Determinant of $12\times12$ matrix

Question

Problem :

Let $A_{12\times12}$=$[a_{ij}]$ satisfies

$$a_{ij}=0\quad (i=j)\\ a_{ij}=1\quad(i\neq j)$$

Evaluate $\det(A)$.

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My Attempt

Consider $\det(A+I)$. Since $\det(A+I)=0$, $\lambda = -1$ is eigenvalue of $A$.

And I know $Rank(A+I)=1$. This says geometric multiplicity of $\lambda = -1$ is $11$.

This implies algebraic multiplicity of $\lambda = -1$ is also $11$ because $A$ is diagonalizable (A is symmetric).

Finally the last eigenvalue is $\lambda_{12}=11$ because $tr(A)=0=-11+\lambda_{12}$.

So, $\det(A)=(-1)^{11}\times 11 = -11$.

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Main question is :

1. Is my solution valid?
2. Is there any method which has better speed?

Thank you.

Answer 1

Your proof is correct.

Regarding your second question, let $\mathbf{1}$ be a vector of all ones. Then $ A = \mathbf{1}\mathbf{1}^T-I. $ Now use Cauchy's determinant formula $$ \det(\mathbf{1}\mathbf{1}^T-I)=(-1)^n\det(I-\mathbf{1}\mathbf{1}^T)=(-1)^n(\det I-\mathbf{1}^T(\operatorname{adj}I)\mathbf{1})=(-1)^n(1 - n). $$