Let $A=\left[\begin{array}{rrr}1 & -1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right] .$ Find $A^{T}, B^{T}$ and verify that
(i) $(A+B)^{T}=A^{\top}+B^{\top}$
(ii) $(A B)^{T}=B^{T} A^{T}$
(iii) $(2 A)^{\top}=2 A^{\top}$.
Given : $A=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$A^{T}=\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 2 \\ 0 & 3 & 1\end{array}\right]$ and $B^{T}=\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1\end{array}\right]$
(i)
$A+B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{array}\right]+\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc}1+1 & -1+2 & 0+3 \\ 2+2 & 1+1 & 3+3 \\ 1+0 & 2+1 & 1+1\end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{lll}2 & 1 & 3 \\ 4 & 2 & 6 \\ 1 & 3 & 2\end{array}\right]$
$\Rightarrow(A+B)^{T}=\left[\begin{array}{lll}2 & 4 & 1 \\ 1 & 2 & 3 \\ 3 & 6 & 2\end{array}\right]$ ...(1)
Now,
$A^{T}+B^{T}=\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 2 \\ 0 & 3 & 1\end{array}\right]+\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1\end{array}\right]$
$\Rightarrow A^{T}+B^{T}=\left[\begin{array}{cc}1+1 & 2+2 & 1+0 \\ -1+2 & 1+1 & 2+1 \\ 0+3 & 3+3 & 1+1\end{array}\right]$
$\Rightarrow A^{T}+B^{T}=\left[\begin{array}{lll}2 & 4 & 1 \\ 1 & 2 & 3 \\ 3 & 6 & 2\end{array}\right]$ ...(2)
$\Rightarrow(A+B)^{T}=A^{T}+B^{T}$ [From eqs. (1) and (2)]
(ii)
$A B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{lll}1-2+0 & 2-1+0 & 3-3+0 \\ 2+2+0 & 4+1+3 & 6+3+3 \\ 1+4+0 & 2+2+1 & 3+6+1\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc}-1 & 1 & 0 \\ 4 & 8 & 12 \\ 5 & 5 & 10\end{array}\right]$
$\Rightarrow(A B)^{T}=\left[\begin{array}{ccc}-1 & 4 & 5 \\ 1 & 8 & 5 \\ 0 & 12 & 10\end{array}\right]$ ...(1)
Now,
$B^{T} A^{T}=\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 2 \\ 0 & 3 & 1\end{array}\right]$
$\Rightarrow B^{T} A^{T}=\left[\begin{array}{lll}1-2+0 & 2+2+0 & 1+4+0 \\ 2-1+0 & 4+1+3 & 2+2+1 \\ 3-3+0 & 6+3+3 & 3+6+1\end{array}\right]$
$\Rightarrow B^{T} A^{T}=\left[\begin{array}{ccc}-1 & 4 & 5 \\ 1 & 8 & 5 \\ 0 & 12 & 10\end{array}\right]$ ...(2)
$\Rightarrow(A B)^{T}=B^{T} A^{T}$ [From eqs. (1) and (2)]
(iii)
$2 A=2\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 1\end{array}\right]$
$\Rightarrow 2 A=\left[\begin{array}{ccc}2 & -2 & 0 \\ 4 & 2 & 6 \\ 2 & 4 & 2\end{array}\right]$
$\Rightarrow(2 A)^{T}=\left[\begin{array}{ccc}2 & 4 & 2 \\ -2 & 2 & 4 \\ 0 & 6 & 2\end{array}\right]$ $\ldots(1)$
Now,
$2 A^{T}=2\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 2 \\ 0 & 3 & 1\end{array}\right]$
$\Rightarrow 2 A^{T}=\left[\begin{array}{ccc}2 & 4 & 2 \\ -2 & 2 & 4 \\ 0 & 6 & 2\end{array}\right]$ ...(2)
$\Rightarrow(2 A)^{T}=2 A^{T}$ [From eqs. (1) and (2)]