Determinant of Large Matrix $A=\left[\begin{array}{ccccc}{-2} & {-1} & {} & {\cdots} & {-1} \\ {-1} & {-2} & {-1} & {\cdots} & {-1} \\ {} & {} & {\ddots} & {} & {} \\ {-1} & {\cdots} & {-1} & {-2} & {-1} \\ {-1} & {\cdots} & {} & {-1} & {-2}\end{array}\right] \in \mathbb{R}^{53 \times 53}$
So we want to find determinant of this big matrix. I tried for some cases I got the pattern like for even dimension determinant is $n+1$ and and for odd dimension it is $-n-1$ so answer should be $-54$ ;I guess. But what is formal method to do this calculation ; idea I have in mind is to find eigenvalue and then product will give me determinant.
 A: We can work with general dimension $n$. The wanted determinant is $(-1)^n$ times the determinant of 
$$
B=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & { 2}\end{array}\right].
$$
Do the combination $C_n\leftarrow \sum_{i=1}^nC_i$. Then $\det(B)=\det(C)$, where 
$$
C=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 2+n-1} \\ { 1} & { 2} & { 1} & {\cdots} & { 2+n-1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & {2+n- 1} \\ { 1} & {\cdots} & {} & { 1} & { 2+n-1}\end{array}\right],
$$
hence 
$$\det(A)=(-1)^n(n-1)\det \left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & {1}\end{array}\right].
$$
Finally, do the substitutions $C_i\leftarrow C_i-C_n$, $1\leqslant i\leqslant n$ to get that the last determinant is one.
A: Notice matrix $A = -(I_{53}+ uu^T)$ where $u$ is the $53 \times 1$ matrix full of $1$.
By matrix determinant lemma,
$$\begin{align}\det A 
&= (-1)^{53} \det(I_{53} + uu^T) = (-1)^{53}\det(I_{53})(1 + u^T I_{53}^{-1} u)\\
&= -(1 + 53) = -54\end{align}$$
An alternative approach is treat any $53\times 1$ matrix as a vector in $\mathbb{R}^{53}$. 
$e_1 = \frac{1}{\sqrt{53}} u$ becomes an unit vector in $\mathbb{R}^{53}$. Extend $e_1$ to an orthronormal basis $e_1, \ldots, e_{53}$ of $\mathbb{R}^{53}$. i.e. take another
$52$ vectors so that
$$e_i \cdot e_j = e_i^T e_j = \begin{cases}
1, & i = j\\
0, & \text{ otherwise }\end{cases}$$
We will have
$\displaystyle\;I_{53} = \sum_{k=1}^{53} e_i e_i^T\;$
and $uu^T = 53 e_1 e_1^T$. This leads to
$$A = -54 e_1 e_1^T - \sum_{i=2}^{53} e_i e_i^T$$
In this basis, $A$ is diagonal with diagonal entry $-54,-1,-1,\ldots,-1$. This means $A$ has a simple eigenvalue $-54$ and an eigenvalue $-1$ with multiplicity $52$. From this, we obtain (again)
$$\det A = (-54)(-1)^{52} = -54$$
