# Question 7, Exercise 2.1

Solutions of Question 7 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 $A=\begin{bmatrix}1 & 0 & -1 & 2 \\3 & 1 & 2 & \quad 5 \\0 & -2 & 1 & 6\end{bmatrix}$ and $B=\begin{bmatrix} 2 & -1 & 3 & 1 \\1 & 3 & -1 & 4 \\3 & 1 & 2 & -1 \end{bmatrix}$. Then show that $( A+B )^t=A^t+B^t$.

Given $A=\left[ \begin{matrix}1 & 0 & -1 & 2 \\3 & 1 & 2 & \quad 5 \\0 & -2 & 1 & 6 \\\end{matrix} \right]$ and $B=\left[ \begin{matrix}2 & -1 & 3 & 1 \\1 & 3 & -1 & 4 \\3 & 1 & 2 & -1 \\ \end{matrix} \right]$. Then $$A^t=\left[ \begin{matrix}1 & 3 & 0 \\0 & 1 & -2 \\-1 & 2 & 1 \\2 & 5 & 6 \\ \end{matrix} \right]$$ and $$B^t=\left[ \begin{matrix}2 & 1 & 3 \\-1 & 3 & 1 \\3 & -1 & 2 \\1 & \,4 & -1 \\ \end{matrix} \right].$$ Now \begin{align}A+B=\left[ \begin{matrix}1 & 0 & -1 & 2 \\ 3 & 1 & 2 & \quad 5 \\ 0 & -2 & 1 & 6 \\\end{matrix} \right]+\left[ \begin{matrix}2 & -1 & 3 & 1 \\1 & 3 & -1 & 4 \\3 & 1 & 2 & -1 \\\end{matrix} \right]\\ &=\left[ \begin{matrix}1+2 & 0-1 & -1+3 & 2+1 \\3+1 & 1+3 & 2-1 & \quad 5+4 \\0+3 & -2+1 & 1+2 & 6-1 \\ \end{matrix} \right]\\ &=\left[ \begin{matrix}3 & -1 & 2 & 3 \\4 & 4 & 1 & \quad 9 \\3 & -1 & 3 & 5 \\\end{matrix} \right]\end{align} Now \begin{align}(A+B)^t=\left[ \begin{matrix}3 & 4 & 3 \\-1 & 4 & -1\\ 2 & 1 & 3 \\ 3 & 9 & 5 \\ \end{matrix} \right] ... (1)\end{align} Also \begin{align}A^t+B^t&=\left[ \begin{matrix}1 & 3 & 0 \\0 & 1 & -2 \\-1 & 2 & 1 \\2 & 5 & 6 \\ \end{matrix} \right]+\left[ \begin{matrix}2 & 1 & 3 \\-1 & 3 & 1 \\3 & -1 & 2 \\1 & \,4 & -1 \\ \end{matrix} \right] \\ &=\left[ \begin{matrix}1+2 & 3+1 & 0+3 \\0-1 & 1+3 & -2+1 \\-1+3 & 2-1 & 1+2 \\2+1 & 5+4 & 6-1 \\ \end{matrix} \right]\\ \implies A^t+B^t&=\left[ \begin{matrix}3 & 4 & 3 \\-1 & 4 & -1 \\2 & 1 & 3 \\ 3 & 9 & 5 \\ \end{matrix} \right]...(2) \end{align} From (1) and (2), we have $(A+B)^t=A^t+B^t.$