\displaystyle \textbf{Question 1:} 

\displaystyle \text{Let } A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}. \text{Verify that }

\displaystyle \text{(i) } (2A)^T = 2A^T                            \displaystyle \text{(ii) } (A + B)^T = A^T + B^T

\displaystyle \text{(iii) } (A - B)^T = A^T - B^T           \displaystyle \text{(iv) } (AB)^T = B^T A^T

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} \Rightarrow  A^{T} = \begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix}

\displaystyle B = \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix} \Rightarrow  B^{T} = \begin{bmatrix} 1 & 2 \\ 0 & -4 \end{bmatrix}

\displaystyle (i)\quad (2A)^{T} = 2A^{T}

\displaystyle \Rightarrow \left( 2 \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} \right)^{T} = 2 \begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix}

\displaystyle \Rightarrow \begin{bmatrix} 4 & -6 \\ -14 & 10 \end{bmatrix}^{T} = \begin{bmatrix} 4 & -14 \\ -6 & 10 \end{bmatrix}

\displaystyle \Rightarrow \begin{bmatrix} 4 & -14 \\ -6 & 10 \end{bmatrix} = \begin{bmatrix} 4 & -14 \\ -6 & 10 \end{bmatrix}

\displaystyle \therefore \text{LHS} = \text{RHS}

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\displaystyle (ii)\ (A + B)^T = A^T + B^T

\displaystyle \Rightarrow \left(\begin{bmatrix}2 & -3 \\-7 & 5\end{bmatrix}+\begin{bmatrix}1 & 0 \\2 & -4\end{bmatrix}\right)^T= \begin{bmatrix}2 & -7 \\-3 & 5\end{bmatrix}+\begin{bmatrix}1 & 2 \\0 & -4\end{bmatrix}

\displaystyle \Rightarrow\left(\begin{bmatrix}2+1 & -3+0 \\-7+2 & 5-4\end{bmatrix}\right)^T=\begin{bmatrix}2+1 & -7+2 \\-3+0 & 5-4 \end{bmatrix}

\displaystyle \Rightarrow\left(\begin{bmatrix}3 & -3 \\-5 & 1\end{bmatrix}\right)^T=\begin{bmatrix}3 & -5 \\-3 & 1\end{bmatrix}

\displaystyle \Rightarrow\begin{bmatrix}3 & -5 \\-3 & 1\end{bmatrix}=\begin{bmatrix}3 & -5 \\-3 & 1\end{bmatrix}

\displaystyle \therefore \text{LHS} = \text{RHS}

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\displaystyle\text{(iii) } (A-B)^T = A^T - B^T

\displaystyle\Rightarrow\left(\begin{bmatrix}2 & -3\\-7 & 5\end{bmatrix}-\begin{bmatrix}1 & 0\\2 & -4\end{bmatrix}\right)^T= \begin{bmatrix}2 & -7\\-3 & 5\end{bmatrix}-\begin{bmatrix}1 & 2\\0 & -4\end{bmatrix}

\displaystyle\Rightarrow\left(\begin{bmatrix}2-1 & -3-0\\-7-2 & 5+4\end{bmatrix}\right)^T=\begin{bmatrix}2-1 & -7-2\\-3-0 & 5+4 \end{bmatrix}

\displaystyle\Rightarrow\begin{bmatrix}1 & -3\\-9 & 9\end{bmatrix}^T=\begin{bmatrix}1 & -9\\-3 & 9\end{bmatrix}

\displaystyle\Rightarrow\begin{bmatrix}1 & -9\\-3 & 9\end{bmatrix}=\begin{bmatrix}1 & -9\\-3 & 9\end{bmatrix}

\displaystyle\therefore \ \text{LHS} = \text{RHS}

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\displaystyle\text{(iv)}\ (AB)^T = B^T A^T

\displaystyle\Rightarrow\left(\begin{bmatrix}2 & -3\\-7 & 5\end{bmatrix}\begin{bmatrix}1 & 0\\2 & -4\end{bmatrix}\right)^T= \begin{bmatrix}1 & 2\\0 & -4\end{bmatrix}\begin{bmatrix}2 & -7\\-3 & 5\end{bmatrix}

\displaystyle\Rightarrow\left(\begin{bmatrix}2-6 & 0+12\\-7+10 & 0-20\end{bmatrix}\right)^T=\begin{bmatrix}2-6 & -7+10\\ 0+12 & 0-20\end{bmatrix}

\displaystyle\Rightarrow\begin{bmatrix}-4 & 12\\3 & -20\end{bmatrix}^T=\begin{bmatrix}-4 & 3\\12 & -20\end{bmatrix}

