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{\bf Question}

Show that if all the elements of one column of a $3 \times 3$
determinant are multiplied by a constant $k$, then the value of
the determinant is multiplied by $k$. Give a geometrical
interpretation in terms of volumes.

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{\bf Answer}

Let my determinant be $\bigtriangleup$ where

\begin{eqnarray*} \bigtriangleup & = & \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|\\ & = & 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|\end{eqnarray*}

Now multiply 1st row by $k$

\begin{eqnarray*} \bigtriangleup_1 & = & \left|\begin{array}{ccc} ka_1
& kb_1 & kc_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{array}
\right|\\ & = & ka_1\left|\begin{array} {cc} b_2 & c_2\\ b_3 & c_3
\end{array} \right| -kb_1\left|\begin{array} {cc} a_2 & c_2\\ a_3 & c_3
\end{array} \right|+kc_1\left|\begin{array} {cc} a_2 & b_2\\ a_3 & b_3
\end{array} \right|\\ & = & k\left\{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|\right\}\\ & = & k\bigtriangleup \rm{as\ required} \end{eqnarray*}

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