First slide

Introduction to Determinants

Question

If $|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}| = (a - b) (b - c) (c - a) (a + b + c)$ , where a, b, c are all different, then the determinant $|\begin{array}{ccc} 1 & 1 & 1 \\ (x - a)^{2} & (x - b)^{2} & (x - c)^{2} \\ (x - b) (x - c) & (x - c) (x - a) & (x - a) (x - b) \end{array}|$ vanishes when

Difficult

A

a + b + c = 0
B

$x = \frac{1}{3} (a + b + c)$
C

$x = \frac{1}{2} (a + b + c)$
D

x = a + b + c

Solution

We have
$|\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c)$ (1)
Also, $|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}|$
$= abc |\begin{array}{ccc} \frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ 1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \end{array}|$ (taking a, b, c common from R₁, R₂, R₃)
$\begin{array}{l} = |\begin{array}{ccc} bc & ac & ab \\ 1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \end{array}| (Multiplying R_{1} by abc) \\ = |\begin{array}{ccc} 1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \\ bc & ac & ab \end{array}| \end{array}$
$\begin{matrix} Then D = |\begin{array}{ccc} 1 & 1 & 1 \\ (x - a)^{2} & (x - b)^{2} & (x - c)^{2} \\ (x - b) (x - c) & (x - c) (x - a) & (x - a) (x - b) \end{array}| \\ = (a - b) (b - c) (c - a) (3 x - a - b - c) \end{matrix}$
Now, given that a, b and c are all different. So, D = 0 when
$x = \frac{1}{3} (a + b + c)$

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