Question
Let $A=\left[\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right], C=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$ Show that $(A-B)^{\top}=A^{\top}-B^{\top}$

Answer

$
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 2 \\
1 & 3
\end{array}\right], B=\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right], C=\left[\begin{array}{ll}
2 & 0 \\
1 & 2
\end{array}\right] \\
& A-B=\left[\begin{array}{ll}
1 & 2 \\
1 & 3
\end{array}\right]-\left[\begin{array}{ll}
4 & 0 \\
1 & 5
\end{array}\right] \\
& =\left[\begin{array}{ll}
-3 & 2 \\
0 & 2
\end{array}\right] \\
& (A-B)^{\top}=\left[\begin{array}{ll}
-3 & 0 \\
2 & -2
\end{array}\right] \ldots(1) \\
& A^{\top}=\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right] \\
& B^{\top}=\left[\begin{array}{ll}
4 & 1 \\
0 & 5
\end{array}\right] \\
& A^{\top}-B^{\top}=\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right]-\left[\begin{array}{ll}
4 & 1 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{ll}
-3 & 0 \\
2 & -2
\end{array}\right] \ldots(2)
\end{aligned}
$

From (1) and (2) we get
$(A-B)^{\top}=A^{\top}-B^{\top}$

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