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The value of the determinant
$\left|\begin{array}{lll}265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{array}\right|$ is
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$\left|\begin{array}{lll}265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181\end{array}\right|$ is
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Verified Answer
The correct answer is:
0
Applying $\mathrm{C}_{1}-\mathrm{C}_{2}$ and $\mathrm{C}_{2}-\mathrm{C}_{3},$ we get
$\begin{aligned}
\text { Det. } &=\left|\begin{array}{ccc}
25 & 21 & 219 \\
15 & 27 & 198 \\
21 & 17 & 181
\end{array}\right|=\left|\begin{array}{ccc}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right| \\
=&\left[\begin{array}{llc}
4 & 21 & 9 \\
0 & 90 & -45 \\
0 & -4 & 2
\end{array} \mid\left[\mathrm{By} \mathrm{R}_{2}+3 \mathrm{R}_{1}, \mathrm{R}_{3}-\mathrm{R}_{1}\right]\right.\\
=4(180-180)=0
\end{aligned}$
$\begin{aligned}
\text { Det. } &=\left|\begin{array}{ccc}
25 & 21 & 219 \\
15 & 27 & 198 \\
21 & 17 & 181
\end{array}\right|=\left|\begin{array}{ccc}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right| \\
=&\left[\begin{array}{llc}
4 & 21 & 9 \\
0 & 90 & -45 \\
0 & -4 & 2
\end{array} \mid\left[\mathrm{By} \mathrm{R}_{2}+3 \mathrm{R}_{1}, \mathrm{R}_{3}-\mathrm{R}_{1}\right]\right.\\
=4(180-180)=0
\end{aligned}$
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