The value of the determinant ka k2+a2 1kb k2+b2 1kc k2+c2 1, is

A
$k (a + b) (b + c) (c + a)$
B
$k a b c (a^{2} + b^{2} + c^{2})$
C
$k (a - b) (b - c) (c - a)$
D
$k (a + b - c) (b + c - a) (c + a - b)$

Solution:

We have,

$Δ = |\begin{array}{lll} k a k^{2} + a^{2} 1 \\ k b k^{2} + b^{2} 1 \\ k c k^{2} + c^{2} 1 \end{array}| = |\begin{array}{lll} k a k^{2} 1 \\ k b k^{2} 1 \\ k c k^{2} 1 \end{array}| + |\begin{array}{lll} k a a^{2} 1 \\ k b b^{2} 1 \\ k c c^{2} 1 \end{array}|$

$\Rightarrow Δ = 0 + k |\begin{array}{lll} a a^{2} 1 \\ b b^{2} 1 \\ c c^{2} 1 \end{array}| = k (a - b) (b - c) (c - a) .$

The value of the determinant $|\begin{array}{lll} k a k^{2} + a^{2} 1 \\ k b k^{2} + b^{2} 1 \\ k c k^{2} + c^{2} 1 \end{array}|,$ is

Solution:

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