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Question: If \(x, y, z\) are all different and not equal to zero and \(\left| {\begin{array}{*{20}{c}} {1 ...

If x,y,zx, 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{x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}} is equal to
A. xyzxyz
B. x−1+y−1+z−1{x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}}
C. −x−y−z - x - y - z
D. −1 - 1

Explanation

Solution

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 xyzxyz to find the value of x−1+y−1+z−1{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 R1→R1−R2{R_1} \to {R_1} - {R_2} and R2→R2−R3{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(y(1+z)+z)+y(0+z)=0 ⇒xy(1+z)+xz+yz=0 ⇒xy+xyz+xz+yz=0 ⇒xy+yz+xz=−xyz  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 xyzxyz, we will get,
x−1+y−1+z−1=−1{x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}} = - 1
Hence, the value of x−1+y−1+z−1{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.