यदि A = begin{bmatrix} i & 1 1 & 0 end{bmatr | vedclass

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(D) दिया गया है $A = \begin{bmatrix} i & 1 \ 1 & 0 \end{bmatrix}$.$A^2 = \begin{bmatrix} i & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} i & 1 \ 1 & 0

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$-\operatorname{adj} A$

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(D) दिया गया है $A = \begin{bmatrix} i & 1 \ 1 & 0 \end{bmatrix}$.$A^2 = \begin{bmatrix} i & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} i & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} i^2+1 & i \ i & 1 \end{bmatrix} = \begin{bmatrix} 0 & i \ i & 1 \end{bmatrix}$.$A^3 = A^2 	imes A = \begin{bmatrix} 0 & i \ i & 1 \end{bmatrix} \begin{bmatrix} i & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix} = iI$.अतः $A^6 = (A^3)^2 = (iI)^2 = i^2 I = -I$.अब,$B = A^{2029} = A^{6 	imes 338 + 1} = (A^6)^{338} 	imes A = (-I)^{338} 	imes A = I 	imes A = A$.चूँकि $B = A$,हमें $B^{-1} = A^{-1}$ ज्ञात करना है।$|A| = (i)(0) - (1)(1) = -1$.$A^{-1} = \frac{1}{|A|} \operatorname{adj} A = \frac{1}{-1} \operatorname{adj} A = -\operatorname{adj} A$.अतः,$B^{-1} = -\operatorname{adj} A$.

यदि $A = \begin{bmatrix} i & 1 \\ 1 & 0 \end{bmatrix}$ जहाँ $i = \sqrt{-1}$ और $B = A^{2029}$ है,तो $B^{-1} =$

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