# Using the property of determinants and without expanding, prove that $\begin{vmatrix} x&a&x+a \\y&b&y+b \\ z&c&z+c \end{vmatrix} = 0$

## 1 Answer

Toolbox:
• If any two rows or column of a determinant are identical (all corresponding elements are same)then the value of the determinant is zero.
Prove that$\begin{vmatrix}x & a& x+a\\y & b &y+b\\z & c & z+c\end{vmatrix}=0$

Now let us split the determinant ,by using the property that 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 two)determinant.

Hence $\begin{vmatrix}x & a & x+a\\y & b &y+b\\z & c &z+c\end{vmatrix}=\begin{vmatrix}x & a & x\\y & a & y\\z & c & z\end{vmatrix}+\begin{vmatrix}x & a & a\\y & b & b\\z & c & c\end{vmatrix}$

Since the elements in the 1st column and 2nd column are identical in $\begin{vmatrix}x & a & x\\y & a & y\\z & c & z\end{vmatrix}$

The value of determinant is 0.

Similarly the 2nd and 3rd column of the determinant $\begin{vmatrix}x & a & a\\y & b & b\\z & c & c\end{vmatrix}$ are identical ,the value of the determinant is 0.

Therefore $\begin{vmatrix}x & a & x+a\\y & b & y+b\\z & c & z+c\end{vmatrix}=0+0=0$

answered Feb 21, 2013

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1 answer