Suppose $A$ is a $3 \times 3$ matrix. If for every column vector $X$ , $X^{T} A X = 0$ and $a_{23} = - 1008$ , then the sum of the digits of $a_{32}$ is:

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Solution

Use the properties of transpose of a matrix
Given, $X^{T} A X = 0$ and $a_{23} = - 1008$
Let $A$ be any $3 \times 3$ square matrix, such that
$A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]$
Then
$X^{T} A X = {(X^{T} A X)}^{T} = X^{T} A^{T} X = - X^{T} A X$
$\Rightarrow X^{T} A X = 0$
Therefore, $X^{T} A^{T} X = - X^{T} A X$
$A^{T} = - A$
$a_{23} = - a_{32}$
or
$a_{32} = - a_{23}$
$a_{32} = - (- 1008) = 1008$
Therefore, the sum of digits $= 1 + 0 + 0 + 8 = 9$
Hence, the sum of digits of $a_{32}$ is $9$

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Suppose A is a 3×3 matrix. If for every column vector X, XTAX=0 and a23=−1008 , then the sum of the digits of a32 is:

Suppose $A$ is a $3 \times 3$ matrix. If for every column vector $X$ , $X^{T} A X = 0$ and $a_{23} = - 1008$ , then the sum of the digits of $a_{32}$ is: