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Problem 34: Prove that

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

Problem 34: Prove that

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

Solution:

Applying $C_1\rightarrow C_1+C_2+C_3+C_4$

\begin{align*} \Delta =\left| \begin{array}{cccc} 1+a+b+c+d& b& c& d\\ 1+a+b+c+d& 1+b& c& d\\ 1+a+b+c+d& b& 1+c& d\\ 1+a+b+c+d& b& c& 1+d\\ \end{array}\right| \end{align*} By taking $x+3a$ common from the $C_1$, \begin{align*} \Delta =(1+a+b+c+d)\left| \begin{array}{cccc} 1& b& c& d\\ 1& 1+b& c& d\\ 1& b& 1+c& d\\ 1& b& c& 1+d\\ \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 =(1+a+b+c+d)\left| \begin{array}{cccc} 1& b& c& 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 =(1+a+b+c+d)\left| \begin{array}{ccc} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ \end{array}\right|\\ \Delta = 1+a+b+c+d. \end{align*}

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