====== Question 9, Exercise 2.1 ======
Solutions of Question 9 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 9(i)=====
If $A=\begin{bmatrix}2 & -1 & 3  \\1 & \quad 0 & 1 \end{bmatrix},$
$B=\begin{bmatrix}1 & 2  \\2 & 2  \\ 3 & 0 \end{bmatrix}$, show that $( AB )^t=B^tA^t$.
====Solution====
$$A=\left[  \begin{matrix}
   2 & -1 & 3  \\
   1 & \quad 0 & 1  \\
\end{matrix}  \right],$$
$$B=\left[  \begin{matrix}
   1 & 2  \\
   2 & 2  \\
   3 & 0  \\
\end{matrix}  \right]$$
$$A^t=\left[  \begin{matrix}
   \quad2 & 1  \\
   -1 & 0  \\
   \quad 3 & 1  \\
\end{matrix}  \right]$$
$$B^t=\left[  \begin{matrix}
   1 & 2 & 3  \\
   2 & 2 & 0  \\
\end{matrix}  \right]$$
$$AB=\left[  \begin{matrix}
   2 & -1 & 3  \\
   1 & \quad 0 & 1  \\
\end{matrix}  \right]\left[ \begin{matrix}
   1 & 2  \\
   2 & 2  \\
   3 & 0  \\
\end{matrix}  \right]$$ 
$$AB=\left[ \begin{matrix}
   2-2+9 & 4-2+0  \\
   1+0+3 & 2+0+0  \\
\end{matrix} \right]$$
$$AB=\left[ \begin{matrix}
   9 & 2  \\
   4 & 2  \\
\end{matrix} \right]$$
$$( AB )^t=\left[ \begin{matrix}
   9 & 4  \\
   2 & 2  \\
\end{matrix} \right]$$
$$B^tA^t=\left[  \begin{matrix}
   1 & 2 & 3  \\
   2 & 2 & 0  \\
\end{matrix}  \right]\left[  \begin{matrix}
   \,\,\,2 & 1  \\
   -1 & 0  \\
   \quad 3 & 1  \\
\end{matrix}  \right]$$
$$B^tA^t=\left[ \begin{matrix}
   2-2+9 & 1+0+3  \\
   4-2+0 & 2+0+0  \\
\end{matrix} \right]$$
$$B^tA^t=\left[ \begin{matrix}
   9 & 4  \\
   2 & 2  \\
\end{matrix} \right]$$
$$( AB)^t=B^tA^t$$


=====Question 9(ii)=====
If $A= \begin{bmatrix}\quad 1 & 2 & 0  \\-1 & 1 & 4 \end{bmatrix}$, $B=\begin{bmatrix}1 & 1  \\2 & 3  \\1 & -2 \end{bmatrix}$, show that $( AB)^t=B^tA^t$.
====Solution====
$$A=\left[  \begin{matrix}
   \quad 1 & 2 & 0  \\
   -1 & 1 & 4  \\
\end{matrix}  \right]$$
$$B=\left[ \begin{matrix}
   1 & 1  \\
   2 & 3  \\
   1 & -2  \\
\end{matrix}  \right]$$
$${{A}^{t}}=\left[  \begin{matrix}
   1 & -1  \\
   2 & \quad 1  \\
   0 & \quad 4  \\
\end{matrix} \right]$$
$${{B}^{t}}=\left[  \begin{matrix}
   1 & 2 & \quad 1  \\
   1 & 3 & -2  \\
\end{matrix} \right]$$
$$AB=\left[  \begin{matrix}
   \quad 1 & 2 & 0  \\
   -1 & 1 & 4  \\
\end{matrix} \right]\left[  \begin{matrix}
   1 & 1  \\
   2 & 3  \\
   1 & -2  \\
\end{matrix}  \right]$$
$$AB=\left[ \begin{matrix}
   1+4+0 & 1+6+0  \\
   -1+2+4 & -1+3-8  \\
\end{matrix} \right]$$
$$AB=\left[ \begin{matrix}
   5 & 7  \\
   5 & -6  \\
\end{matrix} \right]$$
$${{\left( AB \right)}^{t}}=\left[ \begin{matrix}
   5 & 5  \\
   7 & -6  \\
\end{matrix} \right]$$
$${{B}^{t}}{{A}^{t}}=\left[ \begin{matrix}
   1 & 2 & \quad 1  \\
   1 & 3 & -2  \\
\end{matrix}  \right]\left[ \begin{matrix}
   1 & -1  \\
   2 & \quad 1  \\
   0 & \quad 4  \\
\end{matrix}  \right]$$
$${{B}^{t}}{{A}^{t}}=\left[ \begin{matrix}
   1+4+0 & -1+2+4  \\
   1+6+0 & -1+3-8  \\
\end{matrix} \right]$$
$${{B}^{t}}{{A}^{t}}=\left[ \begin{matrix}
   5 & 5  \\
   7 & -6  \\
\end{matrix} \right]$$
$${{\left( AB \right)}^{t}}={{B}^{t}}{{A}^{t}}$$



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