A) 2
B) 3
C) 4
D) 5
Correct Answer: A
Solution :
[a] \[\because {{B}^{n}}-A=I\therefore {{B}^{n}}=I+A\] \[{{B}^{n}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 26 & 26 & 18 \\ 25 & 37 & 17 \\ 52 & 39 & 50 \\ \end{matrix} \right]\] \[{{B}^{n}}=\left[ \begin{matrix} 27 & 26 & 18 \\ 25 & 38 & 17 \\ 52 & 39 & 51 \\ \end{matrix} \right]\] or \[{{\left[ \begin{matrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{matrix} \right]}^{n}}=\left[ \begin{matrix} 27 & 26 & 18 \\ 25 & 38 & 17 \\ 52 & 39 & 51 \\ \end{matrix} \right]....(i)\] \[\therefore n\ne 1\] Now put \[n=2,\] then \[{{B}^{2}}={{\left[ \begin{matrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{matrix} \right]}^{2}}=\left[ \begin{matrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 27 & 26 & 18 \\ 25 & 38 & 17 \\ 52 & 39 & 51 \\ \end{matrix} \right]\] Which is equal to R.H.S. of eq. (i). \[\therefore n=2\]
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