91Ó°ÊÓ

The five matrices are there. Four of them are a 3*3 matrix and the rest one is a 4*4 matrix.

A square matrix is said to be orthogonal if and only if the transpose of the matrix is equal to its inverse matrix. i.e If A is a square matrix, then A is called orthogonal if and only if \({{\bf{A}}^{\bf{T}}}{\bf{ = }}{{\bf{A}}^{{\bf{ - 1}}}}\). i.e, \({\bf{A}}{{\bf{A}}^{\bf{T}}}{\bf{ = I}}\) So, there can be concluded that, if the product of the matrix and its transpose matrix is equal to its identity matrix then the matrix is orthogonal.

Let us consider the first matrix\(A = \left( {\begin{align}{}0&1&0\\0&0&1\\1&0&0\end{align}} \right)\)

So, the transpose of the matrix is \({A^T} = \left( {\begin{align}{}0&0&1\\1&0&0\\0&1&0\end{align}} \right)\)

Now,

\(\begin{array}{c}A{A^T} = \left[ {\begin{array}{}0&1&0\\0&0&1\\1&0&0\end{array}} \right] \times \left[ {\begin{array}{}0&0&1\\1&0&0\\0&1&0\end{array}} \right]\\ = \left[ {\begin{array}{}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\\ = I\end{array}\)

Thus, the product of the matrix and its transpose matrix is equal to identity matrix. So, it is orthogonal.

b.

Let us consider the first matrix\(B = \left( {\begin{align}{}{0.8}&0&{0.6}\\{ - 0.6}&0&{0.8}\\0&{ - 1}&0\end{align}} \right)\)

So, the transpose of the matrix is \({B^T} = \left( {\begin{align}{}{0.8}&{0.6}&0\\0&0&{ - 1}\\{0.6}&{0.8}&0\end{align}} \right)\)

Now,

\(\begin{array}{c}A{A^T} = \left[ {\begin{array}{}{0.8}&0&{0.6}\\{ - 0.6}&0&{0.8}\\0&{ - 1}&0\end{array}} \right] \times \left[ {\begin{array}{}{0.8}&{0.6}&0\\0&0&{ - 1}\\{0.6}&{0.8}&0\end{array}} \right]\\ = \left[ {\begin{array}{}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\\ = I\end{array}\)

Thus, the product of the matrix and its transpose matrix is equal to identity matrix. So, it is orthogonal.

c.

Let us consider the first matrix\(C = \left( {\begin{align}{}{0.8}&0&{0.6}\\{ - 0.6}&0&{0.8}\\0&{0.5}&0\end{align}} \right)\)

So, the transpose of the matrix is \({C^T} = \left( {\begin{align}{}{0.8}&{0.6}&0\\0&0&{0.5}\\{0.6}&{0.8}&0\end{align}} \right)\)

Now,

\(\begin{array}A{A^T} = \left[ {\begin{array}{}{0.8}&0&{0.6}\\{ - 0.6}&0&{0.8}\\0&{0.5}&0\end{array}} \right] \times \left[ {\begin{array}{}{0.8}&{0.6}&0\\0&0&{0.5}\\{0.6}&{0.8}&0\end{array}} \right]\\ = \left[ {\begin{array}{}1&0&0\\0&1&0\\0&0&{0.25}\end{array}} \right]\\ \ne I\end{array}\)

Thus, the product of the matrix and its transpose matrix is not equal to identity matrix. So, it is not orthogonal.

d).

Let us consider the first matrix\(D = \left( {\begin{align}{}{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}\end{align}} \right)\)

So, the transpose of the matrix is \({D^T} = \left( {\begin{align}{}{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}\end{align}} \right)\)

Now,

\(\begin{array}{c}A{A^T} = \left[ {\begin{array}{}{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}\end{array}} \right] \times \left[ {\begin{array}{}{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{ - \frac{1}{{\sqrt 3 }}}\end{array}} \right]\\ = \left[ {\begin{array}{}1&{ - \frac{1}{3}}&{ - \frac{1}{3}}\\{ - \frac{1}{3}}&1&{ - \frac{1}{3}}\\{ - \frac{1}{3}}&{ - \frac{1}{3}}&1\end{array}} \right]\\ \ne I\end{array}\)

Thus, the product of the matrix and its transpose matrix is not equal to identity matrix. So, it is not orthogonal.

e).

Let us consider the first matrix\(E = \left( {\begin{align}{}{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\end{align}} \right)\)

So, the transpose of the matrix is \({E^T} = \left( {\begin{align}{}{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\end{align}} \right)\)

Now,

\(\begin{array}{c}A{A^T} = \left[ {\begin{array}{}{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\end{array}} \right] \times \left[ {\begin{array}{}{\frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{ - \frac{1}{2}}\end{array}} \right]\\ = \left[ {\begin{array}{}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}} \right]\\ = I\end{array}\)

Thus, the product of the matrix and its transpose matrix is equal to identity matrix. So, it is orthogonal.