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 \(x\), \(y\) and \(z\) are non-zero real numbers, prove that \[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-1. \]

Hint:

Factoring out variables to create fraction forms like \(\frac{1}{x}\) is a useful technique for solving determinants that contain cyclic variables on their main diagonal.

Answer 1

Concept: To evaluate the determinant efficiently, we use elementary row and column operations. These operations simplify the determinant without changing its value (except when a common factor is taken out). The simplified determinant can then be expanded easily.

Step 1: Factor out \(x\), \(y\) and \(z\) from the three rows.
The given determinant is \[ \Delta= \begin{vmatrix} 1+x & 1 & 1\\ 1 & 1+y & 1\\ 1 & 1 & 1+z \end{vmatrix}. \] Since \(x,y,z\neq0\), factor out \(x\), \(y\) and \(z\) from the first, second and third rows respectively: \[ \Delta = xyz \begin{vmatrix} 1+\dfrac1x & \dfrac1x & \dfrac1x\\ \dfrac1y & 1+\dfrac1y & \dfrac1y\\ \dfrac1z & \dfrac1z & 1+\dfrac1z \end{vmatrix}. \]

Step 2: Apply a row operation.
Perform \[ R_1\rightarrow R_1+R_2+R_3. \] Then, \[ \Delta = xyz \begin{vmatrix} 1+\dfrac1x+\dfrac1y+\dfrac1z & 1+\dfrac1x+\dfrac1y+\dfrac1z & 1+\dfrac1x+\dfrac1y+\dfrac1z\\ \dfrac1y & 1+\dfrac1y & \dfrac1y\\ \dfrac1z & \dfrac1z & 1+\dfrac1z \end{vmatrix}. \] Factor the common quantity from the first row: \[ \Delta = xyz\left(1+\frac1x+\frac1y+\frac1z\right) \begin{vmatrix} 1 & 1 & 1\\ \dfrac1y & 1+\dfrac1y & \dfrac1y\\ \dfrac1z & \dfrac1z & 1+\dfrac1z \end{vmatrix}. \]

Step 3: Apply column operations.
Now perform \[ C_2\rightarrow C_2-C_1, \qquad C_3\rightarrow C_3-C_1. \] This gives \[ \Delta = xyz\left(1+\frac1x+\frac1y+\frac1z\right) \begin{vmatrix} 1 & 0 & 0\\ \dfrac1y & 1 & 0\\ \dfrac1z & 0 & 1 \end{vmatrix}. \]

Step 4: Evaluate the determinant.
The remaining determinant is triangular, so its value is \[ 1\times1\times1=1. \] Hence, \[ \boxed{ \Delta = xyz\left(1+\frac1x+\frac1y+\frac1z\right). } \]

Step 5: Prove that \(x^{-1}+y^{-1}+z^{-1}=-1\).
Given \[ \Delta=0, \] therefore, \[ xyz\left(1+\frac1x+\frac1y+\frac1z\right)=0. \] Since \(x\), \(y\) and \(z\) are non-zero, \[ xyz\neq0. \] Hence, \[ 1+\frac1x+\frac1y+\frac1z=0. \] Therefore, \[ \boxed{ \frac1x+\frac1y+\frac1z=-1. } \] This completes the proof.