Linear Algebra: Determinant question

• December 25th 2008, 01:26 AM
Isomorphism
Linear Algebra: Determinant question
Hello,

Problem:

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

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

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

My effort:

So I observed that:

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

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

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

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

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 $\text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1$.

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

Thank you,
Srikanth
• December 25th 2008, 01:47 AM
Constatine11
Quote:

Originally Posted by Isomorphism
I have also used it while setting $\text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1$.

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

Thank you,
Srikanth

The matrix (assume its square):

$\begin{pmatrix} I_n & 0_{m,n} \\ X & I_m \end{pmatrix}$

is lower triangular, and so its determinant is the product of the elements on the diagonal, but these are all 1 so the determinant is 1 without having to assume what was to be proven.
• December 25th 2008, 02:18 AM
Constatine11
Assume $A,\ B,\ C,\ D$ are all square and multiplicativly conformant (I believe that is implied by the conditions in the problem)

$\det\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} =\det \left[\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & D^{-1} \end{pmatrix}\right] \det(D)$ $
=\text{det}\begin{pmatrix} A-BCD^{-1} & BD^{-1} \\ 0 & I \end{pmatrix} \det(D)
$

expanding the first determinant in the last expression, from the bottom right gives:

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

etc...