JEE Main 2021 — Mathematics Algebra

We have,

{a}_{ij}={\begin{matrix}{(-1)}^{j-i}\mathrm{if}i<j \\ 2\mathrm{if}i=j \\ {(-1)}^{i+j}\mathrm{if}i>j\end{matrix}

The,

$$A=[\begin{matrix}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{matrix}]$$A=[2−11​−12−1​1−12​]

$$\Rightarrow |A|=4$$⇒∣A∣=4

Now,

$$|3adj(2{A}^{-1})|={3}^{3}|adj(2{A}^{-1})|$$∣3adj(2A−1)∣=33∣adj(2A−1)∣

$$\because |kadj({A}^{-1})|={k}^{n}|adj({A}^{-1})|$$∵∣kadj(A−1)∣=kn∣adj(A−1)∣

$$\Rightarrow |3adj(2{A}^{-1})|={3}^{3}\cdot {2}^{6}{|{A}^{-1}|}^{2}$$⇒∣3adj(2A−1)∣=33⋅26∣A−1∣2

$$\because |adj(k{A}^{-1})|={({k}^{(n-1)})}^{n}|adj({A}^{-1})|={({k}^{(n-1)})}^{n}{|{A}^{-1}|}^{n-1}$$∵∣adj(kA−1)∣=(k(n−1))n∣adj(A−1)∣=(k(n−1))n∣A−1∣n−1

$$\Rightarrow |3adj(2{A}^{-1})|={12}^{3}{|{A}^{-1}|}^{2}=\frac{{12}^{3}}{|A{|}^{2}}=\frac{{12}^{3}}{16}=108$$⇒∣3adj(2A−1)∣=123∣A−1∣2=∣A∣2123​=16123​=108