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  • Using properties of determinants, prove that det x x^2 1+px^3 y y^2 1+py^3 z z^2 1+pz^3 = (1+pxyz)(x-y)(y-z)(z-x), where p is any scalar.

Q : 12        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),  where p is any scalar.

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Given the determinant \triangle = \begin{vmatrix} x & x^2&1+px^3 \\ y& y^2& 1+py^3\\ z&z^2 & 1+pz^3 \end{vmatrix}

Applying the row transformations; R_{2} \rightarrow R_{2} - R_{1}  and R_{3} \rightarrow R_{3} - R_{1} then we have;

\triangle = \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}

Applying row transformation R_{3} \rightarrow R_{3} - R_{2} we have then;

\triangle =(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}

Now we can expand the remaining determinant to get the result;

\triangle =(y-x )(z-x)(z-y)[(-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.

Posted by

Divya Prakash Singh

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