Using the property of determinants and without expanding, prove that: \(\begin{vmatrix}2&7&65\\3&8&75\\5&9&86\end{vmatrix}\)=0
\(\begin{vmatrix}2&7&65\\3&8&75\\5&9&86\end{vmatrix}\)
=\(\begin{vmatrix}2&7&63+2\\3&8&72+3\\5&9&81+5\end{vmatrix}\)
=\(\begin{vmatrix}2&7&63\\3&8&72\\5&9&81\end{vmatrix}+\begin{vmatrix}2&7&2\\3&8&3\\5&9&5\end{vmatrix}\)
=\(\begin{vmatrix}2&7&9(7)\\3&8&9(8)\\5&9&9(9)\end{vmatrix}+0\) [Two columns are identical]
=\(9\begin{vmatrix}2&7&7\\3&8&8\\5&9&9\end{vmatrix}\) [Two columns are identical]
=0
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Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
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Read More: Properties of Determinants