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  4. The determinant D=|1 x x2 x2 1 x x x2 1| is equal to

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Q. The determinant $D=\begin{vmatrix}1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1\end{vmatrix}$ is equal to

Determinants

Solution:

Using $\quad C _1 \rightarrow C _1+ C _2+ C _3$
$D=\left(1+x+x^2\right)\begin{vmatrix}1 & x & x^2 \\ 1 & 1 & x \\ 1 & x^2 & 1\end{vmatrix}=\left(x^2+x+1\right)\begin{vmatrix}0 & x-1 & x(x-1) \\ 0 & -\left(x^2-1\right) & x-1 \\ 1 & x^2 & 1\end{vmatrix}$
$=(x-1)^2\left(x^2+x+1\right)\begin{vmatrix}0 & 1 & x \\ 0 & -(x+1) & 1 \\ 1 & x^2 & x\end{vmatrix}=(x-1)^2\left(x^2+x+1\right)[1+x(x+1)]$
$=\left(x^2+x+1\right)^2(x-1)^2=\left(x^3-1\right)^2$