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

∣∣∣a -1 -1∣∣∣
∣∣∣-1 b -1∣∣∣ = 0
∣∣∣-1 -1 c∣∣∣
R3 → R3 - R2
∣∣∣a -1 -1∣∣∣
∣∣∣-1 b -1∣∣∣
∣∣∣0 -1-b c+1∣∣∣
R2 → R2 - R1
∣∣∣a -1 -1∣∣∣
∣∣∣0 b+1 0∣∣∣
∣∣∣0 -1-b c+1∣∣∣
⇒∣∣∣a -1 -1∣∣∣
∣∣∣0 b+1 0∣∣∣ = 0
∣∣∣0 -1-b c+1∣∣∣
(a)(b+1)(c+1) - (a)(0)(-1-b) + (-1)(0)(-1-b) - (-1)(b+1)(0) = 0
(a)(b+1)(c+1) = 0
a(bc + b + c + 1) = 0
Since a ≠ 0, bc + b + c + 1 = 0
If x = 1, y = 1, z = 1 is not the solution of above equation
The given equation is
ax = y + z
by = z + x
cz = x + y
Adding the equations:
ax + by + cz = 2(x + y + z)
If x = y = z = 1, then a + b + c = 2
If the system has a non-trivial solution, then
∣∣∣a -1 -1∣∣∣ = 0
∣∣∣-1 b -1∣∣∣
∣∣∣-1 -1 c∣∣∣
Expanding the determinant, we get:
a(bc - 1) + 1(-c - 1) - 1(1 + b) = 0
abc - a - c - 1 - b - 1 = 0
abc - a - b - c = 2
abc - (a + b + c) = 2
Since abc - (a + b + c) = 0, we have abc = a + b + c
However, if we consider the case where x = y = z = k, we have:
ak = 2k
bk = 2k
ck = 2k
This implies a = b = c = 2, which is not always true.
Let's use the determinant:
a(bc - 1) - (-c - 1) - (1 + b) = 0
abc - a + c + 1 - 1 - b = 0
abc - a - b - c = 0
abc = a + b + c
Therefore, ab + bc + ca = -1