mathbf{M})$.

Now, performing elementary operations on rows (i.e. swithching two rows, multiplying selected row by a non-zero scalar, or adding to a selected row another one multiplyied by a scalar) of $\mathbf{M}^{‘}$, reduce it to the form

$$\mathbf{M}^{”} = \begin{pmatrix}\mathbf{B}^{‘} & 0 \\ 0 & \mathbf{A}^{‘}\end{pmatrix},$$

where $\mathbf{B}^{‘}$ and $\mathbf{A}^{‘}$ are upper-triangular matrices. Note that $\mathbf{M}^{”}$ is upper-triangular itself with a diagonal $[b_1,\ldots,b_k,a_1\ldots,a_k]$.

Obviously $det(\mathbf{M}^{‘})=C_1C_1b_1\ldots b_k a_1\ldots a_k$ where $C_1$ is a number generated by the operations on the first $k$ rows, and $C_2$ is a number generated by the operations on the rows $k+1,\ldots k+k$. It is clear that $det(\mathbf{B})=C_1b_1\ldots b_k$ and $\det(\mathbf{A})=C_2a_1\ldots a_k$.

Thus $det(\mathbf{M})=(-1)^kC_1C_1b_1\ldots b_k a_1\ldots a_k=(-1)^kdet(\mathbf{A})det(\mathbf{B})$.

I think julien has answered your question in a beautiful way. Here I will put your question in a slightly more general context. Note that for a partitioned matrix

$$

M=\pmatrix{P&Q\\ R&S},

$$

where the four submatrices are square and have the same size, **if** $R$ commutes with $S$ (i.e. $RS=SR$), we have the block determinant formula

$$\det M=\det(PS-QR)\tag{1}$$

that is analogous to the determinant formula for $2\times2$ matrices. So, if we put $P=S=0,\ Q=A$ and $R=B$, we get $\det M=\det(-AB)=(-1)^k\det(AB)=(-1)^k\det(A)\det(B)$.

Certainly you shouldn’t do your exercise using this formula, because an exercise like yours are often intended to be solved in a more elementary way, and $(1)$ is more difficult to prove than your problem statement. Nonetheless it’s good to know some nice formulae. Yet formula $(1)$ should be applied with care: in general, we have

$$

\det M=

\begin{cases}

\det(PS-QR) & \text{ if } RS=SR,\\

\det(SP-RQ) & \text{ if } PQ=QP,\\

\det(SP-QR) & \text{ if } QS=SQ,\\

\det(PS-RQ) & \text{ if } PR=RP.

\end{cases}\tag{2}

$$

Since the four determinants $\det(PS-QR),\,\det(SP-RQ),\,\det(SP-QR)$ and $\det(PS-RQ)$ are in general different, the pair of submatrices that commute as well as the order of the four blocks do matter.

I would use the method that is used in Hoffmann and Kunze, in chapter 5. Namely define a function $D(A, B) = det(M)$ (we are roaming along possible $A, B$). Show $D$ is a “determinant function.” in $A$ and $B$ (alternating and $n$-linear, etc). Then iteratively use that for a determinant function $D(A) =$ det$(A)D(I)$.

It’d probably help if you expanded it out a little.

We can safely say $$det(M)=(0)(0)-AB$$

So you just have to show what -AB is.

$$\begin{bmatrix}a_{11} & … & a_{1k}\\ \vdots & & \vdots \\ a_{k1} & … & a_{kk}\end{bmatrix}\begin{bmatrix}b_{11} & … & b_{1k}\\ \vdots & & \vdots\\b_{k1}& …& b_{kk}\end{bmatrix}$$

Start multiplying it out and see if you start to see a pattern. Remember, you can work in both directions of the proof. So you can start from the $(-1)^{k}det{A}det{B}$ side or det(M) side. If the two meet up at some point then you are done (I usually rewrite it to flow better though).

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