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If A = \begin{bmatrix} \frac{1}{3} & \frac{2}{3} &a \\ \frac{2}{3}&\frac{1}{3} &b \\ \frac{2}{3}& -\frac{2}{3} & c \end{bmatrix} is orthogonal, then find a,b,c

  • Option 1)

    \left(\pm \frac{1}{3}, \pm \frac{2}{3},\pm \frac{2}{3}\right )

  • Option 2)

    \left(\pm \frac{2}{3}, \pm \frac{1}{3},\pm \frac{1}{3}\right )

  • Option 3)

    \left(\pm \frac{2}{3}, \pm \frac{2}{3},\pm \frac{1}{3}\right )

  • Option 4)

    \left(\pm \frac{2}{3}, \pm \frac{1}{3},\pm \frac{2}{3}\right )

 

Answers (1)

As we have learnt,

 

Orthogonal matrix -

AA{}'=I

- wherein

A{}' is transpose matrix of matrix A and I is identity matrix 

 

 For Orthogonal Matrices, AA' = I

\begin{bmatrix} \frac{1}{3} & \frac{2}{3} &a \\ \frac{2}{3}&\frac{1}{3} &b \\ \frac{2}{3}& -\frac{2}{3} & c \end{bmatrix} \begin{bmatrix} \frac{1}{3} & \frac{2}{3} &\frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} &-\frac{2}{3} \\ a& b & c \end{bmatrix} = \begin{bmatrix} 1 & 0 &0 \\ 0 & 1& 0\\ 0 & 0 & 1 \end{bmatrix}

On comparing 

\frac{1}{}9 + \frac{4}{}9 + a^2 = 1 ; \frac{2}{}9 +\frac{2}{}9 + ab = 0 \\*\Rightarrow a^2 = \frac{4}{9} \Rightarrow a = \pm \frac{2}{3} \;and \;b = \mp \frac{2}{}3 \; and \; ||ly \; c =\pm \frac{1}{3} 


Option 1)

\left(\pm \frac{1}{3}, \pm \frac{2}{3},\pm \frac{2}{3}\right )

Option 2)

\left(\pm \frac{2}{3}, \pm \frac{1}{3},\pm \frac{1}{3}\right )

Option 3)

\left(\pm \frac{2}{3}, \pm \frac{2}{3},\pm \frac{1}{3}\right )

Option 4)

\left(\pm \frac{2}{3}, \pm \frac{1}{3},\pm \frac{2}{3}\right )

Posted by

Vakul

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