Determinant of a $58 \times 58$ block matrix

Question

Given a $58 \times 58$ block matrix : $\mspace{20mu}A=\begin{bmatrix} 0 & 0 & R\\ 0 & Q & T\\ P & S & U \end{bmatrix} \in \mathbb{R}^{58\times 58} $,

$P\in \mathbb{R}^{11\times 11}\mspace{10mu},\mspace{10mu}Q\in \mathbb{R}^{28\times 28}\mspace{10mu},\mspace{10mu}R\in \mathbb{R}^{19\times 19},$

$\det(P)=p \mspace{10mu},\mspace{10mu}\det(Q)=q \mspace{10mu},\mspace{10mu}\det(r)=r \\ \\ \det(A)=??$

I have no idea how to find the determinant.

Answer 1

First note that your matrix can be expressed as a permutation of an upper-triangular block matrix.

$$ A = \Pi A' = \begin{bmatrix} 0 & 0 & I_{19}\\ 0 & I_{28} & 0\\ I_{11} & 0 & 0 \end{bmatrix} \begin{bmatrix} P & S & U \\ 0 & Q & T \\ 0 & 0 & R \end{bmatrix} = \begin{bmatrix} 0 & 0 & R\\ 0 & Q & T\\ P & S & U \end{bmatrix}. $$

Now use the following facts:

1. The determinant of an upper triangular of a block matrix is the product of the determinants of the blocks in the diagonal.
2. The determinant of a product is the product of the determinants.
3. The elementary row operation of swapping two rows flips the sign of the determinant. Since $\Pi$ is obtained by performing an odd number of such operations on the identity, $det(\Pi)=-1$.

This yields $$ det(A)=det(\Pi)det(A')=-pqr. $$