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Question Number 116523 by zakirullah last updated on 04/Oct/20

Answered by Olaf last updated on 04/Oct/20

$$(1+a)\left|\begin{array}{cc}1+b& 1\\ 1& 1+c\end{array}\right|-\left|\begin{array}{cc}1& 1\\ 1& 1+c\end{array}\right|+\left|\begin{array}{cc}1& 1\\ 1+b& 1\end{array}\right|$$$$$$$$=(1+a)[(1+b)(1+c)-1]$$$$-(1+c-1)+(1-1-b)$$$$$$$$=(1+a)(b+c+bc)-c-b$$$$=b+c+bc+ab+ac+abc-c-b$$$$=abc+bc+ac+ab$$$$=abc(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$

Commented by zakirullah last updated on 04/Oct/20

$$thankyousir$$

Answered by $@y@m last updated on 04/Oct/20

$$=\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right|$$$$=abc\left|\begin{array}{ccc}\frac{1}{a}+1& \frac{1}{a}& \frac{1}{a}\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$$$$=abc\left|\begin{array}{ccc}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1& \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1& \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|by{R}_1\to R_1+R_2+R_3$$$$=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{b}& \frac{1}{b}+1& \frac{1}{b}\\ \frac{1}{c}& \frac{1}{c}& \frac{1}{c}+1\end{array}\right|$$$$=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)\left|\begin{array}{ccc}0& 0& 1\\ -1& 1& \frac{1}{b}\\ 0& -1& \frac{1}{c}+1\end{array}\right|by{R}_1\to R_1-R_2\mathrm{\&}{R}_2\to R_2-R_3$$$$=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)$$$$$$

Commented by zakirullah last updated on 04/Oct/20

$$\mathrm{thanks}\mathrm{alot}$$

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