Question

For non zero, a, b, c if $\Delta =\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=0$, then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=$

• A. $abc$
• B. $\frac{1}{abc}$
• C. $-a(a+b+c)$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$\Delta =\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|$
$=abc\left|\begin{array}{ccc}\frac{1}{a}+1& \frac{1}{b}& \frac{1}{c}\\ \frac{1}{a}& \frac{1}{b}+1& \frac{1}{c}\\ \frac{1}{a}& \frac{1}{b}& \frac{1}{c}+1\end{array}\right|$, by $\begin{array}{cc}& C_1\to \frac{1}{a}{C}_1\\ & C_2\to \frac{1}{b}{C}_2\\ & C_3\to \frac{1}{c}{C}_3\end{array}=abc\left|\begin{array}{ccc}(1+\sum \frac{1}{a})& \frac{1}{b}& \frac{1}{c}\\ (1+\sum \frac{1}{a})& \frac{1}{b}+1& \frac{1}{c}\\ (1+\sum \frac{1}{a})& \frac{1}{b}& \frac{1}{c}+1\end{array}\right|$
by $C_1\to C_1+C_2+C_3$
$=abc(1+\sum \frac{1}{a})\left|\begin{array}{ccc}1& \frac{1}{b}& \frac{1}{c}\\ 1& \frac{1}{b}+1& \frac{1}{c}\\ 1& \frac{1}{b}& \frac{1}{c}+1\end{array}\right|$
[By taking $(1+\sum \frac{1}{a})$ as common]
$\Delta =abc(1+\sum \frac{1}{a})\left|\begin{array}{ccc}1& \frac{1}{b}& \frac{1}{c}\\ 1& \frac{1}{b}+1& \frac{1}{c}\\ 1& \frac{1}{b}& \frac{1}{c}+1\end{array}\right|,by\begin{array}{cc}& R_2\to R_2-R_1\\ & R_3\to R_3-R_1\end{array}=abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).1,$ (by expanding along $C_1$)
Therefore $\Delta =0\Rightarrow$ Either abc = 0 or $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
But a,b,c are non-zero and hence the product abc cannot be zero. So the only alternative is that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-1$