\displaystyle\Rightarrow\begin{bmatrix}-4 & 3\\12 & -20\end{bmatrix}=\begin{bmatrix}-4 & 3\\12 & -20\end{bmatrix}

\displaystyle\therefore\ \text{LHS} = \text{RHS}

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\displaystyle \textbf{Question 2:} 

\displaystyle \text{If } A=\begin{bmatrix}3\\5\\2\end{bmatrix} \text{ and } B=\begin{bmatrix}1&0&4\end{bmatrix}, \text{ verify that } (AB)^T = B^T A^T. \hspace{2.0cm} \text{[CBSE 2002]}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A=\begin{bmatrix} 3\\5\\2 \end{bmatrix} \Rightarrow A^{T}=\begin{bmatrix} 3&5&2 \end{bmatrix}

\displaystyle B=\begin{bmatrix} 1&0&4\end{bmatrix} \Rightarrow B^{T}=\begin{bmatrix} 1\\0\\4\end{bmatrix}

\displaystyle \text{Now,}

\displaystyle (AB)^{T}=\begin{bmatrix}3&5&2 \\ 0&0&0 \\ 12&20&8 \end{bmatrix}\quad (1)

\displaystyle B^{T}A^{T}=\begin{bmatrix} 1 \\ 0 \\ 4 \end{bmatrix}\begin{bmatrix} 3&5&2 \end{bmatrix}

\displaystyle B^{T}A^{T}=\begin{bmatrix} 3&5&2 \\ 0&0&0 \\ 12&20&8 \end{bmatrix}\quad (2)

\displaystyle\therefore (AB)^{T}=B^{T}A^{T}\quad \text{[From (1) and (2)]}

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\displaystyle \textbf{Question 3:} 

\displaystyle \text{Let } A=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix} \text{ and } B=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}. \text{Find } A^T,\, B^T \text{ and verify that }

\displaystyle \text{(i) }(A+B)^T = A^T + B^T          \displaystyle \text{(ii) }(AB)^T = B^T A^T          \displaystyle \text{(iii) }(2A)^T = 2A^T 

\displaystyle \text{Answer:}

\displaystyle \text{(i) } A+B = \begin{bmatrix} 1+1 & -1+2 & 0+3 \\ 2+2 & 1+1 & 3+3\\ 1+0 & 2+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 3\\ 4 & 2 & 6\\ 1 & 3 & 2 \end{bmatrix}  

\displaystyle (A+B)^T = \begin{bmatrix} 2 & 4 & 1\\ 1 & 2 & 3\\ 3 & 6 & 2 \end{bmatrix} \quad \text{(1)}  

\displaystyle A^T+B^T = \begin{bmatrix} 1+1 & 2+2 & 1+0 \\ -1+2 & 1+1 & 2+1 \\ 0+3 & 3+3 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 1\\ 1 & 2 & 3\\ 3 & 6 & 2 \end{bmatrix} \quad \text{(2)}  

\displaystyle \therefore\ (A+B)^T = A^T + B^T \quad \text{[From (1) and (2)]} 

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\displaystyle \text{(ii) } AB=\begin{bmatrix} 3\\5\\2 \end{bmatrix} \begin{bmatrix} 1&0&4 \end{bmatrix}

\displaystyle AB=\begin{bmatrix} 3&0&12 \\ 5&0&20\\ 2&0&8 \end{bmatrix}

\displaystyle (AB)^{T}= \begin{bmatrix} 3&5&2\\ 0&0&0 \\ 12&20&8 \end{bmatrix} \quad (1)

\displaystyle B^{T}A^{T}= \begin{bmatrix} 1\\0\\4 \end{bmatrix} \begin{bmatrix} 3&5&2 \end{bmatrix}

\displaystyle B^{T}A^{T}= \begin{bmatrix} 3&5&2 \\ 0&0&0 \\ 12&20&8 \end{bmatrix} \quad (2)

\displaystyle \therefore (AB)^{T}=B^{T}A^{T} \quad \text{[From (1) and (2)]} 

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\displaystyle \text{(iii) } 2A = 2 \begin{bmatrix} 1 & -1 & 0\\ 2 & 1 & 3\\ 1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 0\\ 4 & 2 & 6\\ 2 & 4 & 2 \end{bmatrix}

\displaystyle (2A)^T = \begin{bmatrix} 2 & 4 & 2\\ -2 & 2 & 4\\ 0 & 6 & 2 \end{bmatrix} \quad \ldots (1) 

\displaystyle A^T = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 2 \\ 0 & 3 & 1 \end{bmatrix}

\displaystyle 2A^T = 2 \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 2 \\ 0 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 2 \\ -2 & 2 & 4 \\ 0 & 6 & 2 \end{bmatrix} \quad \ldots (2) 

\displaystyle \therefore\ (2A)^T = 2A^T \quad \text{[From (1) and (2)]} 

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\displaystyle \textbf{Question 4:} 

\displaystyle \text{If } A = \begin{bmatrix} -2\\ 4\\ 5 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}, \text{ verify that } (AB)^T = B^T A^T.

