Determinants Matrices Question 111
Question: Let $ A= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} $ be a square matrix of order 3. Then for any positive integer n, what is $ A^{n} $ equal to ?
Options:
A) A
B) $ 3^{n}A $
C) $ ({3^{n-1}})A $
D) 3A
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Given matrix is $ A= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} $ So, $ A^{2} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} $
$ = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \\ \end{bmatrix} =3 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} $ and $ A^{3}=3 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} $
$ =3 \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \\ \end{bmatrix} =9 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} =3^{2}A $ Similarly $ A^{4}=3^{3}A $ . Hence, $ A^{n}={3^{n-1}}A $
BETA
