JEE Main 2023 — Mathematics Algebra

Given:

$$A=[\begin{matrix}1 & {\mathrm{log}}_{x}y & {\mathrm{log}}_{x}z \\ {\mathrm{log}}_{y}x & 2 & {\mathrm{log}}_{y}z \\ {\mathrm{log}}_{z}x & {\mathrm{log}}_{z}y & 3\end{matrix}]$$A=[1logy​xlogz​x​logx​y2logz​y​logx​zlogy​z3​]

$$\Rightarrow A=[\begin{matrix}1 & \frac{\mathrm{log}y}{\mathrm{log}x} & \frac{\mathrm{log}z}{\mathrm{log}x} \\ \frac{\mathrm{log}x}{\mathrm{log}y} & 2 & \frac{\mathrm{log}z}{\mathrm{log}y} \\ \frac{\mathrm{log}x}{\mathrm{log}z} & \frac{\mathrm{log}y}{\mathrm{log}z} & 3\end{matrix}]$$⇒A=[1logylogx​logzlogx​​logxlogy​2logzlogy​​logxlogz​logylogz​3​]

$$\Rightarrow A=\frac{1}{\mathrm{log}x\mathrm{log}y\mathrm{log}z}[\begin{matrix}\mathrm{log}x & \mathrm{log}y & \mathrm{log}z \\ \mathrm{log}x & 2\mathrm{log}y & \mathrm{log}z \\ \mathrm{log}x & \mathrm{log}y & 3\mathrm{log}z\end{matrix}]$$⇒A=logxlogylogz1​[logxlogxlogx​logy2logylogy​logzlogz3logz​]

$$\Rightarrow |A|=\frac{1}{\mathrm{log}x\cdot \mathrm{log}y\cdot \mathrm{log}z}|\begin{matrix}\mathrm{log}x & \mathrm{log}y & \mathrm{log}z \\ \mathrm{log}x & 2\mathrm{log}y & \mathrm{log}z \\ \mathrm{log}x & \mathrm{log}y & 3\mathrm{log}z\end{matrix}|$$⇒∣A∣=logx⋅logy⋅logz1​∣logxlogxlogx​logy2logylogy​logzlogz3logz​∣

$$\Rightarrow |A|=\frac{\mathrm{log}x\cdot \mathrm{log}y\cdot \mathrm{log}z}{\mathrm{log}x\cdot \mathrm{log}y\cdot \mathrm{log}z}|\begin{matrix}1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3\end{matrix}|$$⇒∣A∣=logx⋅logy⋅logzlogx⋅logy⋅logz​∣111​121​113​∣

$$\Rightarrow |A|=|\begin{matrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 2\end{matrix}|$$⇒∣A∣=∣111​010​002​∣

$$\Rightarrow |A|=2$$⇒∣A∣=2

Now,

$$|adj(adj{A}^{2})|={|adj{A}^{2}|}^{2}={|{A}^{2}|}^{4}={|A|}^{8}={2}^{8}$$∣adj(adjA2)∣=∣adjA2∣2=∣A2∣4=∣A∣8=28