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

Step 1: Expand the determinant.
\[ \Delta = (1 + x) \begin{vmatrix} 1 + y & 1 \\ 1 & 1 + z \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 + z \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 + y \\ 1 & 1 \end{vmatrix} \] Step 2: Evaluate the smaller determinants.
First determinant:
\[ (1 + y)(1 + z) - 1 \] \[ = yz + y + z \] Second determinant:
\[ 1(1 + z) - 1 = z \] Third determinant:
\[ 1 - (1 + y) = -y \] Step 3: Substitute back.
\[ \Delta = (1 + x)(yz + y + z) - z - y \] Expand:
\[ = xyz + xy + xz + yz + y + z - z - y \] \[ = xyz + xy + xz + yz \] Thus
\[ \Delta = xy + yz + zx + xyz \] Step 4: Use the condition \( \Delta = 0 \).
\[ xy + yz + zx + xyz = 0 \] Divide both sides by \( xyz \) (since none are zero):
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0 \] Thus
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1 \] Step 5: Final result.
Hence proved.
Final Answer:
\[ \Delta = xy + yz + zx + xyz \] and \[ \boxed{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1} \]