Question:
If
\[
A = \frac{1}{3}
\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{bmatrix}
\]
is an orthogonal matrix, then:
If
\[ A = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \] is an orthogonal matrix, then:
\[ A = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \] is an orthogonal matrix, then:
Show Hint
Orthogonal matrices have mutually orthogonal rows of unit length.
Updated On: Mar 23, 2026
- \(a=-2,b=-1\)
- \(a=2,b=1\)
- \(a=2,b=-1\)
- a=-2,b=1
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1:
For an orthogonal matrix, rows are mutually perpendicular.
Step 2: Dot product of first and third row:
\[ 1 \cdot a + 2 \cdot 2 + 2 \cdot b = 0 \quad \implies \quad a + 4 + 2b = 0 \quad (1) \]
Step 3: Dot product of second and third row:
\[ 2 \cdot a + 1 \cdot 2 + (-2) \cdot b = 0 \quad \implies \quad 2a + 2 - 2b = 0 \quad (2) \]
Step 4: Solving (1) and (2):
\[ a = 2, \quad b = -1 \]
For an orthogonal matrix, rows are mutually perpendicular.
Step 2: Dot product of first and third row:
\[ 1 \cdot a + 2 \cdot 2 + 2 \cdot b = 0 \quad \implies \quad a + 4 + 2b = 0 \quad (1) \]
Step 3: Dot product of second and third row:
\[ 2 \cdot a + 1 \cdot 2 + (-2) \cdot b = 0 \quad \implies \quad 2a + 2 - 2b = 0 \quad (2) \]
Step 4: Solving (1) and (2):
\[ a = 2, \quad b = -1 \]
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