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Using properties of determinants prove that:

contexto.140

Solution:

\begin{array}{l} \left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ =\left|\begin{array}{ccc} a^{2}-1 & a-1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right|\left[R_{1}^{\prime}=R_{1}-R_{2} \& R_{2}^{\prime}=R_{2}-R_{3}\right] \\ =\left|\begin{array}{ccc} a^{2}-1 & a-1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right| \\ =(a-1)^{2}\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right|\left[R_{1}^{\prime}=R_{1} /(a-1) \& R_{2}^{\prime}=R_{2} /(a-1)\right] \\ =(a-1)^{2}[a+1-0-2][e x p a n \operatorname{sion} b y \text { first row }] \\ =(a-1)^{3} \end{array}