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\begin{document}

{\bf Question}
\begin{description}
\item[(i)]
$\left|\begin{array}{ccc} 4 &amp; 6 &amp; 7\\ 2 &amp; 3 &amp; 8\\ 1 &amp; 9 &amp; 0
\end{array} \right|$

\item[(ii)]
$\left|\begin{array}{ccc} 2 &amp; -3 &amp; 0\\ -4 &amp; 5 &amp; 1\\ 0 &amp; 6 &amp; 8
\end{array} \right|$

\item[(iii)]
$\left|\begin{array}{ccc} 1 &amp; 2 &amp; 3\\ 4 &amp; 5 &amp; 6\\ 7 &amp; 8 &amp; 9
\end{array} \right|$

\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
\begin{eqnarray*} &amp; &amp; \left|\begin{array}{ccc} 4 &amp; 6 &amp; 7\\ 2 &amp; 3 &amp; 8\\ 1 &amp; 9 &amp; 0
\end{array} \right|\\ &amp; = &amp; 4\left|\begin{array}{cc} 3 &amp; 8\\ 9 &amp; 0
\end{array}\right|-6\left|\begin{array}{cc} 2 &amp; 8\\ 1 &amp; 0
\end{array}\right|+7\left|\begin{array}{cc} 2 &amp; 3\\ 1 &amp; 9
\end{array}\right|\\ &amp; &amp; NB: \left|\begin{array}{cc} \alpha &amp; \beta\\ \gamma &amp; \delta
\end{array}\right|=\alpha\delta-\beta\gamma\\ &amp; &amp; \rm{Using\ minors,\ expanding\ by\
the\ 1st\ row}\\ &amp; &amp;  \rm{and\ remembering\ the\ +-+\ sign\
pattern.}\\ &amp; = &amp; 4(3 \times 0 - 9 \times 8)-6(2 \times 0 - 8
\times 1)+7(2 \times 9-1 \times 3)\\ &amp; = &amp;
4(0-72)-6(0-8)+7(18-3)\\ &amp; = &amp; -288+48+105\\ &amp; = &amp; \un{-135}
\end{eqnarray*}

\newpage
\item[(ii)]
\begin{eqnarray*} &amp; &amp; \left|\begin{array}{ccc} 2 &amp; -3 &amp; 0\\ -4 &amp; 5 &amp; 1\\ 0 &amp; 6 &amp; 8
\end{array} \right|\\ &amp; = &amp; 2\left|\begin{array}{cc} 5 &amp; 1\\ 6 &amp; 8
\end{array}\right|-(-3)\left|\begin{array}{cc} -4 &amp; 1\\ 0 &amp; 8
\end{array}\right|+0\left|\begin{array}{cc} -4 &amp; 5\\ 0 &amp; 6
\end{array}\right|\\ &amp; &amp; \rm{Using\ minors,\ expanding\ by\
the\ 1st\ row}\\ &amp; &amp;  \rm{and\ remembering\ the\ +-+\ sign\
pattern.}\\ &amp; = &amp; 2(40-6)+3(-32-0)+0(-24-0)\\ &amp; = &amp; 78-96+0\\ &amp; =
&amp; \un{-18}
\end{eqnarray*}

\item[(iii)]
\begin{eqnarray*} &amp; &amp; \left|\begin{array}{ccc} 1 &amp; 2 &amp; 3\\ 4 &amp; 5 &amp; 6\\ 7 &amp; 8 &amp; 9
\end{array} \right|\\ &amp; = &amp; 1\left|\begin{array}{cc} 5 &amp; 6\\ 8 &amp; 9
\end{array}\right|-2\left|\begin{array}{cc} 4 &amp; 6\\ 7 &amp; 9
\end{array}\right|+3\left|\begin{array}{cc} 4 &amp; 5\\ 7 &amp; 8
\end{array}\right|\\ &amp; &amp; \rm{Using\ minors,\ expanding\ by\
the\ 1st\ row}\\ &amp; &amp;  \rm{and\ remembering\ the\ +-+\ sign\
pattern.}\\ &amp; = &amp; 1(45-48)-2(36-42)+3(32-35)\\ &amp; = &amp; -3+12-9\\ &amp; =
&amp; \un{0}
\end{eqnarray*}

\end{description}
Note that we could have evaluated (iii) in the following way:
\begin{eqnarray*} \left|\begin{array}{ccc} 1 &amp; 2 &amp; 3\\ 4 &amp; 5 &amp; 6\\ 7 &amp; 8 &amp; 9
\end{array} \right| &amp; = &amp; \left|\begin{array}{ccc} 1 &amp; 2 &amp; 3\\ 3 &amp; 3 &amp; 3\\ 7 &amp; 8 &amp; 9
\end{array} \right|\ \rm{Rule\ 4:\ with\ row\ 2\ -\ row\ 1}\\ &amp; &amp;
\rm{(i.e.,\ multiple} = -1 \times \rm{row\ (1)})\\ &amp; = &amp;
3\left|\begin{array}{ccc} 1 &amp; 2 &amp; 3\\ 1 &amp; 1 &amp; 1\\ 7 &amp; 8 &amp; 9
\end{array} \right|\ \rm{Rule\ 4:\ with} k=3\\ &amp; = &amp;
3\left|\begin{array}{ccc} 1 &amp; 2 &amp; 3\\ 1 &amp; 1 &amp; 1\\ 1 &amp; 2 &amp; 3
\end{array} \right|\ \rm{Rule\ 4:\ with\ row\ 3} -7\times\rm{\ row\
2}\\ &amp; &amp; \rm{(i.e.,\ multiple} = -7 \times \rm{row\ (2))}\\ &amp; = &amp;
0 \ \rm{Rule\ 2:\ row\ 1\ \&amp;\ row\ 3\ are\ \un{identical}}
\end{eqnarray*}

\end{document}
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