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# Using properties of determinants, evaluate $A=\begin{vmatrix}43 & 1 & 6\\35 & 7 & 4\\17 & 3 &2\end{vmatrix}$

$\begin{array}{1 1} 7 \\ 0 \\ -7 \\ 1 \end{array}$

Toolbox:
• If each element of a column consists of m terms,then the determinant can be expressed as the sum of m determinants.
Let $\Delta=\begin{vmatrix}43 & 1 & 6\\35 & 7 & 4\\17 & 3 & 2\end{vmatrix}$
This can be written as
$\Delta=\begin{vmatrix}7\times 6+1 & 1 & 6\\7\times 4+7 & 7 & 4\\7\times 2+3 & 3 & 2\end{vmatrix}$
This can be now splitted as
$\quad=\begin{vmatrix}7\times 6 & 1 & 6\\7\times 4 & 7 & 4\\7\times 2 & 3 & 2\end{vmatrix}+\begin{vmatrix}1 & 1 & 6\\7 & 7 & 4\\3 &3 & 2\end{vmatrix}$
Taking the common factor 7 from $\Delta_1$
$\quad=7\begin{vmatrix} 6 & 1 & 6\\ 4 & 7 & 4\\ 2 & 3 & 2\end{vmatrix}+\begin{vmatrix}1 & 1 & 6\\7 & 7 & 4\\3 &3 & 2\end{vmatrix}$
But since two columns are identical in $\Delta_1$ and $\Delta_2$,then determinant value is zero.
$\Delta=7\times 0+0=0.$