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Problem 33: Prove that

\begin{align*} \Delta =\left| \begin{array}{cccc} x& a& a& a\\ a& x& a& a\\ a& a& x& a\\ a& a& a& x\\ \end{array}\right| = (x+3a)(x-a)^3. \end{align*}

Problem 33: Prove that

\begin{align*} \Delta =\left| \begin{array}{cccc} x& a& a& a\\ a& x& a& a\\ a& a& x& a\\ a& a& a& x\\ \end{array}\right| = (x+3a)(x-a)^3. \end{align*}

Solution:

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

\begin{align*} \Delta =\left| \begin{array}{cccc} x+3a& a& a& a\\ x+3a& x& a& a\\ x+3a& a& x& a\\ x+3a& a& a& x\\ \end{array}\right| \end{align*} By taking $x+3a$ common from the $C_1$, \begin{align*} \Delta =(x+3a)\left| \begin{array}{cccc} 1& a& a& a\\ 1& x& a& a\\ 1& a& x& a\\ 1& a& a& x\\ \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 =(x+3a)\left| \begin{array}{cccc} 1& a& a& a\\ 0& x-a& 0& 0\\ 0& 0& x-a& 0\\ 0& 0& 0& x-a\\ \end{array}\right|\\ \end{align*} On exapanding the determinant along the $C_1$ \begin{align*} \Delta =(x+3a)\left| \begin{array}{ccc} x-a& 0& 0\\ 0& x-a& 0\\ 0& 0& x-a\\ \end{array}\right|\\ = (x+3a)(x-a)^3 \end{align*}
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