====== 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.

=====Question 7=====
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 $( A+B )^t=A^t+B^t$.
====Solution====
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
$(A+B)^t=A^t+B^t.$


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