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: Definition of an Orthogonal Matrix.
\n\nA matrix \( A \) is orthogonal if:\n\[\nAA^T = I\n\]\nwhere \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.\n\n
Step 2: Problem Setup.\n\nGiven matrix:\n\[\nA = \begin{pmatrix}\n0 & a & a
\n2b & b & -b
\nc & -c & c\n\end{pmatrix}\n\]\n\nCalculate \( AA^T \) and equate it to the identity matrix \( I \).\n\n
Step 3: Calculate \( AA^T \).\n\nRow 1 ⋅ Row 1:\n\[\n0^2 + a^2 + a^2 = 2a^2\n\]\n\nRow 2 ⋅ Row 2:\n\[\n% Option\n(2b)^2 + b^2 + (-b)^2 = 4b^2 + b^2 + b^2 = 6b^2\n\]\n\nRow 3 ⋅ Row 3:\n\[\nc^2 + (-c)^2 + c^2 = 3c^2\n\]\n\nSince this must be an identity matrix:\n\[\n2a^2 = 1, \quad 6b^2 = 1, \quad 3c^2 = 1\n\]\n\nTherefore:\n\[\na^2 = \frac{1}{2}, \quad b^2 = \frac{1}{6}, \quad c^2 = \frac{1}{3}\n\]\n\nTaking square roots:\n\[\na = \pm \frac{1}{\sqrt{2}}, \quad b = \pm \frac{1}{\sqrt{6}}, \quad c = \pm \frac{1}{\sqrt{3}}\n\]\n\n
Step 4: Verify Orthogonality.
\n\nVerify that cross terms are zero (e.g., Row 1 ⋅ Row 2 = 0), which confirms orthogonality for these values.\n\nThus, the answer is option (B).