Question
If $A=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -3 & 2 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ -3 & 2 \\ -1 & 3\end{array}\right]$, prove that $\left(A+B^{\top}\right)^{\top}=$
$
A^{\top}+B
$

Answer

$
\begin{aligned}
& A=\left(\begin{array}{rrr}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right], B=\left[\begin{array}{rr}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right] \\
& \therefore A ^{ T }=\left[\begin{array}{rr}
-1 & -3 \\
2 & 2 \\
1 & -3
\end{array}\right], B ^{ T }=\left(\begin{array}{rrr}
2 & -3 & -1 \\
1 & 2 & 3
\end{array}\right) \\
& \therefore A + B ^{ r }=\left(\begin{array}{rrr}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right)+\left(\begin{array}{rrr}
2 & -3 & -1 \\
1 & 2 & 3
\end{array}\right) \\
& =\left[\begin{array}{rrr}
-1+2 & 2-3 & 1-1 \\
-3+1 & 2+2 & -3+3
\end{array}\right] \\
& =\left(\begin{array}{rrr}
1 & -1 & 0 \\
-2 & 4 & 0
\end{array}\right) \\
& \therefore\left( A + B ^{ T }\right)^{ T }=\left(\begin{array}{rr}
1 & -2 \\
-1 & 4 \\
0 & 0
\end{array}\right) \\
& A^T+B=\left(\begin{array}{rr}
-1 & -3 \\
2 & 2 \\
1 & -3
\end{array}\right)+\left(\begin{array}{rr}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right) \\
& =\left(\begin{array}{rr}
-1+2 & -3+1 \\
2-3 & 2+2 \\
1-1 & -3+3
\end{array}\right]=\left[\begin{array}{rr}
1 & -2 \\
-1 & 4 \\
0 & 0
\end{array}\right) \\
&
\end{aligned}
$
From (1) and (2),
$
\left(A+B^{\top}\right)^{\top}=A^{\top}+B
$

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