\displaystyle \textbf{Question 1: }\text{Let }A=\begin{bmatrix}2&-3\\-7&5\end{bmatrix}\text{ and }B=\begin{bmatrix}1&0\\2&-4\end{bmatrix}.
\displaystyle \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^TA^T
\displaystyle \text{Answer:}
\displaystyle 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 \text{(i) LHS}=(2A)^T
\displaystyle =\left(2\begin{bmatrix}2&-3\\-7&5\end{bmatrix}\right)^T
\displaystyle =\begin{bmatrix}4&-6\\-14&10\end{bmatrix}^T
\displaystyle =\begin{bmatrix}4&-14\\-6&10\end{bmatrix}
\displaystyle \text{RHS}=2A^T
\displaystyle =2\begin{bmatrix}2&-7\\-3&5\end{bmatrix}
\displaystyle =\begin{bmatrix}4&-14\\-6&10\end{bmatrix}
\displaystyle \therefore\ (2A)^T=2A^T.
\displaystyle \\

\displaystyle \text{(ii) LHS}=(A+B)^T
\displaystyle =\left(\begin{bmatrix}2&-3\\-7&5\end{bmatrix}+\begin{bmatrix}1&0\\2&-4\end{bmatrix}\right)^T
\displaystyle =\begin{bmatrix}3&-3\\-5&1\end{bmatrix}^T
\displaystyle =\begin{bmatrix}3&-5\\-3&1\end{bmatrix}
\displaystyle \text{RHS}=A^T+B^T
\displaystyle =\begin{bmatrix}2&-7\\-3&5\end{bmatrix}+\begin{bmatrix}1&2\\0&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}3&-5\\-3&1\end{bmatrix}
\displaystyle \therefore\ (A+B)^T=A^T+B^T.
\displaystyle \\

\displaystyle \text{(iii) LHS}=(A-B)^T
\displaystyle =\left(\begin{bmatrix}2&-3\\-7&5\end{bmatrix}-\begin{bmatrix}1&0\\2&-4\end{bmatrix}\right)^T
\displaystyle =\begin{bmatrix}1&-3\\-9&9\end{bmatrix}^T
\displaystyle =\begin{bmatrix}1&-9\\-3&9\end{bmatrix}
\displaystyle \text{RHS}=A^T-B^T
\displaystyle =\begin{bmatrix}2&-7\\-3&5\end{bmatrix}-\begin{bmatrix}1&2\\0&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-9\\-3&9\end{bmatrix}
\displaystyle \therefore\ (A-B)^T=A^T-B^T.
\displaystyle \\

