Prove that AAT = I
Given:
\[ A = \begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{bmatrix} \]
where \((l_i, m_i, n_i)\) are direction cosines of three mutually perpendicular unit vectors.
Properties Used:
\[ l_i^2 + m_i^2 + n_i^2 = 1 \]
\[ l_1l_2 + m_1m_2 + n_1n_2 = 0 \] (similarly for other pairs)
Step 1: Find AT
\[ A^T = \begin{bmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{bmatrix} \]
Step 2: Compute AAT
\[ AA^T = \begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{bmatrix} \begin{bmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end{bmatrix} \]
\[ AA^T = \begin{bmatrix} l_1^2 + m_1^2 + n_1^2 & l_1l_2 + m_1m_2 + n_1n_2 & l_1l_3 + m_1m_3 + n_1n_3 \\ l_2l_1 + m_2m_1 + n_2n_1 & l_2^2 + m_2^2 + n_2^2 & l_2l_3 + m_2m_3 + n_2n_3 \\ l_3l_1 + m_3m_1 + n_3n_1 & l_3l_2 + m_3m_2 + n_3n_2 & l_3^2 + m_3^2 + n_3^2 \end{bmatrix} \]
Step 3: Apply Properties
\[ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Conclusion:
\[ AA^T = I \]
Hence Proved. Matrix A is orthogonal.