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A=\begin{bmatrix}-2\\4\\5\end{bmatrix},\quad A^{T}=\begin{bmatrix}-2&4&5\end{bmatrix}

\displaystyle B=\begin{bmatrix}1&3&-6\end{bmatrix},\quad B^{T}=\begin{bmatrix}1\\3\\-6\end{bmatrix}

\displaystyle \text{Now, }  AB=\begin{bmatrix}-2\\4\\5\end{bmatrix} \begin{bmatrix}1&3&-6\end{bmatrix}

\displaystyle \Rightarrow AB=\begin{bmatrix} -2 & -6 & 12\\ 4 & 12 & -24\\ 5 & 15 & -30 \end{bmatrix}

\displaystyle \Rightarrow (AB)^T= \begin{bmatrix} -2 & 4 & 5\\ -6 & 12 & 15\\ 12 & -24 & -30 \end{bmatrix} \quad (1)

\displaystyle \text{Also, } B^TA^T= \begin{bmatrix}1\\3\\-6\end{bmatrix} \begin{bmatrix}-2&4&5\end{bmatrix}

\displaystyle \Rightarrow B^TA^T= \begin{bmatrix} -2 & 4 & 5\\ -6 & 12 & 15\\ 12 & -24 & -30 \end{bmatrix} \quad (2)

\displaystyle \therefore\ (AB)^T = B^TA^T \ \text{[From (1) and (2)]}

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\displaystyle \textbf{Question 5:} 

\displaystyle \text{If } A=\begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix} \text{ and }  B=\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}, \text{ find } (AB)^T.

\displaystyle \text{Answer:}

\displaystyle \text{Here,}\\[6pt]AB=\begin{bmatrix}2 & 4 & -1\\-1 & 0 & 2\end{bmatrix}\begin{bmatrix}3 & 4\\-1 & 2\\2 & 1\end{bmatrix}

\displaystyle \Rightarrow AB=\begin{bmatrix}6-4-2 & 8+8-1\\-3-0+4 & -4+0+2\end{bmatrix}\\[8pt]\Rightarrow AB=\begin{bmatrix}0 & 15\\1 & -2\end{bmatrix}

\displaystyle \Rightarrow (AB)^T=\begin{bmatrix}0 & 1\\15 & -2\end{bmatrix}

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\displaystyle \textbf{Question 6:} 

\displaystyle \text{(i) If } A=\begin{bmatrix}2 & 1 & 3 \\ 4 & 1 & 0\end{bmatrix} \text{ and } B=\begin{bmatrix}1 & -1 \\ 0 & 2 \\ 5 & 0\end{bmatrix}, \text{ verify that } (AB)^T = B^T A^T.

\displaystyle \text{(ii) For the matrices } A \text{ and } B, \text{ verify that } (AB)^T = B^T A^T, \text{ where}  A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, \quad \\ B = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{(i)}\\[6pt] \text{Given: } A=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}, \quad A^T=\begin{bmatrix} 2&4 \\ 1&1 \\ 3&0 \end{bmatrix}

\displaystyle B=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}, \quad B^T=\begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix}

\displaystyle \text{Now, } AB=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix} \begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}

\displaystyle \Rightarrow AB= \begin{bmatrix} 2+0+15 & -2+2+0\\ 4+0+0 & -4+2+0 \end{bmatrix} = \begin{bmatrix} 17&4\\ 0&-2 \end{bmatrix}

\displaystyle \Rightarrow (AB)^T= \begin{bmatrix} 17&4\\ 0&-2 \end{bmatrix} \qquad (1)

\displaystyle \text{Now, } B^TA^T= \begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix} \begin{bmatrix}2&4\\1&1\\3&0\end{bmatrix}

\displaystyle \Rightarrow B^TA^T= \begin{bmatrix} 2+0+15 & 4+0+0\\ -2+2+0 & -4+2+0 \end{bmatrix} = \begin{bmatrix} 17&4\\ 0&-2 \end{bmatrix} \qquad (2)

\displaystyle \therefore\ (AB)^T=B^TA^T \quad \text{[From (1) and (2)]}

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\displaystyle \text{(ii) Given: } A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ \ \ \quad A^{T}=\begin{bmatrix}1&2\\3&4\end{bmatrix}

\displaystyle B=\begin{bmatrix}1&4\\2&5\end{bmatrix},\ \ \ \quad B^{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}

