JEE Main 2021 — Mathematics Algebra

Given, $A=[\begin{matrix}2 & 3 \\ 0 & -1\end{matrix}]$A=[20​3−1​]

$$\Rightarrow {A}^{2}=[\begin{matrix}2 & 3 \\ 0 & -1\end{matrix}][\begin{matrix}2 & 3 \\ 0 & -1\end{matrix}]=[\begin{matrix}4 & 3 \\ 0 & 1\end{matrix}]$$⇒A2=[20​3−1​][20​3−1​]=[40​31​]

$$\Rightarrow {A}^{4}=[\begin{matrix}4 & 3 \\ 0 & 1\end{matrix}][\begin{matrix}4 & 3 \\ 0 & 1\end{matrix}]=[\begin{matrix}16 & 15 \\ 0 & 1\end{matrix}]$$⇒A4=[40​31​][40​31​]=[160​151​]

$$\Rightarrow {A}^{5}=[\begin{matrix}16 & 15 \\ 0 & 1\end{matrix}][\begin{matrix}2 & 3 \\ 0 & -1\end{matrix}]=[\begin{matrix}32 & 33 \\ 0 & -1\end{matrix}]$$⇒A5=[160​151​][20​3−1​]=[320​33−1​]

$$\Rightarrow {A}^{10}=[\begin{matrix}32 & 33 \\ 0 & -1\end{matrix}][\begin{matrix}32 & 33 \\ 0 & -1\end{matrix}]=[\begin{matrix}1024 & 1023 \\ 0 & 1\end{matrix}]....(i)$$⇒A10=[320​33−1​][320​33−1​]=[10240​10231​]....(i)

Now, $2A=[\begin{matrix}4 & 6 \\ 0 & -2\end{matrix}]$2A=[40​6−2​]

$$\Rightarrow \mathrm{Adj}(2A)={[\begin{matrix}-2 & 0 \\ -6 & 4\end{matrix}]}^{T}$$⇒Adj(2A)=[−2−6​04​]T

$$\Rightarrow \mathrm{Adj}(2A)=[\begin{matrix}-2 & -6 \\ 0 & 4\end{matrix}]$$⇒Adj(2A)=[−20​−64​]

$$\Rightarrow {(\mathrm{Adj}(2A))}^{2}=[\begin{matrix}-2 & -6 \\ 0 & 4\end{matrix}][\begin{matrix}-2 & -6 \\ 0 & 4\end{matrix}]=[\begin{matrix}4 & -12 \\ 0 & 16\end{matrix}]$$⇒(Adj(2A))2=[−20​−64​][−20​−64​]=[40​−1216​]

$$\Rightarrow {(\mathrm{Adj}(2A))}^{4}=[\begin{matrix}4 & -12 \\ 0 & 16\end{matrix}][\begin{matrix}4 & -12 \\ 0 & 16\end{matrix}]=[\begin{matrix}16 & -240 \\ 0 & 256\end{matrix}]$$⇒(Adj(2A))4=[40​−1216​][40​−1216​]=[160​−240256​]

$$\Rightarrow {(\mathrm{Adj}(2A))}^{5}=[\begin{matrix}16 & -240 \\ 0 & 256\end{matrix}][\begin{matrix}-2 & -6 \\ 0 & 4\end{matrix}]=[\begin{matrix}-32 & -1056 \\ 0 & 1024\end{matrix}]$$⇒(Adj(2A))5=[160​−240256​][−20​−64​]=[−320​−10561024​]

$$\Rightarrow {(\mathrm{Adj}(2A))}^{10}=[\begin{matrix}-32 & -1056 \\ 0 & 1024\end{matrix}][\begin{matrix}-32 & -1056 \\ 0 & 1024\end{matrix}]=[\begin{matrix}1024 & -1047552 \\ 0 & 1048576\end{matrix}].....(ii)$$⇒(Adj(2A))10=[−320​−10561024​][−320​−10561024​]=[10240​−10475521048576​].....(ii)

From equation (i)&(ii)

$$\Rightarrow {A}^{10}-{(\mathrm{Adj}(2A))}^{10}=[\begin{matrix}1024 & 1023 \\ 0 & 1\end{matrix}]-[\begin{matrix}1024 & -1047552 \\ 0 & 1048576\end{matrix}]=[\begin{matrix}0 & 1048575 \\ 0 & -1048575\end{matrix}]$$⇒A10−(Adj(2A))10=[10240​10231​]−[10240​−10475521048576​]=[00​1048575−1048575​]

$$\Rightarrow \mathrm{det}({A}^{10}-{(\mathrm{Adj}(2A))}^{10})=|\begin{matrix}0 & 1048575 \\ 0 & -1048575\end{matrix}|=0$$⇒det(A10−(Adj(2A))10)=∣00​1048575−1048575​∣=0

Let $E=det({A}^{4})+det({A}^{10}-(Adj(2A){)}^{10})$E=det(A4)+det(A10−(Adj(2A))10)

$$\Rightarrow E={(det(A))}^{4}+0$$⇒E=(det(A))4+0

$$\Rightarrow E={(-2-0)}^{4}$$⇒E=(−2−0)4

$$\Rightarrow E=16$$⇒E=16