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  • Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 10

Need solution for RD Sharma Maths Class 12 Chapter Algebra of Matrices Excercise 4.4 Question 10

Answers (1)

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, theA^{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.

Posted by

infoexpert27

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