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# Using the properties of determinants, Prove that$\begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix}\;=\;4xyz$

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Toolbox:
• If each element of a row (or a column)of a determinant is multiplied by a constant k,then |A|=k|A|.
• If A is a square matrix such that each element of a row (or a column) of A is expressed as a sum of two or more terms,then the determinant of A can be expressed as the sum of the determinants of two or more matrices of the same order.
Let $\Delta=\begin{vmatrix}y+z & z & y\\z & z+x & x\\y & x & x+y\end{vmatrix}$

Apply $R_1\rightarrow R_1+R_2+R_3$

$\Delta=\begin{vmatrix}2y+2z & 2x+2z & 2y+2x\\z & z+x & x\\y & x & x+y\end{vmatrix}$

Taking 2 as the common factor from $R_1$,

$\Delta=2\begin{vmatrix}y+z & x+z & y+x\\z & z+x & x\\y & x & x+y\end{vmatrix}$

Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$

$\Delta=2\begin{vmatrix}y & 0 & y\\z-y & z & -y\\y & x & x+y\end{vmatrix}$

Apply $C_1\rightarrow C_1-C_3$

$\Delta=2\begin{vmatrix}0 & 0& y\\z & z & -y\\-x & x & x+y\end{vmatrix}$

Expanding along $R_1$ we get,

$\Delta=2[0-0+y(xz+xz)]$

Therefore $\Delta=2\times 2xyz.$

$\Delta=4xyz.$

Hence proved.

answered Mar 15, 2013
edited Mar 15, 2013

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