# Determinant of block matrix with null row vector

I'm a bit confused on a problem. I've been given an $(n+1)\times(n+1)$ square matrix, which is written in the form of a block matrix with the following dimensions

$\begin{bmatrix} (1x1) & null \\ (nx1) & (nxn) \end{bmatrix}$

where I'm assuming "null" means a "null row vector of size $1xn$". I need to compute the determinant, but the null has me confused. First off, does the "null" mean you can't treat it as a square matrix?

I've tried to understand what is shown here https://en.wikipedia.org/wiki/Determinant#Block_matrices on how to solve this but I'm not sure how it works in my case. In fact, I'm not even sure how to treat a null? Is it a zero or a one? Please explain how I would solve this type of problem. Thanks.

Your assumption is correct! Here "null" means a row vector of length $n$ with each component equal to zero. Personally, I would have written the matrix as $$A = \begin{pmatrix} \alpha & 0^T \\ v & B \end{pmatrix},$$ but that is because I prefer to deal with scalars, and column vectors rather than row vectors. By expanding the determinant along the first row of your matrix you find that $$\text{det}(A) = \alpha \text{det}(B).$$
• There would not be any shortcuts.You would have no choice but to compute the $LU$ factorization of $A$, i.e. $PA = LU$ where $P$ is a permutation matrix, $L$ is unit lower triangular, $U$ is upper triangular, and compute $\text{det}(A) =(1/\text{det}(P)) \text{det}(U)$, exploiting that $\text{det}(L)=1$. – Carl Christian Feb 5 '16 at 23:00