Determinants — Class 12 Maths Solution

We have, $\left| {\begin{array}{cccccccccccccccccccc}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}} \right| = 0$

Applying ${C_1} \to {C_1} - {C_3}$ and ${C_2} \to {C_2} - {C_3}$,

$\Rightarrow$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0&1\\0&y&1\\{ - z}&{ - z}&{1 + z}\end{array}} \right| = 0$

Expanding along ${R_1}$,
$x[y(1 + z) + z] - 0 + 1(yz) = 0$

$\Rightarrow$ $x(y + yz + z) + yz = 0$

$\Rightarrow$ $xy + xyz + xz + yz = 0$

$\Rightarrow$ $\frac{{xy}}{{xyz}} + \frac{{xyz}}{{xyz}} + \frac{{xz}}{{xyz}} + \frac{{yz}}{{xyz}} = 0$

[on dividing $(xyz)$

from both sides]
$\Rightarrow$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0$

$\Rightarrow$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = - 1$

$\therefore {x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}} = - 1$