# Using the property of determinants and without expanding, prove that $\begin{vmatrix} b+c&q+r&y+z \\ c+a&r+p&z+x \\ a+b&p+q&x+y \end{vmatrix} = 2 \begin{vmatrix} a&p&x \\ b&q&y \\ c&r&z \end{vmatrix}$

Toolbox:
• 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.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
By using the property ,the given determinant can be expressed as sum of two determinant say $\bigtriangleup=\bigtriangleup_1+\bigtriangleup_2.$

Therefore $\bigtriangleup=\begin{vmatrix}b+c & q+r & y+z\\c+a & r+p & z+x\\a+b & p+q & x+y\end{vmatrix}$

$\qquad=\begin{vmatrix}b & q & y\\c & r & z\\a & p & x\end{vmatrix}+\begin{vmatrix}c & r& z\\a & p & x\\b & q & y\end{vmatrix}$

By interchanging the $R_3$ as $R_1$ and $R_1$ as $R_3$ in $\bigtriangleup_1$ (i.e)$R_1\leftrightarrow R_3$ and $R_3\leftrightarrow R_1$

$\bigtriangleup_1=\begin{vmatrix}a & p & x\\b & q & y\\c & r & z\end{vmatrix}$

Similarly by interchanging the $R_2$ as $R_1$ and $R_1$ as $R_3$ in $\bigtriangleup_2$ (i.e)$R_1\leftrightarrow R_2$ and $R_1\leftrightarrow R_3$

$\bigtriangleup_2=\begin{vmatrix}a & p & x\\b & q & y\\c & r & z\end{vmatrix}$

Now adding $\bigtriangleup_1$ and $\bigtriangleup_2$

We get $\bigtriangleup=\begin{vmatrix}a & p & x\\b & q & y\\c & r & z\end{vmatrix}+\begin{vmatrix}a & p & x\\b & q & y\\c & r & z\end{vmatrix}$

$\qquad\qquad=2\begin{vmatrix}a & p & x\\b & q & y\\c & r & z\end{vmatrix}$

Hence proved.

edited Feb 24, 2013