\displaystyle \text{(iv) LHS}=(AB)^T
\displaystyle =\left(\begin{bmatrix}2&-3\\-7&5\end{bmatrix}\begin{bmatrix}1&0\\2&-4\end{bmatrix}\right)^T
\displaystyle =\begin{bmatrix}2-6&0+12\\-7+10&0-20\end{bmatrix}^T
\displaystyle =\begin{bmatrix}-4&12\\3&-20\end{bmatrix}^T
\displaystyle =\begin{bmatrix}-4&3\\12&-20\end{bmatrix}
\displaystyle \text{RHS}=B^TA^T
\displaystyle =\begin{bmatrix}1&2\\0&-4\end{bmatrix}\begin{bmatrix}2&-7\\-3&5\end{bmatrix}
\displaystyle =\begin{bmatrix}2-6&-7+10\\0+12&0-20\end{bmatrix}
\displaystyle =\begin{bmatrix}-4&3\\12&-20\end{bmatrix}
\displaystyle \therefore\ (AB)^T=B^TA^T.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }A=\begin{bmatrix}3\\5\\2\end{bmatrix}\text{ and }B=\begin{bmatrix}1&0&4\end{bmatrix},
\displaystyle \text{verify that }(AB)^T=B^TA^T.\qquad[\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}3\\5\\2\end{bmatrix},\qquad B=\begin{bmatrix}1&0&4\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}3&5&2\end{bmatrix},\qquad B^T=\begin{bmatrix}1\\0\\4\end{bmatrix}
\displaystyle AB=\begin{bmatrix}3\\5\\2\end{bmatrix}\begin{bmatrix}1&0&4\end{bmatrix}
\displaystyle =\begin{bmatrix}3&0&12\\5&0&20\\2&0&8\end{bmatrix}
\displaystyle \therefore\ (AB)^T=\begin{bmatrix}3&5&2\\0&0&0\\12&20&8\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{Also,}
\displaystyle B^TA^T=\begin{bmatrix}1\\0\\4\end{bmatrix}\begin{bmatrix}3&5&2\end{bmatrix}
\displaystyle =\begin{bmatrix}3&5&2\\0&0&0\\12&20&8\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \text{From (i) and (ii), we observe that }(AB)^T=B^TA^T.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }A=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\text{ and}
\displaystyle 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^TA^T
\displaystyle \text{(iii) }(2A)^T=2A^T
\displaystyle \text{Answer:}
\displaystyle A^T=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix},\qquad B^T=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}
\displaystyle \text{(i) LHS}=(A+B)^T
\displaystyle =\left(\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}+\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}\right)^T
\displaystyle =\begin{bmatrix}2&1&3\\4&2&6\\1&3&2\end{bmatrix}^T
\displaystyle =\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{RHS}=A^T+B^T
\displaystyle =\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}+\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}
\displaystyle =\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \therefore\ (A+B)^T=A^T+B^T.
\displaystyle \\
\displaystyle \text{(ii) }AB=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&1&0\\4&8&12\\5&5&10\end{bmatrix}
\displaystyle \therefore\ (AB)^T=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{Also,}
\displaystyle B^TA^T=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \therefore\ (AB)^T=B^TA^T.
\displaystyle \\
\displaystyle \text{(iii) LHS}=(2A)^T
\displaystyle =\left(2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\right)^T
\displaystyle =\begin{bmatrix}2&-2&0\\4&2&6\\2&4&2\end{bmatrix}^T
\displaystyle =\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{RHS}=2A^T
\displaystyle =2\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}
\displaystyle =\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \therefore\ (2A)^T=2A^T.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }A=\begin{bmatrix}-2\\4\\5\end{bmatrix}\text{ and }B=\begin{bmatrix}1&3&-6\end{bmatrix},
\displaystyle \text{verify that }(AB)^T=B^TA^T.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}-2\\4\\5\end{bmatrix},\qquad B=\begin{bmatrix}1&3&-6\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}-2&4&5\end{bmatrix},\qquad B^T=\begin{bmatrix}1\\3\\-6\end{bmatrix}
\displaystyle AB=\begin{bmatrix}-2\\4\\5\end{bmatrix}\begin{bmatrix}1&3&-6\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&-6&12\\4&12&-24\\5&15&-30\end{bmatrix}
\displaystyle \therefore\ (AB)^T=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{Also,}
\displaystyle B^TA^T=\begin{bmatrix}1\\3\\-6\end{bmatrix}\begin{bmatrix}-2&4&5\end{bmatrix}
\displaystyle =\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \text{From (i) and (ii), we observe that }(AB)^T=B^TA^T.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix}\text{ and }
\displaystyle B=\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix},\text{ find }(AB)^T.
\displaystyle \text{Answer:}
\displaystyle AB=\begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix}\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}
\displaystyle =\begin{bmatrix}6-4-2&8+8-1\\-3+0+4&-4+0+2\end{bmatrix}
\displaystyle =\begin{bmatrix}0&15\\1&-2\end{bmatrix}
\displaystyle \therefore\ (AB)^T=\begin{bmatrix}0&1\\15&-2\end{bmatrix}
\displaystyle \\

\displaystyle \textbf{Question 6:}
\displaystyle \text{(i) If }A=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}\text{ and }
\displaystyle B=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix},\text{ verify that }(AB)^T=B^TA^T.
\displaystyle \text{(ii) For the matrices }A\text{ and }B,\text{ verify that }(AB)^T=B^TA^T,
\displaystyle \text{where }A=\begin{bmatrix}1&3\\2&4\end{bmatrix}\text{ and }B=\begin{bmatrix}1&4\\2&5\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }A=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}\Rightarrow A^T=\begin{bmatrix}2&4\\1&1\\3&0\end{bmatrix}
\displaystyle B=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}\Rightarrow B^T=\begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix}
\displaystyle AB=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}
\displaystyle =\begin{bmatrix}2+0+15&-2+2+0\\4+0+0&-4+2+0\end{bmatrix}
\displaystyle =\begin{bmatrix}17&0\\4&-2\end{bmatrix}
\displaystyle \therefore\ (AB)^T=\begin{bmatrix}17&4\\0&-2\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{Also,}
\displaystyle B^TA^T=\begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix}\begin{bmatrix}2&4\\1&1\\3&0\end{bmatrix}
\displaystyle =\begin{bmatrix}2+0+15&4+0+0\\-2+2+0&-4+2+0\end{bmatrix}
\displaystyle =\begin{bmatrix}17&4\\0&-2\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \text{From (i) and (ii), we observe that }(AB)^T=B^TA^T.
\displaystyle \\

