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The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to
$\left| {\,\begin{array}{*{20}{c}}2&1&1\\2&2&3\\2&3&6\end{array}\,} \right|$
$\left| {\,\begin{array}{*{20}{c}}2&1&1\\3&2&3\\4&3&6\end{array}\,} \right|$
$\left| {\begin{array}{*{20}{c}}1&2&1\\1&5&3\\1&9&6\end{array}} \right|$
$\left| {\,\begin{array}{*{20}{c}}3&1&1\\6&2&3\\{10}&3&6\end{array}} \right|\,$
Solution
(a) $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|\, = \,\left| {\,\begin{array}{*{20}{c}}2&1&1\\3&2&3\\4&3&6\end{array}\,} \right|$ by ${C_1} \to {C_1} + {C_2}$
= $\left| {\,\begin{array}{*{20}{c}}
1&2&1\\
1&5&3\\
1&9&6
\end{array}\,} \right|\,$ by ${C_2} \to {C_2} + {C_3}$
= $\left| {\,\begin{array}{*{20}{c}}3&1&1\\6&2&3\\{10}&3&6\end{array}\,} \right|$, by ${C_1} \to {C_1} + {C_2} + {C_3}$.
But $ \ne \left| {\,\begin{array}{*{20}{c}}2&1&1\\2&2&3\\2&3&6\end{array}\,} \right|$.