JEE Main 2021 — Mathematics Algebra

We have,$A=[\begin{matrix}2 & 3 \\ a & 0\end{matrix}],a\in R$A=[2a​30​],a∈R

Then,

$${A}^{T}=[\begin{matrix}2 & a \\ 3 & 0\end{matrix}]$$AT=[23​a0​]

Now,

$$P=(\frac{A+{A}^{T}}{2})$$P=(2A+AT​)

$$\Rightarrow P=\frac{1}{2}{[\begin{matrix}2 & 3 \\ a & 0\end{matrix}]+[\begin{matrix}2 & a \\ 3 & 0\end{matrix}]}$$⇒P=21​[2a​30​]+[23​a0​]

$$\Rightarrow P=[\begin{matrix}2 & \frac{3+a}{2} \\ \frac{a+3}{2} & 0\end{matrix}]$$⇒P=[22a+3​​23+a​0​]

And,

$$Q=\frac{A-{A}^{T}}{2}$$Q=2A−AT​

$$\Rightarrow Q=\frac{1}{2}{[\begin{matrix}2 & 3 \\ a & 0\end{matrix}]-[\begin{matrix}2 & a \\ 3 & 0\end{matrix}]}$$⇒Q=21​[2a​30​]−[23​a0​]

$$\Rightarrow Q=[\begin{matrix}0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0\end{matrix}]$$⇒Q=[02a−3​​23−a​0​]

As,

$$det(Q)=9$$det(Q)=9

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

$$\Rightarrow a=3\pm 6$$⇒a=3±6

$$\therefore a=9,-3$$∴a=9,−3

Now,

$$det(P)=-{(\frac{3+a}{2})}^{2}$$det(P)=−(23+a​)2

For $a=-3,det(P)=0$a=−3,det(P)=0

For $a=9$a=9, $det(P)=-{(\frac{3+9}{2})}^{2}=-36$det(P)=−(23+9​)2=−36

$\therefore$∴ Modulus of the sum of all possible values of $\mathrm{det}(P)$det(P)

$$=|-36|+|0|=36$$=∣−36∣+∣0∣=36