Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c\end{array}\right|=a^{3}$
(A) Let $\Delta = \left|\begin{array}{ccc}a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c\end{array}\right|$.
Applying operations $R_{2} \rightarrow R_{2}-2R_{1}$ and $R_{3} \rightarrow R_{3}-3R_{1}$,we get:
$\Delta = \left|\begin{array}{ccc}a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b\end{array}\right|$.
Now,applying $R_{3} \rightarrow R_{3}-3R_{2}$,we get:
$\Delta = \left|\begin{array}{ccc}a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a\end{array}\right|$.
Expanding along the first column $C_{1}$,we obtain:
$\Delta = a \left|\begin{array}{cc}a & 2a+b \\ 0 & a\end{array}\right| - 0 + 0 = a(a^{2} - 0) = a(a^{2}) = a^{3}$.
Thus,the determinant is equal to $a^{3}$.
Applying operations $R_{2} \rightarrow R_{2}-2R_{1}$ and $R_{3} \rightarrow R_{3}-3R_{1}$,we get:
$\Delta = \left|\begin{array}{ccc}a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b\end{array}\right|$.
Now,applying $R_{3} \rightarrow R_{3}-3R_{2}$,we get:
$\Delta = \left|\begin{array}{ccc}a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a\end{array}\right|$.
Expanding along the first column $C_{1}$,we obtain:
$\Delta = a \left|\begin{array}{cc}a & 2a+b \\ 0 & a\end{array}\right| - 0 + 0 = a(a^{2} - 0) = a(a^{2}) = a^{3}$.
Thus,the determinant is equal to $a^{3}$.
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