if x,y,z are diff (det) A = x x2 1+x3 y y2 1+y3 = 0 , - Maths - Determinants

Question

if x,y,z are diff & (det) A = x x2
1+x3

y y2 1+y3 = 0 , then show that 1+xyz = 0
?
z z2 1+z3

Ankush Jain · Accepted Answer

We have,

$| \begin{matrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{matrix} | = | \begin{matrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{matrix} | + | \begin{matrix} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{matrix} | [Since each element of third column is sum of two elements] = | \begin{matrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{matrix} | + x y z | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} | [Taking x, y and z {common from C}_{1}, C_{2} and C_{3} in second det.]$

$= - | \begin{matrix} x & 1 & x^{2} \\ y & 1 & y^{2} \\ z & 1 & z^{2} \end{matrix} | + x y z | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} | [{Interchanging C}_{2} and C_{3} in first det.] = | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} | + x y z | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} | [{Interchanging C}_{1} and C_{2} in first det.] = | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} | (1 + x y z) = | \begin{matrix} 1 & x & x^{2} \\ 0 & y - x & y^{2} - x^{2} \\ 0 & z - x & z^{2} - x^{2} \end{matrix} | (1 + x y z) [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = (y - x) (z - x) | \begin{matrix} 1 & x & x^{2} \\ 0 & 1 & y + x \\ 0 & 1 & z + x \end{matrix} | (1 + x y z) [Taking (y - x) and (z - x) common R_{2} and R_{3} resp.]$

$= (y - x) (z - x) .1. | \begin{matrix} 1 & y + x \\ 1 & z + x \end{matrix} | (1 + x y z) [{Expanding along C}_{1}] = (y - x) (z - x) (z + x - y - x) (1 + x y z) = (y - x) (z - x) (z - y) (1 + x y z) = (x - y) (y - z) (z - x) (1 + x y z) ∴ | \begin{matrix} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{matrix} | = 0$

$\Rightarrow (x - y) (y - z) (z - x) (1 + x y z) = 0 \Rightarrow 1 + x y z = 0 [∵ x \neq y \neq z \Rightarrow x - y \neq 0, y - z \neq 0 and z - x \neq 0]$

if x,y,z are diff. & (det) A = x x2 1+x3 y y2 1+y3 = 0 , then show that 1+xyz = 0 ? z z2 1+z3

if x,y,z are diff. & (det) A = x x² 1+x³

y y² 1+y³ = 0 , then show that 1+xyz = 0 ?

z z² 1+z³