12) c...
1) Find the least positive integral value of "k" such that it satisfies-
{\begin{pmatrix} cos\frac{2 \pi}{7}\ -sin\frac{2 \pi}{7}\\ sin\frac{2 \pi}{7}\ \ cos\frac{2 \pi}{7} \end{pmatrix}}^k = \begin{pmatrix} 1 \ 0 \\ 0 \1 \end{pmatrix}
2) If A= {\begin{pmatrix} i \ \ 0 \\ 0 \ \ i \end{pmatrix}}, n ε N, then A4n equals-
a) A b) Null matrix c) Identity matrix d) -A
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11 Answers
39A try at Q1..
Q1. \begin{bmatrix} cos\frac{2\pi}{7} & -sin\frac{2\pi}{7} \\ sin\frac{2\pi}{7} & cos\frac{2\pi}{7} \end{bmatrix}^{k - 1}\begin{bmatrix} cos\frac{2\pi}{7} & -sin\frac{2\pi}{7} \\ sin\frac{2\pi}{7} & cos\frac{2\pi}{7} \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}
The first matrix is the inverse of the second matrix on the LHS, let us say. So now we need to find the inverse of the second matrix on the LHS.
The inverse is \begin{bmatrix} cos\frac{2\pi}{7} & -sin\frac{2\pi}{7} \\ sin\frac{2\pi}{7} & cos\frac{2\pi}{7} \end{bmatrix}^{k - 1} = \begin{bmatrix} cos\frac{2\pi}{7} & sin\frac{2\pi}{7} \\ -sin\frac{2\pi}{7} & cos\frac{2\pi}{7} \end{bmatrix}
Lol I can't get beyond this..
13Hmm... i didn't think of inverse though, lekin aage wakai samajh nahi aa rahaa abhi.
1Q1 find square of the given matrix it follows a pattern
that is after mulitplying the matrix k times
that first term is Cos(2Ï€k/7) likewise for other terms.
for Cos(2Ï€k/7) to be 1
2Ï€k/7=1 => k=7