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(A) Let $\Delta = \left|\begin{array}{lll}2 & 7 & 65 \ 3 & 8 & 75 \ 5 & 9 & 86\end{array}ight|$.We can write the third column as $C_3 = 9C_2 + C_1

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(A) Let $\Delta = \left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$.
We can write the third column as $C_3 = 9C_2 + C_1$:
$\Delta = \left|\begin{array}{lll}2 & 7 & 9(7)+2 \\ 3 & 8 & 9(8)+3 \\ 5 & 9 & 9(9)+5\end{array}\right|$
Using the property of determinants,we can split this into two determinants:
$\Delta = \left|\begin{array}{lll}2 & 7 & 9(7) \\ 3 & 8 & 9(8) \\ 5 & 9 & 9(9)\end{array}\right| + \left|\begin{array}{lll}2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5\end{array}\right|$
In the first determinant,we can take $9$ as a common factor from the third column:
$\Delta = 9 \left|\begin{array}{lll}2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9\end{array}\right| + \left|\begin{array}{lll}2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5\end{array}\right|$
In the first determinant,the second and third columns are identical,so its value is $0$.
In the second determinant,the first and third columns are identical,so its value is $0$.
Therefore,$\Delta = 9(0) + 0 = 0$.

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

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Using the property of determinants and without expanding,prove that:$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$