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Problem 35: Prove that

\begin{align*} \Delta =\left| \begin{array}{cccc} 1+a& 1& 1& 1\\ 1& 1+b& 1& 1\\ 1& 1& 1+c& 1\\ 1& 1& c& 1+d\\ \end{array}\right| \end{align*}

Problem 35: Prove that

\begin{align*} \Delta =\left| \begin{array}{cccc} 1+a& 1& 1& 1\\ 1& 1+b& 1& 1\\ 1& 1& 1+c& 1\\ 1& 1& c& 1+d\\ \end{array}\right| \\= abcd(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}). \end{align*}

Solution:

Taking a,b,c,d common from the $C_1, C_2, C_3$ and $C_4$ res[ectovely. we get

\begin{align*} \Delta =abcd\left| \begin{array}{cccc} \frac{1}{a}+1& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}\\ \frac{1}{a}& \frac{1}{b}+1& \frac{1}{c}& \frac{1}{d}\\ \frac{1}{a}& \frac{1}{b}& \frac{1}{c}+1& \frac{1}{d}\\ \frac{1}{a}& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}+1\\ \end{array}\right| \end{align*} applying $C_1\rightarrow C_1+C_2+C_3+C_4$, \begin{align*} \Delta =(abcd)\left| \begin{array}{cccc} (1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}\\ (1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})& \frac{1}{b}+1& \frac{1}{c}& \frac{1}{d}\\ (1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})& \frac{1}{b}& \frac{1}{c}+1& \frac{1}{d}\\ (1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}+1\\ \end{array}\right|\\ \end{align*} \begin{align*} \Delta =(abcd)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\left| \begin{array}{cccc} 1& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}\\ 1& \frac{1}{b}+1& \frac{1}{c}& \frac{1}{d}\\ 1& \frac{1}{b}& \frac{1}{c}+1& \frac{1}{d}\\ 1& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}+1\\ \end{array}\right|\\ \end{align*} By Applying $R_2 \rightarrow R_2-R_1,R_3 \rightarrow R_3-R_1,R_4 \rightarrow R_4-R_1, $ , \begin{align*} \Delta =(abcd)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\left| \begin{array}{cccc} 1& \frac{1}{b}& \frac{1}{c}& \frac{1}{d}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ \end{array}\right|\\ \end{align*} On exapanding the determinant along the $C_1$ \begin{align*} \Delta =(abcd)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\left| \begin{array}{ccc} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ \end{array}\right|\\ \Delta = (abcd)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}). \end{align*}

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