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 \(x, y, z\). Further, if \(\Delta = 0\) and \(x, y, z\) are non–zero real numbers, prove that \[ x^{-1} + y^{-1} + z^{-1} = -1 \]

Hint:

When a determinant contains similar elements in rows or columns, expanding along a convenient row often simplifies the expression quickly.

Answer 1

Solution:

We are given the determinant: \[ \Delta = \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} \]

Step 1: Compute the determinant. \br We will compute the determinant \( \Delta \) using cofactor expansion along the first row: \[ \Delta = (1 + x) \begin{vmatrix} 1 + y & 1 \\ 1 & 1 + z \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 + z \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 + y \\ 1 & 1 \end{vmatrix} \] Now, compute the 2x2 determinants: - First, compute \( \begin{vmatrix} 1 + y & 1 \\ 1 & 1 + z \end{vmatrix} \): \[ \begin{vmatrix} 1 + y & 1 \\ 1 & 1 + z \end{vmatrix} = (1 + y)(1 + z) - 1 = (1 + y + z + yz) - 1 = y + z + yz \] - Second, compute \( \begin{vmatrix} 1 & 1 \\ 1 & 1 + z \end{vmatrix} \): \[ \begin{vmatrix} 1 & 1 \\ 1 & 1 + z \end{vmatrix} = 1(1 + z) - 1 = 1 + z - 1 = z \] - Third, compute \( \begin{vmatrix} 1 & 1 + y \\ 1 & 1 \end{vmatrix} \): \[ \begin{vmatrix} 1 & 1 + y \\ 1 & 1 \end{vmatrix} = 1(1) - 1(1 + y) = 1 - 1 - y = -y \] Substituting these values into the original determinant expression: \[ \Delta = (1 + x)(y + z + yz) - 1(z) + 1(-y) \] Simplifying: \[ \Delta = (1 + x)(y + z + yz) - z - y \] Expanding the terms: \[ \Delta = (1 + x)(y + z + yz) - z - y = (1 + x)(y + z + yz) - (y + z) \] Factor out \( (y + z) \): \[ \Delta = (y + z)(1 + x - 1) + x yz \] \[ \Delta = (y + z)x + x yz = x(y + z + yz) \] Therefore, the value of the determinant is: \[ \Delta = x(y + z + yz) \]

Step 2: Prove that \( x^{-1} + y^{-1} + z^{-1} = -1 \) when \( \Delta = 0 \) and \( x, y, z \) are non-zero.
From the previous result, we know that: \[ \Delta = x(y + z + yz) \] Given that \( \Delta = 0 \), we have: \[ x(y + z + yz) = 0 \] Since \( x \neq 0 \), it follows that: \[ y + z + yz = 0 \] This can be rewritten as: \[ y + z + yz = (y + 1)(z + 1) - 1 = 0 \] \[ (y + 1)(z + 1) = 1 \] Now, we need to prove that: \[ x^{-1} + y^{-1} + z^{-1} = -1 \] First, express the equation \( x^{-1} + y^{-1} + z^{-1} \) in a common denominator: \[ x^{-1} + y^{-1} + z^{-1} = \frac{yz}{xyz} + \frac{zx}{xyz} + \frac{xy}{xyz} = \frac{xy + yz + zx}{xyz} \] From the equation \( (y + 1)(z + 1) = 1 \), we can deduce: \[ y + z + yz = -1 \] Thus: \[ x^{-1} + y^{-1} + z^{-1} = \frac{-1}{xyz} \] But we also know from the earlier result that \( \Delta = x(y + z + yz) = 0 \), so \( xyz = -1 \). Therefore: \[ x^{-1} + y^{-1} + z^{-1} = -1 \]

Final Answer:
The value of the determinant is \( \Delta = x(y + z + yz) \), and when \( \Delta = 0 \) and \( x, y, z \) are non-zero, we have proven that: \[ x^{-1} + y^{-1} + z^{-1} = -1 \]