Question
If $A=\left[\begin{array}{cc}5 & -3 \\ 4 & -3 \\ -2 & 1\end{array}\right]$, prove that $(2 A)^{\top}=2 A^{\top}$
$\therefore \quad 2 A=2\left[\begin{array}{cc}5 & -3 \\ 4 & -3 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}10 & -6 \\ 8 & -6 \\ -4 & 2\end{array}\right]$
$\therefore \quad(2 \mathrm{~A})^{\mathrm{T}}=\left[\begin{array}{ccc}10 & 8 & -4 \\ -6 & -6 & 2\end{array}\right]$
$\ldots$ (i)
$A^T=\left[\begin{array}{ccc}5 & 4 & -2 \\ -3 & -3 & 1\end{array}\right]$
$\therefore \quad 2 A^{\top}=2\left[\begin{array}{ccc}5 & 4 & -2 \\ -3 & -3 & 1\end{array}\right]$
$=\left[\begin{array}{rrr}10 & 8 & -4 \\ -6 & -6 & 2\end{array}\right]$
$\ldots$ (ii)
From (i) and (ii), we get
$(2 \mathrm{~A})^{\top}=2 \mathrm{~A}^{\top}$
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R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
(i) the first card drawn is kept aside.
ii. the first card drawn is replaced in the pack.