Question

For what values of k , the matrix A = [1/2 k ]
[-k 1/2] is an orthogonal matrix ?​

Answer 1

Answer:

1/3

Step-by-step explanation:

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Answer 2

Answer:

The values of k are  $+\frac{\sqrt{3} }{2}$  0r  $-\frac{\sqrt{3} }{2}$

Step-by-step explanation:

The orthogonal matrix is the product of a square matrix, and its transpose gives an identity matrix.

That is $AA^{T}$ = I

The given matrix is

$A = \left[\begin{array}{22}\frac{1}{2} &k\\-k&\frac{1}{2} \end{array}\right]$

The transpose of a given matrix A is given by

$A^{T} = \left[\begin{array}{22}\frac{1}{2} &-k\\k&\frac{1}{2} \end{array}\right]$

They have given that matrix A is an orthogonal matrix

Therefore

$AA^{T}= \left[\begin{array}{22}\frac{1}{2} &k\\-k&\frac{1}{2} \end{array}\right] \left[\begin{array}{22}\frac{1}{2} &-k\\k&\frac{1}{2} \end{array}\right]$$=\left[\begin{array}{22}1&0\\0&1\end{array}\right]$

         $=\left[\begin{array}{22}\frac{1+4k^{2} }{4} &0\\0&\frac{1+4k^{2} }{4} \end{array}\right]$$=\left[\begin{array}{22}1&0\\0&1\end{array}\right]$

1 + 4$k^{2}$ = 1

$k=+\frac{\sqrt{3} }{2}$  or $k=-\frac{\sqrt{3} }{2}$

Therefore the k values are  $k=+\frac{\sqrt{3} }{2}$  or $k=-\frac{\sqrt{3} }{2}$