Question

Using properties of determinants,prove that:
\(\begin{vmatrix} x & x^2 & 1+px^3\\ y & y^2 & 1+py^3\\z&z^2&1+pz^3 \end{vmatrix}\)\(=(1+pxyz)(x-y)(y-z)(z-x)\)

Answer 1

\(Δ=\)\(\begin{vmatrix} x & x^2 & 1+px^3\\ y & y^2 & 1+py^3\\z&z^2&1+pz^3 \end{vmatrix}\)
Applying \(R_2\rightarrow R_2-R_1\) and \(R_3\rightarrow R_3-R_1\),we have
=\(\begin{vmatrix} x & x^2 & 1+px^3\\ y-x & y^2-x^2 & p(y^3-x^3)\\z-x&z^2-x^2&p(z^3-x^3)\end{vmatrix}\)
\(=(y-x)(z-x)\)\(\begin{vmatrix} x & x^2 & 1+px^3\\ 1 & y+x & p(y^2+x^2+xy)\\1&z+x&p(z^2+x^2+xz)\end{vmatrix}\)

Applying \(R_3\rightarrow R_3-R_2\),we have:
\( Δ=(y-x)(z-x)\)\(\begin{vmatrix} x & x^2 & 1+px^3\\ 1& y+x & p(y^2+x^2+xy)\\0&z-y&p(z-y)(x+y+z)\end{vmatrix}\)
\(=(y-x)(z-x)(z-y)\)\(\begin{vmatrix} x & x^2 & 1+px^3\\ 1& y+x & p(y^2+x^2+xy)\\0&1&p(x+y+z)\end{vmatrix}\)

Expanding along \(R_3\),we have:
\(Δ=(x-y)(y-z)(z-x)[(-1)(p)(xy^2+x^3+x^2y)+1+px^3+p(x+y+z)(xy)]\)
\(=(x-y)(y-z)(z-x)[-pxy^2-px^3-px^2y+1+px^3+px^2y+pxy^2+pxyz]\)
\(=(x-y)(y-z)(z-x)(1+pxyz)\)

Hence,the given result is proved.