Determinants Matrices Question 141
Question: If $ B^{n}-A=I $ and $ A= \begin{bmatrix} 26 & 26 & 18 \\ 25 & 37 & 17 \\ 52 & 39 & 50 \\ \end{bmatrix} ,B= \begin{bmatrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{bmatrix} $ then n =
Options:
A) 2
B) 3
C) 4
D) 5
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ \because B^{n}-A=I\therefore B^{n}=I+A $
$ B^{n}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} + \begin{bmatrix} 26 & 26 & 18 \\ 25 & 37 & 17 \\ 52 & 39 & 50 \\ \end{bmatrix} $
$ B^{n}= \begin{bmatrix} 27 & 26 & 18 \\ 25 & 38 & 17 \\ 52 & 39 & 51 \\ \end{bmatrix} $ or $ {{ \begin{bmatrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{bmatrix} }^{n}}= \begin{bmatrix} 27 & 26 & 18 \\ 25 & 38 & 17 \\ 52 & 39 & 51 \\ \end{bmatrix} ….(i) $
$ \therefore n\ne 1 $ Now put $ n=2, $ then $ B^{2}={{ \begin{bmatrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{bmatrix} }^{2}}= \begin{bmatrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{bmatrix} \begin{bmatrix} 1 & 4 & 2 \\ 3 & 5 & 1 \\ 7 & 1 & 6 \\ \end{bmatrix} $
$ = \begin{bmatrix} 27 & 26 & 18 \\ 25 & 38 & 17 \\ 52 & 39 & 51 \\ \end{bmatrix} $ Which is equal to R.H.S. of eq. (i).
$ \therefore n=2 $
BETA
