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Suppose $A=\left(a_{i j}\right)_{n\times n}$ is an invertible real symmetric matrix. Prove: matrix of the quadratic form

\begin{align}f\left(x_1,\text{...},x_n\right)=\left| \begin{array}{cccc} 0 & x_1 & \text{...} & x_n \\ -x_1 & a_{11} & \text{...} & a_{1 n} \\ \text{...} & \text{...} & \text{...} & \text{...} \\ -x_n & a_{n 1} & \text{...} & a_{n n} \\ \end{array} \right|\end{align}

is the adjungate matrix $adj (A)$ of $A$.

$\text{adj}(A)$is a (n)-matrix, matrix of quadratic form is n+1 matrix, isn't it?

How to do?

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The determinant of the $(n+1)\times (n+1)$-matrix you are given is used to define a function $f(x_1,\dots,x_n)$ of $n$ real variables. This function is a quadratic form, and therefore can be described by means of an $n\times n$ matrix. It's this latter matrix that must be shown to be $\operatorname{adj}A$.

Unwinding the above, you will see that the coefficient of $x_i x_j$ in the quadratic form comes from one (if $i=j$) or two (if $i\ne j$) terms in the determinant. Such a term includes $x_i$ and $x_j$, and the rest is the determinant of the submatrix of $A$ with omitted $i$th row and $j$th column (or vice versa). This agrees with the definition of $\operatorname{adj}A$ (but you should check the sign carefully).

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