Determinants Matrices Question 105
Question: If $ A= \begin{bmatrix} 1 & 0 \\ -1 & 7 \\ \end{bmatrix} $ and $ I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $ , then the value of k so that $ A^{2}=8A+kI $ is
Options:
A) $ k=7 $
B) $ k=-7 $
C) $ k=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] We have, $ A^{2}= \begin{bmatrix} 1 & 0 \\ -1 & 7 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 7 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -8 & 49 \\ \end{bmatrix} $ and $ 8A+kI=8 \begin{bmatrix} 1 & 0 \\ -1 & 7 \\ \end{bmatrix} +k \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $
$ = \begin{bmatrix} 8 & 0 \\ -8 & 56 \\ \end{bmatrix} + \begin{bmatrix} k & 0 \\ 0 & k \\ \end{bmatrix} = \begin{bmatrix} 8+k & 0 \\ -8 & 56+k \\ \end{bmatrix} $ Thus, $ A^{2}=8A+kI\Rightarrow \begin{bmatrix} 1 & 0 \\ -8 & 49 \\ \end{bmatrix} = \begin{bmatrix} 8+k & 0 \\ -8 & 56+k \\ \end{bmatrix} $
$ \Rightarrow 1=8+k $ and $ 56+k=49\Rightarrow k=-7 $
BETA
