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

Question

The value of the determinant |ka&k2+a2&1 kb&k2+b2&1 kc&k2+c2&1| is (A) k(a+b)(b+c)(c+a) (B) kabc (a2+b2+c2) (C) k(a-b)(b-c)(c-a) (D) k(a+b-c)(b+c-a)(c+a-b)

Tardigrade Admin · Accepted Answer

Q. The value of the determinant $∣ ∣ ka kb k c k^{2} + a^{2} k^{2} + b^{2} k^{2} + c^{2} 111 ∣ ∣$ is

$k (a + b) (b + c) (c + a)$

$kab c (a^{2} + b^{2} + c^{2})$

$k (a - b) (b - c) (c - a)$

$k (a + b - c) (b + c - a) (c + a - b)$

We have, $∣ ∣ ka kb k c k^{2} + a^{2} k^{2} + b^{2} k^{2} + c^{2} 111 ∣ ∣ = ∣ ∣ ka kb k c k^{2} k^{2} k^{2} 111 ∣ ∣ + ∣ ∣ ka kb k c a^{2} b^{2} c^{2} 111 ∣ ∣$
$= 0 + k ∣ ∣ a b c a^{2} b^{2} c^{2} 111 ∣ ∣$
$= k (a - b) (b - c) (c - a)$

Q. The value of the determinant ∣∣​kakbkc​k2+a2k2+b2k2+c2​111​∣∣​ is

k(a+b)(b+c)(c+a)

kabc(a2+b2+c2)

k(a−b)(b−c)(c−a)

k(a+b−c)(b+c−a)(c+a−b)

We have, ∣∣​kakbkc​k2+a2k2+b2k2+c2​111​∣∣​=∣∣​kakbkc​k2k2k2​111​∣∣​+∣∣​kakbkc​a2b2c2​111​∣∣​ =0+k∣∣​abc​a2b2c2​111​∣∣​ =k(a−b)(b−c)(c−a)

Q. The value of the determinant $∣ ∣ ka kb k c k^{2} + a^{2} k^{2} + b^{2} k^{2} + c^{2} 111 ∣ ∣$ is

$k (a + b) (b + c) (c + a)$

$kab c (a^{2} + b^{2} + c^{2})$

$k (a - b) (b - c) (c - a)$

$k (a + b - c) (b + c - a) (c + a - b)$

We have, $∣ ∣ ka kb k c k^{2} + a^{2} k^{2} + b^{2} k^{2} + c^{2} 111 ∣ ∣ = ∣ ∣ ka kb k c k^{2} k^{2} k^{2} 111 ∣ ∣ + ∣ ∣ ka kb k c a^{2} b^{2} c^{2} 111 ∣ ∣$
$= 0 + k ∣ ∣ a b c a^{2} b^{2} c^{2} 111 ∣ ∣$
$= k (a - b) (b - c) (c - a)$