3 and 4 .Determinants and Matrices
hard

मान $A =\left[\begin{array}{cc} i & - i \\ - i & i \end{array}\right], i =\sqrt{-1}$ है। तो रैखिक समीकरण निकाय $A ^{8}\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$

A

का अद्वितीय हल है

B

के अनंत हल हैं

C

का कोई हल नहीं है

D

के मात्र दो हल है

(JEE MAIN-2021)

Solution

$A=\left[\begin{array}{cc}i & -i \\ -i & i\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right]=2\left[\begin{array}{cc}-1 & 1 \\ 1 & -1\end{array}\right]$

$A^{4}=2^{2}\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=8\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$

$A^{8}=64\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]=128\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$

$A^{8}\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}8 \\ 64\end{array}\right]$

$\Rightarrow 128\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$

$\Rightarrow \quad 128\left[\begin{array}{c}x-y \\ -x+y\end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$

$\Rightarrow \quad x-y=\frac{1}{16}………(1)$

$\quad-x+y=\frac{1}{2}$   $…..(2)$

$\Rightarrow$ From $(1)$ and $(2)$ No solution.

Standard 12
Mathematics

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