First slide

Introduction to Determinants

Question

$I f x, y, z a r e d i s t i n c t, t h e n |\begin{array}{llr} 1 + x^{3} x^{2} 1 \\ 1 + y^{3} y^{2} 1 \\ 1 + z^{3} z^{2} 1 \end{array}| = 0$ ,when xyz is

Moderate

A

1
B

2
C

-1
D

-2

Solution

Since each element of C₁ is the sum of two elements, putting the determinant as sum of two determinants, we get
$\begin{matrix} Δ = |\begin{array}{lll} x^{3} & x^{2} & x \\ y^{3} & y^{2} & y \\ z^{3} & z^{2} & z \end{array}| + |\begin{array}{lll} 1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z \end{array}| \\ = xyz |\begin{array}{lll} x^{2} & x & 1 \\ y^{2} & y & 1 \\ z^{2} & z & 1 \end{array}| + |\begin{array}{ccc} 1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z \end{array}| \\ = - (xyz + 1) |\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}| \\ = - (xyz + 1) (x - y) (y - z) (z - x) (x + y + z) \end{matrix}$
Since $Δ = 0, x, y, z$ all are distinct, wo have xyz + 1 = 0
or xyz = -1.

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