Determinants Matrices Question 127
Question: Let $ A=( \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \\ \end{matrix} ). $ and 10 $ B=( \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{matrix} ). $ If B is the inverse of matrix A, then $ \alpha $ is
Options:
A) 5
B) -1
C) 2
D) -2
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Here,
$ \Rightarrow B=\frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{bmatrix} $ Also since, $ B={A^{-1}}\Rightarrow AB=I $
$ \Rightarrow \frac{1}{10} \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $
$ \Rightarrow \frac{1}{10} \begin{bmatrix} 10 & 0 & 5-2 \\ 0 & 10 & -5+\alpha \\ 0 & 0 & 5+\alpha \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $
$ \Rightarrow \frac{5-\alpha }{10}=0\Rightarrow \alpha =5 $
BETA
