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.