\displaystyle \text{Now,  } AB=\begin{bmatrix}1&3\\2&4\end{bmatrix} \begin{bmatrix}1&4\\2&5\end{bmatrix} =\begin{bmatrix} 1+6 & 4+15\\ 2+8 & 8+20 \end{bmatrix} =\begin{bmatrix}7&19\\10&28\end{bmatrix}

\displaystyle \Rightarrow (AB)^{T} =\begin{bmatrix}7&10\\19&28\end{bmatrix}

\displaystyle  \text{   Also, } B^{T}A^{T} =\begin{bmatrix}1&2\\4&5\end{bmatrix} \begin{bmatrix}1&2\\3&4\end{bmatrix} =\begin{bmatrix} 1+6 & 2+8\\ 4+15 & 8+20 \end{bmatrix} =\begin{bmatrix}7&10\\19&28\end{bmatrix}

\displaystyle  \therefore (AB)^{T}=B^{T}A^{T} \quad [\text{From (1) and (2)}]

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\displaystyle \textbf{Question 7:} 

\displaystyle \text{If } A^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \text{ and }  B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}, \text{ find } A^T - B^T. \hspace{2.0cm} \text{[CBSE2012]}

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A^T=\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}

\displaystyle B^T=\begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}

\displaystyle \text{Now, } A^T-B^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}

\displaystyle = \begin{bmatrix} 3+1 & 4-1 \\ -1-2 & 2-2 \\ 0-1 & 1-3 \end{bmatrix}

\displaystyle = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}

\displaystyle \therefore\; A^T-B^T= \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}

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\displaystyle \textbf{Question 8:} 

\displaystyle \text{If } A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}, \text{ then verify that } A^T A = I_2.

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \\[8pt] A^T = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \\[10pt] \text{Now,} \\[6pt] A^T A = I_2 \\[10pt] \text{Consider: } LHS = A^T A \\[8pt] = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \\[10pt] = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \\[10pt] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = RHS \\[8pt] \text{Hence proved.}

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\displaystyle \textbf{Question 9:} 

\displaystyle \text{If } A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}, \text{ verify that } A^T A = I_2.

\displaystyle \text{Answer:}

\displaystyle \text{Given: } A=\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix}

\displaystyle A^{T}=\begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix}

\displaystyle \text{Now,}

\displaystyle A^{T}A= \begin{bmatrix} \sin\alpha & -\cos\alpha\\ \cos\alpha & \sin\alpha \end{bmatrix} \begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos\alpha & \sin\alpha \end{bmatrix}

\displaystyle = \begin{bmatrix} (\sin\alpha)(\sin\alpha)+(-\cos\alpha)(-\cos\alpha) & (\sin\alpha)(\cos\alpha)+(-\cos\alpha)(\sin\alpha)\\ (\cos\alpha)(\sin\alpha)+(\sin\alpha)(-\cos\alpha) & (\cos\alpha)(\cos\alpha)+(\sin\alpha)(\sin\alpha) \end{bmatrix}

\displaystyle = \begin{bmatrix} \sin^{2}\alpha+\cos^{2}\alpha & \sin\alpha\cos\alpha-\sin\alpha\cos\alpha\\ \sin\alpha\cos\alpha-\sin\alpha\cos\alpha & \cos^{2}\alpha+\sin^{2}\alpha \end{bmatrix}

\displaystyle = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = I

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\displaystyle \textbf{Question 10:}  \text{If } l_i, m_i, n_i;\ i = 1,2,3 \text{ denote the direction cosines of three mutually} \\ \text{perpendicular vectors in space, prove that }  AA^T = I, \text{ where} \\ A = \begin{bmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end{bmatrix}

\displaystyle \text{Answer:}

\displaystyle \text{Given : } (l_1,m_1,n_1), (l_2,m_2,n_2), (l_3,m_3,n_3) \\ \text{ are the direction cosines of three mutually perpendicular vectors in space.}

\displaystyle l_1^2+m_1^2+n_1^2=1, \quad l_2^2+m_2^2+n_2^2=1, \quad l_3^2+m_3^2+n_3^2=1 \qquad (i)

\displaystyle l_1l_2+m_1m_2+n_1n_2=0, \quad l_2l_3+m_2m_3+n_2n_3=0, \quad l_3l_1+m_3m_1+n_3n_1=0 \qquad (ii)

\displaystyle \text{Let } A=\begin{bmatrix} l_1 & m_1 & n_1\\ l_2 & m_2 & n_2\\ l_3 & m_3 & n_3 \end{bmatrix}.

\displaystyle 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}.

\displaystyle 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}.

\displaystyle \text{Using (i) and (ii),}

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

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