# Determinant of an $n \times n$-matrix [closed]

Can you help me with this $n\times n$ determinant? Can't find what from what i have to substract.. Spent hours of trying...

\begin{vmatrix} 1&\cdots& 1& 1& 4\\ 1 &\cdots&1 & 9& 1 \\ 1 & \cdots &16& 1 &1\\ \vdots & \dots& \vdots&\vdots&\vdots\\ (n+1)^2 &\cdots& 1& 1& 1 \end{vmatrix}

## closed as off-topic by Namaste, Did, Isaac Browne, mathematics2x2life, rtybaseApr 29 '18 at 8:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Did, Isaac Browne, mathematics2x2life, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.

• – Shaun Dec 11 '17 at 19:41
• You should give an account of what you tried in "hours of trying" and what you discovered. It will help Readers respond in a helpful way, and in the future make your question a more valuable post. – hardmath Dec 11 '17 at 20:11

We can rewrite this matrix as $$\pmatrix{0& \cdots & 0 & 0 & 2^2 - 1\\ 0 & \cdots & 0 & 3^2 - 1 & 0\\ 0 & \cdots & 4^2 - 1 & 0 & 0\\ \vdots & \ddots & \vdots & \vdots & \vdots\\ (n+1)^2 - 1 & \cdots & 0 & 0 & 0} + xx^T$$ where $x = (1,1,\dots,1)^T$. From there, it suffices to apply the matrix determinant lemma.