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{\bf Question}

By explicitly expanding the following two determinants, prove the
rule that the sign of a $3 \times 3$ determinant is changed by
exchanging two rows:

$$det\ {\bf{M_1}}=\left|\begin{array} {ccc} a_1 & b_1 & c_1\\ a_2
& b_2 & c_2\\ a_3 & b_3 & c_3 \end{array} \right|\ \ \ \ det\
{\bf{M_2}}=\left|\begin{array} {ccc} a_2 & b_2 & c_2\\ a_1 & b_1 &
c_1\\ a_3 & b_3 & c_3 \end{array} \right|$$

\medskip

{\bf Answer}
\begin{eqnarray*} det({\bf{m_1}}) & = & \left|\begin{array}{ccc}
a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3
\end{array} \right|\\ & & \rm{Using\ minors,\ expanding\ by\
the\ 1st\ row}\\ & &  \rm{and\ remembering\ the\ +-+\ sign\
pattern.}\\& = & a_1\left|\begin{array}{cc} b_2 & c_2\\ b_3 & c_3
\end{array}\right|-b_1\left|\begin{array}{cc} a_2 & c_2\\ a_3 &
c_3
\end{array}\right|+c_1\left|\begin{array}{cc} a_2 & b_2\\ a_3 &
b_3
\end{array}\right|\\ & = &
a_1(b_2c_3-b_3c_2)-b_1(a_2c_3-a_3c_2)+c_1(a_2b_3-a_3b_2)\ \ \ (1)
\end{eqnarray*}

\begin{eqnarray*} det({\bf{m_2}}) & = & \left|\begin{array}{ccc}
a_2 & b_2 & c_2\\ a_1 & b_1 & c_1\\ a_3 & b_3 & c_3
\end{array} \right|\\ & & \rm{Using\ minors,\ expanding\ by\
the\ 1st\ row}\\ & &  \rm{and\ remembering\ the\ +-+\ sign\
pattern.}\\& = & a_2\left|\begin{array}{cc} b_1 & c_1\\ b_3 & c_3
\end{array}\right|-b_2\left|\begin{array}{cc} a_1 & c_1\\ a_3 &
c_3
\end{array}\right|+c_2\left|\begin{array}{cc} a_1 & b_1\\ a_3 &
b_3
\end{array}\right|\\ & = &
a_2(b_1c_3-b_3c_1)-b_2(a_1c_3-a_3c_1)+c_2(a_1b_3-a_3b_1)\ \ \ (2)
\end{eqnarray*}

Now compare $(1)$ and $(2)$

Expand $(1)$

$det
{\bf{m_1}}=(1)=a_1b_2c_3-a_1b_3c_2-a_2b_1c_3+a_3b_1c_2+a_2b_3c_1-a_3b_2c_1$

$det
{\bf{m_2}}=(2)=-a_1b_2c_3+a_1b_3c_2+a_2b_1c_3-a_3b_1c_2-a_2b_3c_1+a_3b_2c_1$

i.e., $(1)=-(2)$ or $det {\bf{m_1}}=det {\bf{m_2}}$.

Since all the $a_i's,\ b_j'\,\ c_k's$ are arbitrary. This rule
holds for all $3 \times 3$ determinants. Hence if you exchange two
rows the sign changes.

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