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Maths-

General

Easy

Question

If the matrix $open square brackets table row 0 cell 2 beta end cell gamma row alpha beta cell negative gamma end cell row alpha cell negative beta end cell gamma end table close square brackets$ is orthogonal, then

$α$ = ± $\frac{1}{\sqrt{2}}$
$β$ = ± $\frac{1}{\sqrt{6}}$
$γ$ = ± $\frac{1}{\sqrt{3}}$
All of these

The correct answer is: All of these

since A is orthogonal,
AA´ = I
$open square brackets table row cell 4 beta to the power of 2 end exponent plus gamma to the power of 2 end exponent end cell cell 2 beta to the power of 2 end exponent minus gamma to the power of 2 end exponent end cell cell negative 2 beta to the power of 2 end exponent plus gamma to the power of 2 end exponent end cell row cell 2 beta to the power of 2 end exponent minus gamma to the power of 2 end exponent end cell cell alpha to the power of 2 end exponent plus beta to the power of 2 end exponent plus gamma to the power of 2 end exponent end cell cell alpha to the power of 2 end exponent minus beta to the power of 2 end exponent minus gamma to the power of 2 end exponent end cell row cell negative 2 beta to the power of 2 end exponent plus gamma to the power of 2 end exponent end cell cell alpha to the power of 2 end exponent minus beta to the power of 2 end exponent minus gamma to the power of 2 end exponent end cell cell alpha to the power of 2 end exponent plus beta to the power of 2 end exponent plus gamma to the power of 2 end exponent end cell end table close square brackets$ = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$
$therefore$ $alpha$ = ± $fraction numerator 1 over denominator square root of 2 end fraction$ , $beta$ = ± $fraction numerator 1 over denominator square root of 6 end fraction$ , $gamma$ = ± $fraction numerator 1 over denominator square root of 3 end fraction$

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