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.

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 $( AB )^t=B^tA^t$.

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

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 $( AB)^t=B^tA^t$.

$$A=\left[ \begin{matrix} \quad 1 & 2 & 0 \\ -1 & 1 & 4 \\ \end{matrix} \right]$$ $$B=\left[ \begin{matrix} 1 & 1 \\ 2 & 3 \\ 1 & -2 \\ \end{matrix} \right]$$ $${{A}^{t}}=\left[ \begin{matrix} 1 & -1 \\ 2 & \quad 1 \\ 0 & \quad 4 \\ \end{matrix} \right]$$ $${{B}^{t}}=\left[ \begin{matrix} 1 & 2 & \quad 1 \\ 1 & 3 & -2 \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} \quad 1 & 2 & 0 \\ -1 & 1 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 \\ 2 & 3 \\ 1 & -2 \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 1+4+0 & 1+6+0 \\ -1+2+4 & -1+3-8 \\ \end{matrix} \right]$$ $$AB=\left[ \begin{matrix} 5 & 7 \\ 5 & -6 \\ \end{matrix} \right]$$ $${{\left( AB \right)}^{t}}=\left[ \begin{matrix} 5 & 5 \\ 7 & -6 \\ \end{matrix} \right]$$ $${{B}^{t}}{{A}^{t}}=\left[ \begin{matrix} 1 & 2 & \quad 1 \\ 1 & 3 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 \\ 2 & \quad 1 \\ 0 & \quad 4 \\ \end{matrix} \right]$$ $${{B}^{t}}{{A}^{t}}=\left[ \begin{matrix} 1+4+0 & -1+2+4 \\ 1+6+0 & -1+3-8 \\ \end{matrix} \right]$$ $${{B}^{t}}{{A}^{t}}=\left[ \begin{matrix} 5 & 5 \\ 7 & -6 \\ \end{matrix} \right]$$ $${{\left( AB \right)}^{t}}={{B}^{t}}{{A}^{t}}$$

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