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सारणिक $\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ का मान है
$0$
$679$
$779$
$1000$
Solution
(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|$
{ ${C_1} \to {C_1} – {C_2};\,{C_2} \to {C_2} – {C_3} $} से
=$\left| {\,\begin{array}{*{20}{c}}4&{21}&9\\{ – 12}&{27}&{ – 72}\\4&{17}&{11}\end{array}\,} \right|$
{ ${C_1} \to {C_1} – {C_2};\,{C_3} \to {C_3} – 10{C_2} $} से
= $\,\left| {\,\begin{array}{*{20}{c}}4&{21}&9\\{ – 12}&{27}&{ – 72}\\0&{ – 4}&2\end{array}\,} \right|$ { ${R_3} \to {R_3} – {R_1}$}
=$4\,\left| {\,\begin{array}{*{20}{c}}1&{21}&9\\{ – 3}&{27}&{ – 72}\\0&{ – 4}&2\end{array}\,} \right| = 4\,\left| {\,\begin{array}{*{20}{c}}1&{21}&9\\0&{90}&{ – 45}\\0&{ – 4}&2\end{array}\,} \right|$
(${R_2} \to 3{R_1} + {R_2}$ के द्वारा) $=4(90 × 2-45 × 4) = 0 $
