Calculate the determinant of $A \in Mat_n$

Question

Given matrix $A$

\begin{bmatrix} a_1b_1+1 &a_1b_2 + 1&...&a_1b_n + 1\\ a_2b_1+1 &a_2b_2 + 1&...&a_2b_n + 1\\ ...&...&...&...\\ a_nb_1+1 &a_nb_2 + 1&...&a_nb_n + 1 \end{bmatrix} Find its determinant.

I was using some properties of $\det$ here.

Suppose we have an arbitrary quadratic $n \times n$ matrix $A$.

Let's denote $A + \alpha$ matrix, where we added $\alpha$ to all elements of matrix $A$. Then $\det(A + \alpha) = \det(A) + \alpha \cdot\sum \limits_{i = 1}^n \sum \limits_{j = 1}^n A_{ij}$, where $A_{ij} = (-1)^{i + j}M_{ij}$.

So I got: \begin{align} & \det \begin{bmatrix} a_1b_1+1 &a_1b_2 + 1&...&a_1b_n + 1\\ a_2b_1+1 &a_2b_2 + 1&...&a_2b_n + 1\\ ...&...&...&...\\ a_nb_1 +1 &a_nb_2 + 1&...&a_nb_n + 1 \end{bmatrix} \\ & = \det \begin{bmatrix} a_1b_1 &a_1b_2 &...&a_1b_n\\ a_2b_1 &a_2b_2 &...&a_2b_n \\ ...&...&...&...\\ a_nb_1 &a_nb_2 &...&a_nb_n \end{bmatrix} + 1\cdot \sum \limits_{i = 1}^n \sum \limits_{j = 1}^n A_{ij}\ \ \ \ (*), \end{align} where $A$ is matrix without $+ 1$'s.

Then $(*) = \det \begin{bmatrix} b_1 &b_2 &...&b_n\\ b_1 &b_2 &...&b_n \\ ...&...&...&...\\ b_1 &b_2 &...&b_n \end{bmatrix} \cdot \prod \limits_{i = 1}^n a_i + \sum \limits_{i = 1}^n \sum \limits_{j = 1}^n A_{ij} = 0 \cdot \prod \limits_{i = 1}^n a_i + 0 = 0$, because we have equal strings everywhere.

So, $\det$ of initial matrix is equal to $0$.

Can you please check my result?

Answer 1

The rank of the matrix $A$ is less than or equal to 2 (because every $3\times 3$ minor is zero). If $n \geq 3$, the determinant is 0.

Answer 2

The matrix has rank at most two, because its column space is spanned by $(a_1, a_2, \ldots, a_n)$ and $(1,1,\ldots,1)$.

So for $n>2$ the determinant is $0$.

You will need a separate analysis for $n\le 2$. Certainly when $n=1$ the determinant will be $a_1b_1+1$ and there's no guarantee that is zero.