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  4. If abc ≠ 0 and | beginmatrix1+a&1&1 1&1+b&1 1&1&1+c endmatrix|=0, then (1/a)+(1/b)+(1/c)=

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Q. If abc $\ne 0$ and $\left|\begin{matrix}1+a&1&1\\ 1&1+b&1\\ 1&1&1+c\end{matrix}\right|=0$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=$

Determinants

Solution:

$\Delta=\left(1+a\right)\left(b+c+bc\right)-c-b=0$
$\Rightarrow \quad abc \left[1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]=0$
Since, abc $\ne0$
So, $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ $\quad$ $\Rightarrow \quad\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1$