Hello,

Problem:

If A,B,C,D are matrices such that C and D commute and D is invertible, then

$\displaystyle \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det}(AD - BC)$

There is a hint that says " multiply on the right by $\displaystyle \begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix}$".

My effort:

So I observed that:

$\displaystyle \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = $ $\displaystyle \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix}\cdot\text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = \text{det}\begin{pmatrix} A+BX & B \\ C+DX & D \end{pmatrix}$

If we choose $\displaystyle X = -D^{-1}C$, then $\displaystyle \text{det}\begin{pmatrix} A+BX & B \\ C+DX & D \end{pmatrix} = \text{det}\begin{pmatrix} A- BD^{-1}C & B \\ 0 & D \end{pmatrix}$

Now since C and D commute, $\displaystyle D^{-1}C = CD^{-1}$. Thus

$\displaystyle \text{det}\begin{pmatrix} A-BD^{-1}C & B \\ 0 & D \end{pmatrix} = \text{det}\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} $$\displaystyle = \text{det}(A-BCD^{-1})\text{det}(D) = \text{det}(AD - BC)$

It looked like a convincing proof until I realised that I have used the hypothesis in the penultimate step. I have also used it while setting $\displaystyle \text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1$.

So how can I fix those steps? Or do we need a different direction?

Thank you,

Srikanth