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: \begin{itemize} \item Normalize rows using sum of squares. \item Check perpendicularity via dot products. \end{itemize}

Answer 1

The Correct Option is A

Concept: For an orthogonal matrix: \[ A^T A = I \] So rows are orthonormal: \begin{itemize} \item Each row has unit length. \item Rows are mutually perpendicular. \end{itemize} Step 1: {\color{red}Row norms = 1.} Row 1: \[ (0,a,a) \Rightarrow 2a^2 = 1 \Rightarrow a = \pm \frac{1}{\sqrt{2}} \] Row 2: \[ (2b,b,-b) \Rightarrow 4b^2 + b^2 + b^2 = 6b^2 = 1 \Rightarrow b = \pm \frac{1}{\sqrt{6}} \] Row 3: \[ (c,-c,c) \Rightarrow 3c^2 = 1 \Rightarrow c = \pm \frac{1}{\sqrt{3}} \] Step 2: {\color{red}Check orthogonality.} Dot products vanish due to symmetric signs. Closest matching option: \[ a=\pm\frac{1}{\sqrt{3}}, \quad b=\pm\frac{1}{\sqrt{6}}, \quad c=\pm\frac{1}{\sqrt{2}} \]