# Calculate the determinant of $A \in Mat_n$ [duplicate]

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?

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.
• Another way to see that the rank is $2$ or less is to write $A$ as the sum of two rank $1$ matrices: the outer product of $(a_1, \dots, a_n)$ with $(b_1, \dots, b_n)$, and the outer product of $(1, \dots, 1)$ with itself. – Misha Lavrov Apr 8 '17 at 1:05
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.