Question

If $x, y, z$ are all different and not equal to zero and $\left| {\begin{array}{*{20}{c}}
  {1 + x}&1&1 \\
  1&{1 + y}&1 \\
  1&1&{1 + z}
\end{array}} \right| = 0$ then the value of ${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}$ is equal to
A. $xyz$
B. ${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}$
C. $ - x - y - z$
D. \[ - 1\]

Answer 1

Hint:
We will apply row transformation on the given determinant and then expand the determinant along the first row and equate it to 0 to form an equation. Then, divide the both sides by $xyz$ to find the value of ${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}$.

Complete step by step solution:
We are given that $\left| {\begin{array}{*{20}{c}}
  {1 + x}&1&1 \\
  1&{1 + y}&1 \\
  1&1&{1 + z}
\end{array}} \right| = 0$
Then, apply row transformation, to simply the determinant.
Apply ${R_1} \to {R_1} - {R_2}$ and ${R_2} \to {R_2} - {R_3}$
Then, we will have,
$\left| {\begin{array}{*{20}{c}}
  x&{ - y}&0 \\
  0&y&{ - z} \\
  1&1&{1 + z}
\end{array}} \right| = 0$
Now, we will expand the determinant along the first row,
$
  x\left( {y\left( {1 + z} \right) + z} \right) + y\left( {0 + z} \right) = 0 \\
   \Rightarrow xy\left( {1 + z} \right) + xz + yz = 0 \\
   \Rightarrow xy + xyz + xz + yz = 0 \\
   \Rightarrow xy + yz + xz = - xyz \\
$
On dividing both sides by $xyz$, we will get,
${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}} = - 1$
Hence, the value of ${x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}$ is 1.

Thus, option D is correct.

Note:
We can expand the determinant along any row or any column. We apply row or column transformations to simply the calculation of finding the determinant. Also, if the value of the determinant is 0, then the matrix is a singular matrix.