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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
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 skew symmetric matrix. If $\operatorname{det}(Q)=9$, then the modulus of the sum of all possible values of determinant of $P$ is equal to:
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Answer

d
$A=\left[\begin{array}{ll}2 & 3 \\ a & 0\end{array}\right], a \in R$

and and $P \frac{A+A^{T}}{2}=\left[\begin{array}{cc}2 & \frac{3+a}{2} \\ \frac{a+3}{2} & 0\end{array}\right]$

and $\operatorname{and} Q \frac{A-A^{T}}{2}=\left[\begin{array}{cc}0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0\end{array}\right]$

As, $\operatorname{det}(Q)=9$

$\Rightarrow(a-3)^{2}=36$

$\Rightarrow a=3 \pm 6$

$\therefore a=9,-3$

$\operatorname{det}(P)=$ $\left|\begin{array}{cc}2 & \frac{3+a}{2} \\ \frac{a+3}{2} & 0\end{array}\right|$

$=0-\frac{(a+3)^{2}}{4}=0, \text { for } a=-3 \Rightarrow \operatorname{det}(P)=0$

$=0-\frac{(a+3)^{2}}{4}=\frac{1}{4}(12)^{2}, \text { for } a=9 \Rightarrow \operatorname{det}(P)=36$

$\therefore$ Modulus of the sum of all possible values of det. $(P)=|36|+|0|=36$

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