Determinants Matrices Question 129
Question: The values of a, b, c if $ \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \\ \end{bmatrix} $ is orthogonal are
Options:
A) $ a=\pm \frac{1}{\sqrt{2}};b=\pm \frac{1}{\sqrt{6}};c=\pm \frac{1}{\sqrt{3}} $
B) $ a=\pm \frac{1}{\sqrt{2}};b=\pm \frac{1}{\sqrt{3}};c=\pm \frac{1}{\sqrt{6}} $
C) $ a=\pm \frac{1}{\sqrt{6}};b=\pm \frac{1}{\sqrt{2}};c=\pm \frac{1}{\sqrt{3}} $
D) $ a=\pm \frac{1}{\sqrt{3}};b=\pm \frac{1}{\sqrt{2}};c=\pm \frac{1}{\sqrt{6}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Let $ A= \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \\ \end{bmatrix} $ , Now, $ A^{T}= \begin{bmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \\ \end{bmatrix} $
$ \because $ A is orthogonal
$ \therefore AA^{T}=I $
$ \Rightarrow \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \\ \end{bmatrix} \begin{bmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $ Equating the corresponding elements, we get $ 4b^{2}+c^{2}=1…(1) $
$ 2b^{2}-c^{2}=0…(2) $
$ a^{2}+b^{2}+c^{2}=1…(3) $ On solving (1), (2) and (3), we get $ a=\pm \frac{1}{\sqrt{2}};b=\pm \frac{1}{\sqrt{6}};c=\pm \frac{1}{\sqrt{3}} $
BETA
