Linear Algebra, Determinant of symmetric matrix

Question

Let $A = [a_{ij}]$ be a square matrix of order n whose entries are given as follows. For $1 \leq i, j \leq n$ we have \begin{equation} a_{ij} = \begin{cases} ij &\text{ if } i\neq j\\ 1+ij &\text{ if } i = j \end{cases} \end{equation} Evaluate the determinant of $ A$( in terms of $n$).

Answer 1

We have the $i^\text{th}$ row of $A$ as $A_i=\begin{bmatrix} i&2i&\ldots&(i-1)i&i^2+1&i(i+1)&\ldots&ni \end{bmatrix}$.

Therefore, $A_i-iA_1=\begin{bmatrix} -i&0&\ldots&0&1&0&\ldots&0 \end{bmatrix}$

Thus we get \begin{align} \text{det}(A)&=\begin{vmatrix} 2 & 2 & \ldots & n\\ -2&\\ -3&\\ \vdots&&I\\ -n\end{vmatrix}\\ &=2+2^2+3^2+\ldots+n^2\\ &=1+\frac{n(n+1)(2n+1)}{6} \end{align}

Answer 2

Your matrix has the form $A=I+vv^T$ with $v_i=i$. A well-known determinant formula is $$ \det(I+ab^T)=1+b^Ta, $$ in this concrete case $$ \det(A)=1+v^Tv=1+\sum_{i=1}^ni^2=1+\frac{n(n+1)(2n+1)}6. $$