Question
If $A=\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$, prove that $(3 A)^{\top}=3 A^{\top}$.
$\therefore \quad 3 A=3\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]=\left[\begin{array}{ccc}3 & 6 & -15 \\ 6 & -9 & 12 \\ -15 & 12 & 27\end{array}\right]$
$\therefore \quad(3 \mathrm{~A})^{\mathrm{T}}=\left[\begin{array}{ccc}3 & 6 & -15 \\ 6 & -9 & 12 \\ -15 & 12 & 27\end{array}\right]$
$\ldots$..(i)
$A^T=\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$
$\therefore \quad 3 A^{\mathrm{T}}=3\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]=\left[\begin{array}{ccc}3 & 6 & -15 \\ 6 & -9 & 12 \\ -15 & 12 & 27\end{array}\right]$
...(ii)
From (i) and (ii), we get
$(3 \mathrm{~A})^{\top}=3 \mathrm{~A}^{\top}$
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