Question

Let $A$ be $a 3 \times 3$ matrix given by $A=\left(a_{i j}\right)_{3 \times 3}$. If for every column vector $X$ satisfies $X^{\prime} A X=0$ and $a_{12}=2008, a_{13}=$ 1010 and $a_{23}=-2012$. Then the value of $a_{21}+a_{31}+a_{32}=$

• A. -6
• B. 2006
• C. -2006
• D. 0

Answer 1

Answer: (C)

Let $X= \begin{bmatrix}x_{1} \\ x_{2} \\ x_{3}\end{bmatrix},\left(x_{1} x_{2} x_{3}\right)(\begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix} \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix}=0$
$a_{11} x_{1}^{2}+a_{22} x_{2}^{2}+a_{33} x_{3}^{2}+\left(a_{12}+a_{21}\right) x_{1} x_{2}+\left(a_{13}+a_{31}\right) x_{1} x_{3}$
$+\left(a_{23}+a_{32}\right) x_{2} x_{3}=0$
It is true for every $x_{1}, x_{2}, x_{3}$,
then $a_{11}=a_{22}=a_{33}=0, a_{12}+a_{21}=0, a_{13}+a_{31}=0, a_{23}+a_{32}=0$
$\therefore A$ is a skew symmetric matrix
$a_{21} =-2008$
$a_{31} =-2010$
$a_{32} =2012$