Let A = {bmatrix} i & -i \\ -i & i {bmatrix}, i = {-1}. Then, the system of linear equations A^{8\{{x ; y}\}} = \{{8 ; 64}\} has :

Correct Answer is : No solution

Correct Answer is : No solution

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JEE Main 2026 Model Paper 1

Question : 53 of 75

Marks: +1, -0

Let

A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}, i = \sqrt{-1}

A^{8\{\text{x ; y}\}} = \{\text{8 ; 64}\}

Solution:

\;A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}

\;A^2 = 2 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}

\;A^4=2^2\begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix}=8\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}

\;A^8=64\{\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}\}=128\{\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\}

\;A^8{\begin{bmatrix} x \\ y \end{bmatrix}} = {\begin{bmatrix} 8 \\ 64 \end{bmatrix}}

\;\Rightarrow 128\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}

\;\Rightarrow 128{\begin{bmatrix} x-y \\ -x+y \end{bmatrix}} = {\begin{bmatrix} 8 \\ 64 \end{bmatrix}}

\;⇒ x-y=\;{1}/{16}

⇒ From (1) & (2) : No solution.

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