Question
Find $A^{\top}$, if
$A=\left[\begin{array}{ccc}2 & -6 & 1 \\ -4 & 0 & 5\end{array}\right]$
$A=\left[\begin{array}{ccc}2 & -6 & 1 \\ -4 & 0 & 5\end{array}\right]$
$\therefore A^{\top}=\left[\begin{array}{cc}2 & -4 \\ -6 & 0 \\ 1 & 5\end{array}\right]$
[Note: Answer given in the textbook is $A^{\top}=\left[\begin{array}{cc}2 & -4 \\ 6 & 0 \\ 1 & 5\end{array}\right]$. However, as per our calculation it
is $\left.A^{\top}=\left[\begin{array}{cc}2 & -4 \\ -6 & 0 \\ 1 & 5\end{array}\right] \cdot\right]$
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$\frac{x}{3}-\frac{y}{2}=0$