Question

$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=$

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

Answer 1

Answer: (A), (B)

The correct options are
A $abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
B $(abc+bc+ca+ab)$

Let $\Delta =\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|$
Since the answer contain abc, then taking a, b and c common from $R_1,R_2,R_3$ respectively, then
$\Delta =abc\left|\begin{array}{ccc}\frac{1}{a}+1& \frac{1}{a}& \frac{1}{a}\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
But answer also contains $(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
then applying $R_1\to R_1+R_2+R_3$
$\therefore \Delta =abc\left|\begin{array}{ccc}1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}& 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}& 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
Taking $(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ common from $R_1$, then
$\Delta =abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
Applying $C_2\to C_2-C_1$, then
$\Delta =abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\left|\begin{array}{ccc}1& 0& 1\\ \frac{1}{b}& 1& \frac{1}{b}\\ \frac{1}{c}& 0& \frac{1}{c}+1\end{array}\right|$
Expanding along $C_2$, then
$\Delta =abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\left|\begin{array}{cc}1& 1\\ \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
Hence $\Delta =abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$