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$A=\left[\begin{array}{ccccc}{-2} & {-1} & {} & {\cdots} & {-1} \\ {-1} & {-2} & {-1} & {\cdots} & {-1} \\ {} & {} & {\ddots} & {} & {} \\ {-1} & {\cdots} & {-1} & {-2} & {-1} \\ {-1} & {\cdots} & {} & {-1} & {-2}\end{array}\right] \in \mathbb{R}^{53 \times 53}$

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.

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    $\begingroup$ First, find the eigenvalues of the matrix $J$, consisting only of $1$s (check out here: math.stackexchange.com/questions/217521/…). Then your matrix is $-J - I$. Working out the determinant should be easy enough from there. $\endgroup$ – Theo Bendit Oct 3 '19 at 7:18
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    $\begingroup$ $A = -I -J$ where $I$ is the identity matrix and $J$ is the matrix full of $1$’s. $J$ has a one-dimensional eigenspace with eigenvalue n, and an $(n-1)$-dimensional eigenspace with eigenvalue $0$. So $A$ has a one-dimensional eigenspace with eigenvalue $-n-1$, and an $(n-1)$-dimensional eigenspace with eigenvalue $-1$. Therefore the determinant of $A$ is $(-n-1)(-1)^{n-1}$ $\endgroup$ – Joppy Oct 3 '19 at 7:22
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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.

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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) $$\det A = (-54)(-1)^{52} = -54$$

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