Question

Using properties of determinants, prove the following:
$\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|=ab+bc+ca+abc$

Answer 1

LHS= $\Delta =\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b1& 1\\ 1& 1& 1+c\end{array}\right|$
Taking out a, b, c common from 1, 2 and 3 row respectively, we get
$\Delta =abc\left|\begin{array}{ccc}\frac{1}{a}+1& \frac{1}{a}& \frac{1}{a}\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
Applying $R_1\to R_1+R_2+R_3$
$\Delta =abc\left|\begin{array}{ccc}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1& \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1& \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
$=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$
Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$, we get
$\Delta =abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)\left|\begin{array}{ccc}1& 0& 0\\ \frac{1}{b}& 1& 0\\ \frac{1}{c}& 0& 1\end{array}\right|$
$=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)\times (1\times 1\times 1)$ ($\therefore$ the determinant of a triangular matrix is the product of its diagonal elements.)
$=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)=abc\left(\frac{bc+ac+ca+abc}{abc}\right)=ab+bc+ca+abc$ = RHS
Hence proved.