Question

If $x, y, z$ are non-zero real numbers and $\begin{vmatrix} 1+x & 1 & 1 \\ 1+y & 1+2y & 1 \\ 1+z & 1+z & 1+3z \end{vmatrix} = 0$, then $-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ is equal to ____.

Answer 1

3

Answer 2

Apply column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$ to the determinant. $\begin{vmatrix} 1+x & 1 & 1 \\ 1+y & 1+2y & 1 \\ 1+z & 1+z & 1+3z \end{vmatrix} = \begin{vmatrix} x & 0 & 1 \\ -y & 2y & 1 \\ 0 & -2z & 1+3z \end{vmatrix}$ Expand along the first row: $x \begin{vmatrix} 2y & 1 \\ -2z & 1+3z \end{vmatrix} - 0 + 1 \begin{vmatrix} -y & 2y \\ 0 & -2z \end{vmatrix} = 0$ $x(2y(1+3z) - (1)(-2z)) + (-y)(-2z) - (2y)(0) = 0$ $x(2y + 6yz + 2z) + 2yz = 0$ $2xy + 6xyz + 2xz + 2yz = 0$ Divide by $2xyz$ (since $x, y, z \neq 0$): $\frac{1}{z} + 3 + \frac{1}{y} + \frac{1}{x} = 0$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -3$ Therefore, $-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = -(-3) = 3$.