Question

If the matrix \[ \begin{pmatrix} 0 & a & a\\ 2b & b & -b\\ c & -c & c \end{pmatrix} \] is orthogonal, then the values of \( a, b, c \) are:

• A. \( a = \pm \frac{1}{\sqrt{3}}, b = \pm \frac{1}{\sqrt{6}}, c = \pm \frac{1}{\sqrt{2}} \)
• B. \( a = \pm \frac{1}{\sqrt{2}}, b = \pm \frac{1}{\sqrt{6}}, c = \pm \frac{1}{\sqrt{3}} \)
• C. \( a = \pm \frac{1}{\sqrt{2}}, b = \pm \frac{-1}{\sqrt{6}}, c = \pm \frac{-1}{\sqrt{3}} \)
• D. \( a = \pm \frac{1}{\sqrt{3}}, b = \pm \frac{1}{\sqrt{6}}, c = \pm \frac{1}{\sqrt{3}} \)

Hint:

For orthogonal matrices, always ensure: 1. Each row (or column) vector must have unit magnitude. 2. Rows (or columns) must be mutually perpendicular (dot product zero).

Answer 1

The Correct Option is B

Step 1: Understand the property of an orthogonal matrix.
A matrix \( A \) is orthogonal if: \[ AA^T = I \] where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.
Step 2: Set up the condition. Given matrix: \[ A = \begin{pmatrix} 0 & a & a
2b & b & -b
c & -c & c \end{pmatrix} \] Find \( AA^T \) and equate it to the identity matrix \( I \).
Step 3: Compute \( AA^T \). First row dot first row: \[ 0^2 + a^2 + a^2 = 2a^2 \] Second row dot second row: \[ % Option (2b)^2 + b^2 + (-b)^2 = 4b^2 + b^2 + b^2 = 6b^2 \] Third row dot third row: \[ c^2 + (-c)^2 + c^2 = 3c^2 \] Since it must be an identity matrix: \[ 2a^2 = 1, \quad 6b^2 = 1, \quad 3c^2 = 1 \] Thus: \[ a^2 = \frac{1}{2}, \quad b^2 = \frac{1}{6}, \quad c^2 = \frac{1}{3} \] Taking positive or negative square roots: \[ a = \pm \frac{1}{\sqrt{2}}, \quad b = \pm \frac{1}{\sqrt{6}}, \quad c = \pm \frac{1}{\sqrt{3}} \]
Step 4: Verify other orthogonality conditions.
Cross terms like:
First row dot second row = 0,
First row dot third row = 0,
Second row dot third row = 0,
can be verified similarly to ensure orthogonality, which they satisfy for these values. Thus, the final answer is option (B).