Q : 12        By using properties of determinants, show that:

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

Answers (1)

Give determinant  \begin{vmatrix} 1 &x &x^2 \\ x^2 &1 &x \\ x &x^2 &1 \end{vmatrix}

Applying column transformation C_{1} \rightarrow C_{1}+C_{2}+C_{3} we get;

\triangle = \begin{vmatrix} 1+x+x^2 &x &x^2 \\ x^2+1+x &1 &x \\ x+x^2+1 &x^2 &1 \end{vmatrix}

= (1+x+x^2)\begin{vmatrix} 1 &x &x^2 \\ 1 &1 &x \\ 1 &x^2 &1 \end{vmatrix}      [after taking the (1+x+x) factor common out.]

Now, applying row transformations, R_{1} \rightarrow R_{1}-R_{2}    and then R_{2} \rightarrow R_{2}-R_{3}.

we have now,

= (1+x+x^2)\begin{vmatrix} 0 &x-1 &x^2-x \\ 0 &1-x^2 &x-1 \\ 1 &x^2 &1 \end{vmatrix}

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

= (1+x+x^2)((x-1)^2-x(x-1)(1-x^2))

= (1+x+x^2)(x-1)(x^3-1) = (x^3-1)^2

As we know \left [\because (1+x+x^2)(x-1) = (x^3-1) \right ]

Hence proved.

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