Let $A = {[a_{i j}]}_{3 \times 3}$ be a matrix such that $a_{i j} = 0 \forall i \neq j$ and $a_{i j} > 0 \forall$ $i = j$ . If $a d j A$ satisfies the equation $x^{3} - 9 x^{2} + p x - 27 = 0, p \in R$ then which of the following is/are not correct
Note: $a d j (P)$ , $t r (P)$ and $d e t (P)$ denote adjoint matrix of matrix $P$ , trace of matrix $P$ and determinant of matrix $P$ respectively]

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$\det . (A) = 3 \sqrt{3}$

$t r (A) = 3 \sqrt{3}$

$\det . (a d j (a d j A)) = 3^{6}$

$t r . (A^{- 1}) = \frac{1}{3 \sqrt{3}}$

answer is C.

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Detailed Solution

$A = (\begin{matrix} d_{1} & 0 & 0 \\ 0 & d_{2} & 0 \\ 0 & 0 & d_{3} \end{matrix}); a d j . A = (\begin{matrix} d_{2} d_{3} & 0 & 0 \\ 0 & d_{1} d_{3} & 0 \\ 0 & 0 & d_{1} d_{2} \end{matrix})$

Let C=adj. A satisfy $x^{3} - 9 x + p x - 27 = 0$
$∴ T r . (C) = 9$ and
det. C=27 $\Rightarrow | a d j . A | = 27$
$\Rightarrow | A^{2} | = 27$ $\Rightarrow | A | = 3 \sqrt{3}$
Also, $| a d j . (a d j . A) | = {| A |}^{4} = 3^{6}$
Now, Tr. $(a d j . A) = 9$ given
$∴ d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1} = 9$ ………..(1)
$A n d | A | = d_{1} d_{2} d_{3} = 3 \sqrt{3}$ ………….(2)
$\Rightarrow \frac{1}{d_{1}} + \frac{1}{d_{2}} + \frac{1}{d_{3}} = \frac{9}{3 \sqrt{3}} = \sqrt{3} = T r (A^{- 1})$
Also, GM of $d_{1}; d_{2}; d_{3}$ is $\sqrt{3}$
and, H.M. of $d_{1} d_{2} d_{3}$ is $\sqrt{3}$
$\Rightarrow T r . (A) = 3 \sqrt{3}$