Question

Let $A$ be $a3\times 3$ matrix given by $A={\left(a_{ij}\right)}_{3\times 3}$. If for every column vector $X$ satisfies $X^{\prime }AX=0$ 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 $X=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right],\left(x_1{x}_2{x}_3\right)(\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=0$
$a_{11}{x}_1^2+a_{22}{x}_2^2+a_{33}{x}_3^2+\left(a_{12}+a_{21}\right)x_1{x}_2+\left(a_{13}+a_{31}\right)x_1{x}_3$
$+\left(a_{23}+a_{32}\right)x_2{x}_3=0$
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
$a_{21}=-2008$
$a_{31}=-2010$
$a_{32}=2012$