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Let $A, B, C$ be matrices with size $m \times m$, $n \times n$, and $n \times m$, respectively. If $\det(A) = 2$ and $\det(B) = 3,$ then find $$\det \begin{pmatrix} 0 & A \\ B & C \end{pmatrix} =\ldots $$

I stuck to solve this problem. I also wonder how can we calculate a determinant of matrix with some matrices in it (submatrices)? Please, anyone help me

Yiorgos S. Smyrlis
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Shane Dizzy Sukardy
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  • Search for determinant of block matrices, there are plenty of relevant posts here. Like this one:https://math.stackexchange.com/questions/75293/determinant-of-a-block-lower-triangular-matrix?noredirect=1&lq=1. – StubbornAtom Feb 10 '18 at 16:31

1 Answers1

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Hints.

Step 1. $$ \det \left(\begin{array}{cc} 0 & A \\ B & C\end{array}\right) =(-1)^m\det \left(\begin{array}{cc} A & 0 \\ C & B\end{array}\right) $$

Step 2. $$ \det \left(\begin{array}{cc} A & 0 \\ C & B\end{array}\right)=\det A\cdot \det B $$

Step 1, is obtained by $m^2$ permutations of rows and as many changes of sign.

Step 2, is obtained using the Jordan forms of $A$ and $B$.

Yiorgos S. Smyrlis
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  • What about if the matrix (which contain block matrices) is not triangular? Can we calculate the determinant? – Shane Dizzy Sukardy Feb 11 '18 at 13:27
  • @ShaneDizzySukardy There exist certain tricks. Nevertheless, most of them for special matrices. For example, if all blocks are square of the same size, you can imitate the treatment of the $2\times 2$ case. – Yiorgos S. Smyrlis Feb 11 '18 at 19:01