Question

Obtain the value of \( \Delta = \begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} \) in terms of \( x, y \) and \( z \). Further, if \( \Delta = 0 \) and \( x, y, z \) are non-zero real numbers, prove that \( x^{-1} + y^{-1} + z^{-1} = -1 \).

Hint:

Factoring out terms to create a common row/column sum is a powerful determinant strategy.

Answer 1

Concept:

• Properties of determinants for simplification.

Step 1: Simplify the determinant
Take \( x, y, z \) common from \( R_1, R_2, R_3 \):
\[ \Delta = xyz \begin{vmatrix} 1/x + 1 & 1/x & 1/x \\ 1/y & 1/y + 1 & 1/y \\ 1/z & 1/z & 1/z + 1 \end{vmatrix} \]
Applying \( R_1 \to R_1 + R_2 + R_3 \):
\[ \Delta = xyz(1 + 1/x + 1/y + 1/z) \begin{vmatrix} 1 & 1 & 1 \\ 1/y & 1/y+1 & 1/y \\ 1/z & 1/z & 1/z+1 \end{vmatrix} \]
Expanding gives \( \Delta = xyz(1 + 1/x + 1/y + 1/z) \).

Step 2: Apply the condition \( \Delta = 0 \)
Since \( x, y, z \neq 0 \), we have \( 1 + 1/x + 1/y + 1/z = 0 \).
\[ x^{-1} + y^{-1} + z^{-1} = -1 \]