1 Using properties of determinants, prove the following: | x y z x2 y2 z2 x3 y3 z3 | - Maths - Determinants

Question

1 Using properties of determinants, prove the following:
| x y z
x2 y2 z2
x3 y3 z3 | = xyz(x - y)(y -
z)(z - x)
2 Using properties of determinants, prove the following :
| x x2 1+px3
y y2 1+py3
z z2 1+pz3 | = (1+ pxyz)(x - y)(y - z)(z -
x)

Ankush Jain · Accepted Answer

1 $T o p r o v e : | \begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{3} & y^{3} & z^{3} \end{matrix} | = x y z (x - y) (y - z) (z - x) \Pr o o f : L H S = | \begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{3} & y^{3} & z^{3} \end{matrix} | = x y z | \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^{2} & y^{2} & z^{2} \end{matrix} | [t a k i n g x, y a n d z c o m m o n f r o m C_{1}, C_{2} a n d C_{3} r e s p e c t i v e l y] = x y z | \begin{matrix} 1 - 1 & 1 - 1 & 1 \\ x - y & y - z & z \\ x^{2} - y^{2} & y^{2} - z^{2} & z^{2} \end{matrix} | [o p e r a t i n g C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}] = x y z | \begin{matrix} 0 & 0 & 1 \\ x - y & y - z & z \\ (x + y) (x - y) & (y + z) (y - z) & z^{2} \end{matrix} | = x y z (x - y) (y - z) | \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & z \\ (x + y) & (y + z) & z^{2} \end{matrix} | [t a k i n g (x - y) a n d (y - z) c o m m o n f r o m C_{1} a n d C_{2}] = x y z (x - y) (y - z) \times 1 | \begin{matrix} 1 & 1 \\ x + y & y + z \end{matrix} | [\exp a n d i n g a l o n g R_{1}] = x y z (x - y) (y - z) \times 1 (y + z - x - y) = x y z (x - y) (y - z) (z - x) = R H S [H e n c e \Pr o v e d]$ 2 To prove:

Proof: Let,

Expanding along R3, we have:

1. Using properties of determinants, prove the following: | x y z x2 y2 z2 x3 y3 z3 | = xyz(x - y)(y - z)(z - x) . 2. Using properties of determinants, prove the following : | x x2 1+px3 y y2 1+py3 z z2 1+pz3 | = (1+ pxyz)(x - y)(y - z)(z - x) .

1. Using properties of determinants, prove the following:

| x y z

x² y² z²

x³ y³ z^{3 | =}xyz(x - y)(y - z)(z - x) .

2. Using properties of determinants, prove the following :

| x x² 1+px³

y y² 1+py³

z z² 1+pz³ | = (1+ pxyz)(x - y)(y - z)(z - x) .