Question

By using properties of determinants, show that: $\begin{vmatrix}1&x&x^2\\\x^2&1&x\\\x&x^2&1\end{vmatrix}$=(1-x3)2

Answer 1

△=$\begin{vmatrix}1&x&x^2\\\x^2&1&x\\\x&x^2&1\end{vmatrix}$

Applying R1 → R1 + R2 + R3, we have:

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

=(1+x+x2)$\begin{vmatrix}1&1&1\\\x^2&1&x\\\x&x^2&1\end{vmatrix}$

Applying C2 → C2 − C1 and C3 → C3 − C1, we have:

△=(1+x+x2)$\begin{vmatrix}1&0&0\\\x^2&1-x^2&x-x^2\\\\-x&x^2-x&1-x\end{vmatrix}$

=(1+x+x2)(1-x)(1-x)I100 x2 1+x x x -x 1I

=(1-x3)(1-x)I100 x2 1+x x x -x 1I

Expanding along R1, we have:

△=(1-x3)(1-x)(1)I1+x x -x 1I

=(1-x3)(1-x)(1+x+x2)

=(1-x3)(1-x3)

=(1-x3)2

Hence, the given result is proved.