JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 5)
If a + x = b + y = c + z + 1, where a, b, c, x, y, z
are non-zero distinct real numbers, then
$$\left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} & {z + c} \cr } } \right|$$ is equal to :
are non-zero distinct real numbers, then
$$\left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} & {z + c} \cr } } \right|$$ is equal to :
y(b – a)
y(a – b)
y(a – c)
0
Explanation
$$\left| {\matrix{
x & {a + y} & {x + a} \cr
y & {b + y} & {y + b} \cr
z & {c + y} & {z + c} \cr
} } \right|$$
C3 $$ \to $$ C3 – C1
= $$\left| {\matrix{ x & {a + y} & a \cr y & {b + y} & b \cr z & {c + y} & c \cr } } \right|$$
C2 $$ \to $$ C2 – C3
= $$\left| {\matrix{ x & y & a \cr y & y & b \cr z & y & c \cr } } \right|$$
R3 $$ \to $$ R3 – R1, R2 $$ \to $$ R2 – R1
= $$\left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr {z - x} & 0 & {c - a} \cr } } \right|$$
= (–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)
C3 $$ \to $$ C3 – C1
= $$\left| {\matrix{ x & {a + y} & a \cr y & {b + y} & b \cr z & {c + y} & c \cr } } \right|$$
C2 $$ \to $$ C2 – C3
= $$\left| {\matrix{ x & y & a \cr y & y & b \cr z & y & c \cr } } \right|$$
R3 $$ \to $$ R3 – R1, R2 $$ \to $$ R2 – R1
= $$\left| {\matrix{ x & y & a \cr {y - x} & 0 & {b - a} \cr {z - x} & 0 & {c - a} \cr } } \right|$$
= (–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)
Comments (0)