\displaystyle \text{(ii) }A=\begin{bmatrix}1&3\\2&4\end{bmatrix}\Rightarrow A^T=\begin{bmatrix}1&2\\3&4\end{bmatrix}
\displaystyle B=\begin{bmatrix}1&4\\2&5\end{bmatrix}\Rightarrow B^T=\begin{bmatrix}1&2\\4&5\end{bmatrix}
\displaystyle AB=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&4\\2&5\end{bmatrix}
\displaystyle =\begin{bmatrix}1+6&4+15\\2+8&8+20\end{bmatrix}
\displaystyle =\begin{bmatrix}7&19\\10&28\end{bmatrix}
\displaystyle \therefore\ (AB)^T=\begin{bmatrix}7&10\\19&28\end{bmatrix}\qquad\cdots\text{(i)}
\displaystyle \text{Also,}
\displaystyle B^TA^T=\begin{bmatrix}1&2\\4&5\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}
\displaystyle =\begin{bmatrix}1+6&2+8\\4+15&8+20\end{bmatrix}
\displaystyle =\begin{bmatrix}7&10\\19&28\end{bmatrix}\qquad\cdots\text{(ii)}
\displaystyle \text{From (i) and (ii), we observe that }(AB)^T=B^TA^T.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }A^T=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}\text{ and }
\displaystyle B=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix},\text{ find }A^T-B^T.\qquad[\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{Given }A^T=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix},\qquad B=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}
\displaystyle \therefore\ B^T=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}
\displaystyle 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}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix},
\displaystyle \text{verify that }A^TA=I_2.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}
\displaystyle A^TA=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}
\displaystyle =\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}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =I_2
\displaystyle \therefore\ A^TA=I_2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }A=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix},
\displaystyle \text{verify that }A^TA=I_2.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{bmatrix}
\displaystyle \therefore\ A^T=\begin{bmatrix}\sin\alpha&-\cos\alpha\\\cos\alpha&\sin\alpha\end{bmatrix}
\displaystyle A^TA=\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^2\alpha+\cos^2\alpha&\sin\alpha\cos\alpha-\cos\alpha\sin\alpha\\\cos\alpha\sin\alpha-\sin\alpha\cos\alpha&\cos^2\alpha+\sin^2\alpha\end{bmatrix}
\displaystyle =\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =I_2
\displaystyle \therefore\ A^TA=I_2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }l_i,m_i,n_i;\ i=1,2,3\text{ denote the direction cosines}
\displaystyle \text{of three mutually perpendicular vectors in space, prove that }AA^T=I_3,
\displaystyle \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{Since }(l_1,m_1,n_1),(l_2,m_2,n_2)\text{ and }(l_3,m_3,n_3)
\displaystyle \text{are the direction cosines of three mutually perpendicular vectors,}
\displaystyle l_1^2+m_1^2+n_1^2=1,\quad l_2^2+m_2^2+n_2^2=1,
\displaystyle l_3^2+m_3^2+n_3^2=1\qquad\cdots\text{(i)}
\displaystyle l_1l_2+m_1m_2+n_1n_2=0,\quad l_2l_3+m_2m_3+n_2n_3=0,
\displaystyle l_3l_1+m_3m_1+n_3n_1=0\qquad\cdots\text{(ii)}
\displaystyle A=\begin{bmatrix}l_1&m_1&n_1\\l_2&m_2&n_2\\l_3&m_3&n_3\end{bmatrix}
\displaystyle \therefore\ A^T=\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&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 =\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_3
\displaystyle \therefore\ AA^T=I_3.
\displaystyle \\

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