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# Using the property of determinants and without expanding, prove that $\begin{vmatrix} 2&7&65 \\ 3&8&75 \\ 5&9&86 \end{vmatrix} = 0$

Toolbox:
• (i)If any two rows or column of a determinant are identical (all corresponding elements are same)then the value of the determinant is zero.
• (ii)If some or all elements of a row or column of a determinant are expressed as sum of two (or more)terms,then the determinant can be expressed as sum of two(or more) determinants.
Prove that $A=\begin{vmatrix}2 & 7 & 65\\3 & 8 & 75\\5 & 9 & 80\end{vmatrix}=0$

The given determinant can be expressed as

$\mid A\mid=\begin{vmatrix}2 & 7 & 63+2\\3 & 8 & 72+3\\5 & 9 & 81+5\end{vmatrix}$

By using the property of determinants,we can split the determinant and write as,

$\mid A\mid=\begin{vmatrix}2 & 7 & 63\\3 & 8 & 72\\5 & 9 & 81\end{vmatrix}+\begin{vmatrix}2 & 7 & 2\\3 & 8 &3\\5 & 9 & 5\end{vmatrix}$

But this can be again written as,

$\mid A\mid=\begin{vmatrix}2 & 7 & 9\times 7\\3 & 8 & 9\times 8\\5 & 9 & 9\times 9\end{vmatrix}+\begin{vmatrix}2 & 7 & 2\\3 & 8 & 3\\5 & 9 & 5\end{vmatrix}$

Taking 9 as the common factor from column 3

$\mid A\mid=\begin{vmatrix}2 & 7 & 7\\3 & 8 & 8\\5 & 9 & 9\end{vmatrix}+\begin{vmatrix}2 & 7 & 2\\3 & 8 & 3\\5 & 9 & 5\end{vmatrix}=0.$

(Because two columns are identical.)