Using the property of determinants and without expanding,prove that $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$.
(N/A) Let $\Delta = \left|\begin{array}{ccc}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|$.
Using the property of determinants,we can split the determinant into two as follows:
$\Delta = \left|\begin{array}{ccc}x & a & x \\ y & b & y \\ z & c & z\end{array}\right| + \left|\begin{array}{ccc}x & a & a \\ y & b & b \\ z & c & c\end{array}\right|$.
In the first determinant,column $1$ and column $3$ are identical $(C_1 = C_3)$. Therefore,its value is $0$.
In the second determinant,column $2$ and column $3$ are identical $(C_2 = C_3)$. Therefore,its value is $0$.
Thus,$\Delta = 0 + 0 = 0$.
Using the property of determinants,we can split the determinant into two as follows:
$\Delta = \left|\begin{array}{ccc}x & a & x \\ y & b & y \\ z & c & z\end{array}\right| + \left|\begin{array}{ccc}x & a & a \\ y & b & b \\ z & c & c\end{array}\right|$.
In the first determinant,column $1$ and column $3$ are identical $(C_1 = C_3)$. Therefore,its value is $0$.
In the second determinant,column $2$ and column $3$ are identical $(C_2 = C_3)$. Therefore,its value is $0$.
Thus,$\Delta = 0 + 0 = 0$.
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