Question

If \(x, y, z\) are all different from zero and \(\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0\), then the value of \(x^{-1} + y^{-1} + z^{-1}\) is

• A. \(xyz\)
• B. \(-1\)
• C. \(1\)
• D. \(0\)

Hint:

Expand to get \(xyz+xy+yz+zx=0\), then divide by \(xyz\).

Answer 1

The Correct Option is B

Step 1: Expand the determinant.
Expand along the first row.
\[ (1+x)\big[(1+y)(1+z)-1\big] - 1\big[(1+z)-1\big] + 1\big[1-(1+y)\big] \] Simplify each bracket: the first is \(y+z+yz\), the second is \(z\), the third is \(-y\).

Step 2: Collect the terms.
\[ (1+x)(y+z+yz) - z - y = xy + xz + xyz + yz \] So the determinant equals \(xyz + xy + yz + zx\).

Step 3: Use the condition.
The determinant is 0, so \(xyz + xy + yz + zx = 0\). Since \(x, y, z \neq 0\), divide by \(xyz\).
\[ 1 + \frac{1}{z} + \frac{1}{x} + \frac{1}{y} = 0 \]

Step 4: Read off the answer.
Hence \(x^{-1}+y^{-1}+z^{-1} = -1\).

Step 5: Check the other options.
Option 1 (\(xyz\)) is not fixed by the condition. Options 3 and 4 would need the sum of reciprocals to be 1 or 0, but the equation gives \(-1\).

Final Answer:
The sum of reciprocals is -1, option 2. \[ \boxed{-1} \]