If A=131 2 22 1 −2a 2 b is an orthogonal matrix, then

A
a 2,b =1
B
a =-2,b = -1
C
a= 2, b =-1
D
a =- 2, b =1

Solution:

Since A is an orthogonal matrix.

$\begin{array}{l} ∴ {AA}^{T} = 1 \\ \Rightarrow \frac{1}{3} [\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ a & 2 & b \end{array}] \cdot \frac{1}{3} [\begin{array}{rrr} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & - 2 & b \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow \frac{1}{9} [\begin{array}{llr} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ a & 2 & b \end{array}] [\begin{array}{lll} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & - 2 & b \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} 9 & 0 & a + 4 + 2 b \\ 0 & 9 & 2 a + 2 - 2 b \\ a + 4 + 2 b & 2 a + 2 - 2 b & a^{2} + 4 + b^{2} \end{array}] = [\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}] \\ \equiv a + 4 + 2 b = 0, 2 a + 2 - 2 b = 0 and a^{2} + 4 + b^{2} = 9 \\ \Rightarrow a + 2 b + 4 = 0, a - b + 1 = 0 and a^{2} + b^{2} = 5 \end{array}$

$\Rightarrow a = - 2, b = - 1$

If $A = \frac{1}{3} [\begin{array}{rrr} 1 2 2 \\ 2 1 - 2 \\ a 2 b \end{array}]$ is an orthogonal matrix, then

Solution:

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