Using properties of determinants, prove that \begin{vmatrix} a^2 + 2a & 2a + 1 & 1\\ 2a +1 &a +2 &1 \\ 3 & 3 & 1 \end{vmatrix} = (a-1)^3

 

 

 

 
 
 
 
 

Answers (1)

\begin{vmatrix} a^2 + 2a & 2a + 1 & 1\\ 2a +1 &a +2 &1 \\ 3 & 3 & 1 \end{vmatrix} = (a-1)^3

R_1 \rightarrow R_1 - R_3\qquad R_2 \rightarrow R_2 - R_3

LHS = \begin{vmatrix} a^2 + 2a - 3 & 2a -2 & 0\\ 2a -2 &a -1 &0 \\ 3 & 3 & 1 \end{vmatrix}

              \Rightarrow \begin{vmatrix} (a+3)(a-1) & 2(a - 1) & 0\\ 2(a -1) &a -1 &0 \\ 3 & 3 & 1 \end{vmatrix}

                \Rightarrow (a-1)^2 \begin{vmatrix} a+3 & 2 & 0\\ 2 &1 &0 \\ 3 & 3 & 1 \end{vmatrix}

                \Rightarrow (a-1)^2 [1(a+3 - 4)]

                \Rightarrow (a-1)^2 [a-1]\Rightarrow (a-1)^3 = RHS

Hence Proved.

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