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Let $A$ be a $3\times 3$ matrix given by $A=[a_{ij}]$ and $B$ be a column vector such that $B^TAB$ is a null matrix for every column vector $B.$If $C=A-A^T$ and $a_{13}=1,a_{23}=-5,a_{21}=15$,then find the value of det(adj $A$)+det(adj$C$).


My Attempt:
Since $C=A-A^T$,therefore $C$ is a skew symmetric matrix,and the determinant of a skew symmetric matrix is $0$.So,$\det C=0$
$\det(\operatorname{adj}C)=(\det C)^2=0$ by using the property $\det(\operatorname{adj}M)=(\det M)^{n-1}$

But i cannot find $\det(\operatorname{adj}A)$.Please help me.Thanks.

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We know that for every column vector $b$ it holds $$b^\top A b =0.\tag 1$$

  • Considering some vectors for $b$, e.g. $b_1 = (1,0,0)^\top, b_2 = (0,1,0)^\top, b_3 = (0,0,1)^\top$ and applying to $(1)$ we take that $a_{11} = 0 , a_{22} = 0, a_{33} = 0$ respectively.

  • Considering the vector $b_4 = (1,1,0)^\top$ we have that $$a_{11} + a_{12} + a_{21} +15 = 0\implies a_{12} = -15.$$ Similarly, considering the vector $b_5 = (1,0,1)^\top$ we have that $$a_{11}+a_{31}+a_{33} + 1 = 0 \implies a_{31} = -1.$$ Similarly, we find that $a_{32} = -5$ and we have defined completely the matrix $A$, i.e. $$A = \begin{bmatrix} 0 & -15 & 1 \\ 15 & 0 & -5\\ -1 & 5 & 0\end{bmatrix},$$ which we observe that it is a $3\times 3$ skew - symmetric matrix. Since its dimension is odd, its determinant is equal to zero.

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