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\begin{document}

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

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

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

\end{description}
\medskip

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

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

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

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

\end{document}