# Determinant of symmetric matrix with unit diagonal

Let $$A_2$$ be a symmetric $$2 \times 2$$-matrix $$A_2 = \begin{pmatrix} 1 & a_{12}\\ a_{12} & 1 \end{pmatrix}.$$ Clearly, we have $$\tag{1} \det(A_2) = 1 - a_{12}^2.$$

Let $$A_3$$ be a symmetric $$3 \times 3$$-matrix $$A_3 = \begin{pmatrix} 1 & a_{12} & a_{13}\\ a_{12} & 1 & a_{23} \\ a_{13} & a_{23} & 1 \end{pmatrix}.$$ We have \begin{align} \det(A_3) = 1 &- a_{12}^2 - a_{13}^2 - a_{23}^2 \\ \tag{2} &+ 2 a_{12}a_{13}a_{23}. \end{align}

Let $$A_4$$ be a symmetric $$4 \times 4$$-matrix $$A_4 = \begin{pmatrix} 1 & a_{12} & a_{13} & a_{14}\\ a_{12} & 1 & a_{23} & a_{24} \\ a_{13} & a_{23} & 1 & a_{34} \\ a_{14} & a_{24} & a_{34} & 1 \end{pmatrix}.$$ By using a general formula, we calculate \begin{align} \det(A_4) = 1 &- a_{12}^2 - a_{13}^2 - a_{14}^2 - a_{23}^2 - a_{24}^2 - a_{34}^2 \\ &+ 2 a_{12}a_{13}a_{23} + 2 a_{12}a_{14}a_{24} + 2 a_{13}a_{14}a_{34} + 2a_{23}a_{24}a_{34} \\ &-2a_{13}a_{14}a_{23}a_{24} - 2a_{12}a_{14}a_{23}a_{34} - 2a_{12}a_{13}a_{24}a_{34} \\ \tag{3} &+a_{14}a_{14} a_{23}a_{23} + a_{13}a_{13} a_{24}a_{24} + a_{12}a_{12} a_{34}a_{34}. \end{align}

What will be a structural generalization of $$(1)$$, $$(2)$$, $$(3)$$ to symmetric $$n \times n$$-matrix $$A_4 = \begin{pmatrix} 1 & a_{12} & \dots & a_{1n}\\ a_{12} & 1 & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \dots & 1 \end{pmatrix}?$$

• This isn't any easier than finding a formula for the determinant of an arbitrary symmetric matrix (indeed, you can always congrue your $A_n$ by a diagonal matrix to make the diagonal entries generic), for which I haven't seen any useful expressions. Mar 1 '20 at 23:01