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Answer: $AA^{T}=I$

Given: If $l_{i},m_{i},n_{i},i=1,2,3$ denotes direction cosines of 3 mutually ⊥ vertices

$A=\begin{bmatrix} l_{1} &m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3} \end{bmatrix}$

Hint: I refers to identity function, the$A^{T}$ of matrix $A$ can be obtained by reflecting the elements along its main diagonal.

Solution:

Given$\left ( l_{1}, m_{1}, n_{1} \right ),\left ( l_{2}, m_{2}, n_{2}\right ), (l_{3}, m_{3}, n_{3})$ are the direction cosines of 3 mutually + vectors in space.

$\begin{Bmatrix} l_{1}^{2}+ m_{1}^{2}+ n_{1}^{2}=1 \\ l_{2}^{2}+ m_{2}^{2}+ n_{2}^{2}=1\\ l_{3}^{2}+m_{3}^{2}+ n_{3}^{2} =1 \end{Bmatrix}$                                                                                                 ……. (1)

$\begin{Bmatrix} l_{1}l_{2}+ m_{1}m_{2}+ n_{1}n_{2}=0 \\ l_{2}l_{3}+ m_{2}m_{3}+ n_{2}n_{3}=0\\ l_{3}l_{1}+m_{3}m_{1}+ n_{3}n_{1} =0 \end{Bmatrix}$                                                                                      ……….. (2)

Let, $A=\begin{bmatrix} l_{1} &m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3} \end{bmatrix}$$,A^{T}=\begin{bmatrix} l_{1} &l_{2} &l_{3} \\ m_{1} & m_{2} & m_{3}\\ n_{1} & n_{2} & n_{3} \end{bmatrix}$

$AA^{T}=\begin{bmatrix} l_{1} &m_{1} &n_{1} \\ l_{2} & m_{2} & n_{2}\\ l_{3} & m_{3} & n_{3} \end{bmatrix}\begin{bmatrix} l_{1} &l_{2} &l_{3} \\ m_{1} & m_{2} & m_{3}\\ n_{1} & n_{2} & n_{3} \end{bmatrix}$

$AA^{T}=\begin{bmatrix} l_{1}^{2}+m_{1}^{2} +n_{1}^{2} & l_{1} l_{2}+m_{1}m_{2}+n_{1}n_{2} & l_{3}l_{1}+m_{3}m_{1}+n_{3}n_{1}\\ l_{1} l_{2}+m_{1}m_{2}+n_{1}n_{2} & l_{2}^{2}+m_{2}^{2} +n_{2}^{2} &l_{2}l_{3}+m_{2}m_{3}+n_{2}n_{3} \\ l_{3}l_{1}+m_{3}m_{1}+n_{3}n_{1} &l_{2}l_{3}+m_{2}m_{3}+n_{2}n_{3} & l_{3}^{2}+m_{3}^{2}+n_{3}^{2} \end{bmatrix}$

From (1) & (2) we get

$AA^{T}=\begin{bmatrix} 1 & 0 &0 \\ 0& 1 & 0\\ 0 &0 &1 \end{bmatrix}=I$

Hence, proved.

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