a
(a) $\left| {{\mkern 1mu} \begin{array}{*{20}{c}}
{265}&{240}&{219}\\
{240}&{225}&{198}\\
{219}&{198}&{181}
\end{array}{\mkern 1mu} } \right|$ $ = \left| {{\mkern 1mu} \begin{array}{*{20}{c}}
{25}&{21}&{219}\\
{15}&{27}&{198}\\
{21}&{17}&{181}
\end{array}{\mkern 1mu} } \right|$
$=\left| {\,\begin{array}{*{20}{c}}4&{21}&9\\{ - 12}&{27}&{ - 72}\\4&{17}&{11}\end{array}\,} \right|$
{Applying ${C_1} \to {C_1} - {C_2};\,{C_3} \to {C_3} - 10{C_2} $}
$= \,\left| {\,\begin{array}{*{20}{c}}4&{21}&9\\{ - 12}&{27}&{ - 72}\\0&{ - 4}&2\end{array}\,} \right|$ {Applying ${R_3} \to {R_3} - {R_1}$}
$=4\,\left| {\,\begin{array}{*{20}{c}}1&{21}&9\\{ - 3}&{27}&{ - 72}\\0&{ - 4}&2\end{array}\,} \right| = 4\,\left| {\,\begin{array}{*{20}{c}}1&{21}&9\\0&{90}&{ - 45}\\0&{ - 4}&2\end{array}\,} \right|$
by ${R_2} \to 3{R_1} + {R_2}$ $=4(90 × 2 - 45 × 4)=0.$