Find the determinant of A $\in \mathbb{M}_n$

Question

Let A be the matrix defined as $\begin{pmatrix} 3 & 2 & 2 & . & . & . &2 \\ 2& 3 & 2 & . & . & . & 2\\ 2& 2 & 3 & . & . & . & 2\\ .& . & . & . & . & . &. \\ .& .& . & . & . & . & .\\ .& .& . & . & . & . &. \\ 2& 2& 2 & .& . & . &3 \end{pmatrix}$ I need to calculate its determinant. I tried to add the 2nd, 3rd,...nth row to the 1st one and then multiplying the first row with $\frac{1}{3+2n-2}$ so I would get the first row full of one and then multiplying the first row by two to get only twos in the first row and then subtracting the first row from all other rows, so in this way I would get a diagonal matrix. But this dosen't seem to work.

Answer 1

The $n$-by-$n$ matrix $\textbf{J}_n$ consisting of only $1$ is of rank $1$, so it has $0$ as an eigenvalue with multiplicity $n-1$. Since the trace of $\mathbf{J}_n$ is $n$, the other eigenvalue of $\textbf{J}_n$ is $n$. Therefore, the characteristic polynomial of $\textbf{J}_n$ is $$\det\left(x\,\textbf{I}_n-\textbf{J}_n\right)=x^{n-1}\left(x-n\right)\,,$$ where $\textbf{I}_n$ is the $n$-by-$n$ identity matrix. You are asking for the determinant of $$\textbf{I}_n+2\,\textbf{J}_n=-2\left(-\frac{1}{2}\,\textbf{I}_n-\textbf{J}_n\right)\,.$$ (This assumes, of course, that the base field is of characteristic not equal to $2$. Nonetheless, in characteristics $2$, we have $\textbf{I}_n+2\,\textbf{J}_n=\textbf{I}_n$.)

Answer 2

You may find the eigenvalues directly, but your method with column reductions work fine. Just subtract the last row from all the others. You'll get $$ \left( \begin{matrix} 1 & 0 & 0 & ... & -1\\ 0 & 1 & 0 & ... & -1\\ 0 & 0 & 1 & ... & -1\\ . & . & . & ... & .\\ 2 & 2 & 2 & ... & 3 \end{matrix} \right) $$ now subtract 2 times each of the first $n-1$ rows from the last to get $$ \left( \begin{matrix} 1 & 0 & 0 & ... & -1\\ 0 & 1 & 0 & ... & -1\\ 0 & 0 & 1 & ... & -1\\ . & . & . & ... & .\\ 0 & 0 & 0 & ... & 1+2n \end{matrix} \right) $$ And the determinant is $1+2n$