Question
$a \mid b$ iff $a c \mid b c$, when $c \neq 0$
Step 1
We are given that $a \mid b$. This means that there exists an integer $k$ such that $b = a k$. Show more…
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Let $a, b, d \in \mathbb{Z}$ with $d \neq 0 .$ Show that $a \mid b$ if and only if $d a \mid d b$.
If $A=\left[\begin{array}{ccc}0 & a & -b \\ -a & 0 & c \\ b & -c & 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}0 & -p & q \\ p & 0 & -q \\ -q & q & 0\end{array}\right]$, then (a) $|\mathrm{A}| \neq|\mathrm{B}|$ (b) $|\mathrm{AB}|=\left|\mathrm{A}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}}\right|$ (c) $|\mathrm{AB}|=\left|\mathrm{A}^{-1} \mathrm{~B}^{-1}\right|$ (d) $\left|\mathrm{A}^{\mathrm{T}}\right|=\left|\mathrm{B}^{\mathrm{T}}\right|$
Prove: $\quad$ If $\frac{a}{b}=\frac{c}{d}$ (where $a, b, c,$ and $d$ are nonzero) $$\text { then } \frac{a}{c}=\frac{b}{d}$$
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