Let A=left[begin{array}{ll}2 & 3 a & 0end{ar | vedclass

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(D) Given $A = \begin{bmatrix} 2 & 3 \ a & 0 \end{bmatrix}$.Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{A + A^T}{2}$ is symm

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$36$

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(D) Given $A = \begin{bmatrix} 2 & 3 \ a & 0 \end{bmatrix}$.Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{A + A^T}{2}$ is symmetric and $Q = \frac{A - A^T}{2}$ is skew-symmetric.$P = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \ a & 0 \end{bmatrix} + \begin{bmatrix} 2 & a \ 3 & 0 \end{bmatrix} ight) = \begin{bmatrix} 2 & \frac{3+a}{2} \ \frac{3+a}{2} & 0 \end{bmatrix}$.$Q = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \ a & 0 \end{bmatrix} - \begin{bmatrix} 2 & a \ 3 & 0 \end{bmatrix} ight) = \begin{bmatrix} 0 & \frac{3-a}{2} \ \frac{a-3}{2} & 0 \end{bmatrix}$.Given $\operatorname{det}(Q) = 9$,we have $0 - \left( \frac{3-a}{2} ight) \left( \frac{a-3}{2} ight) = 9$.$\Rightarrow \frac{(a-3)^2}{4} = 9 \Rightarrow (a-3)^2 = 36 \Rightarrow a-3 = \pm 6$.Thus,$a = 9$ or $a = -3$.Now,$\operatorname{det}(P) = 0 - \left( \frac{3+a}{2} ight)^2 = -\frac{(a+3)^2}{4}$.For $a = 9$,$\operatorname{det}(P) = -\frac{(9+3)^2}{4} = -\frac{144}{4} = -36$.For $a = -3$,$\operatorname{det}(P) = -\frac{(-3+3)^2}{4} = 0$.The sum of all possible values of $\operatorname{det}(P)$ is $-36 + 0 = -36$.The modulus of the sum is $|-36| = 36$.

Let $A=\left[\begin{array}{ll}2 & 3 \\ a & 0\end{array}\right], a \in R$ be written as $P+Q$ where $P$ is a symmetric matrix and $Q$ is a skew-symmetric matrix. If $\operatorname{det}(Q)=9$,then the modulus of the sum of all possible values of the determinant of $P$ is equal to:

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