Let

Question:

Let $A=\left[\begin{array}{rr}2 & -3 \\ -7 & 5\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & 0 \\ 2 & -4\end{array}\right]$, verify that

(i) $(2 A)^{T}=2 A^{T}$

(ii) $(A+B)^{T}=A^{T}+B^{T}$

(iii) $(A-B)^{\top}=A^{\top}-B^{\top}$

(iv) $(A B)^{T}=B^{T} A^{\top}$

Solution:

Given : $A=\left[\begin{array}{cc}2 & -3 \\ -7 & 5\end{array}\right]$

$A^{T}=\left[\begin{array}{cc}2 & -7 \\ -3 & 5\end{array}\right]$

$B=\left[\begin{array}{cc}1 & 0 \\ 2 & -4\end{array}\right]$

$B^{T}=\left[\begin{array}{cc}1 & 2 \\ 0 & -4\end{array}\right]$

(i)

$(2 A)^{T}=2 A^{T}$

$\Rightarrow\left(2\left[\begin{array}{cc}2 & -3 \\ -7 & 5\end{array}\right]\right)^{T}=2\left[\begin{array}{cc}2 & -7 \\ -3 & 5\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}4 & -6 \\ -14 & 10\end{array}\right]^{T}=\left[\begin{array}{cc}4 & -14 \\ -6 & 10\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}4 & -14 \\ -6 & 10\end{array}\right]=\left[\begin{array}{cc}4 & -14 \\ -6 & 10\end{array}\right]$

$\therefore$ LHS $=$ RHS

(iii) $(A-B)^{T}=A^{T}-B^{T}$

$\Rightarrow\left(\left[\begin{array}{cc}2 & -3 \\ -7 & 5\end{array}\right]-\left[\begin{array}{cc}1 & 0 \\ 2 & -4\end{array}\right]\right)^{T}=\left[\begin{array}{cc}2 & -7 \\ -3 & 5\end{array}\right]-\left[\begin{array}{cc}1 & 2 \\ 0 & -4\end{array}\right]$

$\Rightarrow\left(\left[\begin{array}{cc}2-1 & -3-0 \\ -7-2 & 5+4\end{array}\right]\right)^{T}=\left[\begin{array}{cc}2-1 & -7-2 \\ -3-0 & 5+4\end{array}\right]$

$\Rightarrow\left(\left[\begin{array}{cc}1 & -3 \\ -9 & 9\end{array}\right]\right)^{T}=\left[\begin{array}{cc}1 & -9 \\ -3 & 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & -9 \\ -3 & 9\end{array}\right]=\left[\begin{array}{cc}1 & -9 \\ -3 & 9\end{array}\right]$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

$(i v)(A B)^{T}=B^{T} A^{T}$

$\Rightarrow\left(\left[\begin{array}{cc}2 & -3 \\ -7 & 5\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 2 & -4\end{array}\right]\right)^{T}=\left[\begin{array}{cc}1 & 2 \\ 0 & -4\end{array}\right]\left[\begin{array}{cc}2 & -7 \\ -3 & 5\end{array}\right]$

$\Rightarrow\left(\left[\begin{array}{cc}2-6 & 0+12 \\ -7+10 & 0-20\end{array}\right]\right)^{T}=\left[\begin{array}{cc}2-6 & -7+10 \\ 0+12 & 0-20\end{array}\right]$

$\Rightarrow\left(\left[\begin{array}{cc}-4 & 12 \\ 3 & -20\end{array}\right]\right)^{T}=\left[\begin{array}{cc}-4 & 3 \\ 12 & -20\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-4 & 3 \\ 12 & -20\end{array}\right]=\left[\begin{array}{cc}-4 & 3 \\ 12 & -20\end{array}\right]$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

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