# Determinant of a block matrix in which the lower right is the zero matrix

Let $D=\begin{pmatrix} C & A \\ B & 0 \end{pmatrix}$ where $C$ is $k\times n$, $A$ is $k\times k$ and $B$ is $n\times n$. Let $P=\begin{pmatrix}A & C \\ 0 & B\end{pmatrix}$. I'm given that $\operatorname{det}\begin{pmatrix} A & C \\ 0 & B \end{pmatrix}=\operatorname{det}(A)\operatorname{det}(B)$. I want to find $\operatorname{det}D.$

Can I use elementary column operations on the matrix $D$ in order to get it into the same form as $P$? Or is that going to cause a problem with signs?

• Swapping two columns multiplies the determinant by $(-1)$. So as long as you correctly count the number of column flips it would take, this would work. Commented Oct 26, 2016 at 18:13
• @KenDuna So would $\operatorname{det}D=(-1)^m\operatorname{det}P$ where $m$ is the number of column swaps? Commented Oct 26, 2016 at 18:17
• That is correct. Commented Oct 26, 2016 at 18:18

You can use elementary column and row operations to change $D$ to $P$, where $\det(D)=(-1)^m\det(P)$.
Also you can use Laplace Expansion Theorem to show that |A||B| is the only cofactor product left in expanding $D$ into sum of cofactor products of $k\times k$ and $(n-k)\times (n-k)$.