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.

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.
