If a, b, c are non-zero real numbers and if the system of equations. (a - 1)x = y + z, (b - 1)y = z + x, (c - 1)z = x + y, has a non-trivial solution, then ab + bc + ca equals:

Question

If a, b, c are non-zero real numbers and if the system of equations. (a - 1)x = y + z, (b - 1)y = z + x, (c - 1)z = x + y, has a non-trivial solution, then ab + bc + ca equals: (A) a + b + c (B) abc (C) 1 (D) -1

Tardigrade Admin · Accepted Answer

for non-trivial solution D =0 |1- a 1 1 1 1- b 1 1 1 1- c |=0 R 1 arrow R 1- R 3 |- a 0 c 1 1- b 1 1 1 1- c | ⇒ a (1-b)(1-c)-1 +c 1-(1-b) =0 ⇒ a b+a c+b c-a b c=0 ⇒ a b+a c+b c=a b c

Q. If a, b, c are non-zero real numbers and if the system of equations.
$(a - 1) x = y + z$ ,
$(b - 1) y = z + x$ ,
$(c - 1) z = x + y$ ,
has a non-trivial solution, then $ab + b c + c a$ equals:

a + b + c

abc

1

-1

for non-trivial solution $D = 0$
$∣ ∣ 1 - a 11 1 1 - b 1 11 1 - c ∣ ∣ = 0$
$R_{1} \to R_{1} - R_{3}$
$∣ ∣ - a 11 0 1 - b 1 c 1 1 - c ∣ ∣$
$\Rightarrow$ a ${(1 - b) (1 - c) - 1} + c {1 - (1 - b)} = 0$
$\Rightarrow ab + a c + b c - ab c = 0$
$\Rightarrow ab + a c + b c = ab c$

Q. If a, b, c are non-zero real numbers and if the system of equations. (a−1)x=y+z, (b−1)y=z+x, (c−1)z=x+y, has a non-trivial solution, then ab+bc+ca equals:

a + b + c

abc

1

-1

for non-trivial solution D=0 ∣∣​1−a11​11−b1​111−c​∣∣​=0 R1​→R1​−R3​ ∣∣​−a11​01−b1​c11−c​∣∣​ ⇒ a {(1−b)(1−c)−1}+c{1−(1−b)}=0 ⇒ab+ac+bc−abc=0 ⇒ab+ac+bc=abc

Q. If a, b, c are non-zero real numbers and if the system of equations.
$(a - 1) x = y + z$ ,
$(b - 1) y = z + x$ ,
$(c - 1) z = x + y$ ,
has a non-trivial solution, then $ab + b c + c a$ equals:

for non-trivial solution $D = 0$
$∣ ∣ 1 - a 11 1 1 - b 1 11 1 - c ∣ ∣ = 0$
$R_{1} \to R_{1} - R_{3}$
$∣ ∣ - a 11 0 1 - b 1 c 1 1 - c ∣ ∣$
$\Rightarrow$ a ${(1 - b) (1 - c) - 1} + c {1 - (1 - b)} = 0$
$\Rightarrow ab + a c + b c - ab c = 0$
$\Rightarrow ab + a c + b c = ab c$