# calculation of the determinant of a block matrix little help

I need to prove $$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}= \operatorname{det}(DA-CB),$$ where $A,B,C,D \in M_{n\times n}(R)$ with the property that $A$ and $B$ commute and moreover $\operatorname{det}(B) \neq 0,$ using the following hint: $$\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix}.$$

Using the hint, I have: $$\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix} = \begin{pmatrix}0 & B \\ CB-DA & D\\ \end{pmatrix}.$$Then:

$$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\operatorname{det}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix} = \operatorname{det}\begin{pmatrix}0 & B \\ CB-DA & D\\ \end{pmatrix},$$ but $$\operatorname{det}\begin{pmatrix}B & 0 \\ -A & I_n\\ \end{pmatrix} = \operatorname{det}(B)$$ $$\operatorname{det}\begin{pmatrix}0 & B \\ CB-DA & D\\ \end{pmatrix} = (-1)^n\operatorname{det}(CB-DA)\operatorname{det}(B).$$ Then: $$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}\operatorname{det}(B) = (-1)^n\operatorname{det}(CB-DA)\operatorname{det}(B)$$ which can be rewritten as follows:

$$[\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}+(-1)^{n+1}\operatorname{det}(CB-DA)]\operatorname{det}(B) = 0$$

But $\operatorname{det}(B) \neq 0$, hence:$$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix}+(-1)^{n+1}\operatorname{det}(CB-DA) = 0,$$ So

$$\operatorname{det}\begin{pmatrix}A & B \\ C & D\\ \end{pmatrix} = (-1)^n\operatorname{det}(CB-DA).$$

Im stuck on this part, can anyone help me on this step

• I think you almost succeed. Note $CB-DA=-(DA-CB)$. Commented Mar 11, 2015 at 6:53
• @EclipseSun Using that fact, i want to know if $\operatorname{det}(CB-DA) = \operatorname{det}(-(DA-CB)) = -\operatorname{det}((DA-CB))$ holds. How can i get rid of the $(-1)^n$
– okie
Commented Mar 11, 2015 at 7:04
• @RicardoCervantes Look at property 5 under Properties of the determinant: en.wikipedia.org/wiki/Determinant#Properties_of_the_determinant. Commented Mar 11, 2015 at 7:07
• @KimJongUn thank you, then basically I have: $(-1)^n\operatorname{det}(CB-DA) = \operatorname{det}(-1(CB-DA)) = \operatorname{det}(DA-CB)$ and the proof is complete.
– okie
Commented Mar 11, 2015 at 7:10
• @RicardoCervantes Yes you're done! Good job. P.s. to type $\det$, use \det. Commented Mar 11, 2015 at 7:12