सिद्ध कीजिए कि $\Delta=\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$
सिद्ध कीजिए कि $\Delta=\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c\end{array}\right|=a^{3}$
Solution Applying operations $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$ to the given determinant $\Delta$, we have
$\Delta = \left| {\begin{array}{*{20}{c}}
a&{a + b}&{a + b + c} \\
0&a&{2a + b} \\
0&{3a}&{7a + 3b}
\end{array}} \right|$
Now applying $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{2},$ we get
$\Delta=\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 0 & a
\end{array}\right|$
Expanding along $C_{1},$ we obtain
$\Delta = a\left| {\begin{array}{*{20}{c}}
a&{2a + b} \\
0&a
\end{array}} \right| + 0 + 0$
$ = a\left( {{a^2} - 0} \right) = a\left( {{a^2}} \right) = {a^3}$
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