Let

Question:

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) $(2 A)^{\top}=2 A^{\top}$.

Solution:

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)]

 

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