If a + x = b + y = c + z + 1, where a, b, c, x, y, z
are non-zero distinct real numbers, then
\[\left| {xyza+yb+yc+yx+ay+bz+c } \right|\] is equal to :
Solutions · 1
Official solution
\[\left| {xyza+yb+yc+yx+ay+bz+c } \right|\]
C3 \[ \to \] C3 -- C1
= \[\left| {xyza+yb+yc+yabc } \right|\]
C2 \[ \to \] C2 -- C3
= \[\left| {xyzyyyabc } \right|\]
R3 \[ \to \] R3 -- R1, R2 \[ \to \] R2 -- R1
= \[\left| {xy−xz−xy00ab−ac−a } \right|\]
= (--y)[(y -- x) (c -- a) -- (b -- a) (z -- x)]
Given, a + x = b + y = c + z + 1
= (--y)[(a -- b) (c -- a) + (a -- b) (a -- c -- 1)]
= (--y)[(a -- b) (c -- a) + (a -- b) (a -- c) + b -- a)
= --y(b -- a) = y(a -- b)