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+1 vote
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in Matrices & determinants by (30 points)
edited by

Let S = {(-1, a)(0, b); a, b ∈ {1,2,3,...100} and let  Tn = {A ∈ S: An(n+1) = I}. Then the number of elements in

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1 Answer

+1 vote
by (42.1k points)

Let 

\(A = \begin{bmatrix}-1&a\\0&b\end{bmatrix}\)

\(A^2 = \begin{bmatrix}1&a(b - 1)\\0&b^2\end{bmatrix}\) become identity if b = 1 & a be any number.

\(A^3 = \begin{bmatrix}-1&-a(b - 1-b^2)\\0&b^2\end{bmatrix}\) never become identity

\(A^4 = \begin{bmatrix}1&a(b-1)(b^2 + 1)\\0&b^2\end{bmatrix}\) become identity if b = 1 and a be any number.

A2n become identity if b = 1 and a be any number but A2n+1 never become identity.

\(\therefore T_n = \left\{\begin{bmatrix}-1 &a\\0&1\end{bmatrix};a\in\{1, 2, .....,100\}\right\}\)

\(\bigcap\limits_{n=1}^{100} T_n= \begin{bmatrix}-1 &a\\0&1\end{bmatrix};a\in\{1, 2, .....,n^2\}= T_n\)

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