Aspire Faculty ID #13858 · Topic: JAMIA MILLIA ISLAMIA MCA 2021 · Just now
JAMIA MILLIA ISLAMIA MCA 2021

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:

Solution

Expand the determinant: $(1+x)(1+y)(1+z) - (1+x) - (1+y) - (1+z) + 2 = 0$ Simplify: $xyz(x^{-1} + y^{-1} + z^{-1}) + ... = 0$ Final result gives $\boxed{x^{-1} + y^{-1} + z^{-1} = -1}$ $\boxed{\text{Answer: (D) -1}}$

Previous 10 Questions — JAMIA MILLIA ISLAMIA MCA 2021

Nearest first

Next 10 Questions — JAMIA MILLIA ISLAMIA MCA 2021

Ascending by ID
Ask Your Question or Put Your Review.

loading...