Determinants Matrices Question 2
Question: If $ A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \\ \end{bmatrix} $ and I is the unit matrix of order 3, then $ A^{2}+2A^{4}+4A^{6} $ is equal to
Options:
A) $ 7A^{8} $
B) $ 7A^{7} $
C) 8I
D) 6I
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ A^{2}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a^{2}+b & ab & -a -1 \\ \end{bmatrix} $
$ A^{2}=A^{4}=A^{6}=I_3\Rightarrow A^{2}+2A^{4}+4A^{6} $
$ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix} + \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \\ \end{bmatrix} $
$ = \begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \\ \end{bmatrix} =7I_3 \neq 7A^{8} $
BETA
