JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 14)
If the system of linear equations
2x + 2y + 3z = a
3x – y + 5z = b
x – 3y + 2z = c
where a, b, c are non zero real numbers, has more one solution, then :
2x + 2y + 3z = a
3x – y + 5z = b
x – 3y + 2z = c
where a, b, c are non zero real numbers, has more one solution, then :
b – c – a = 0
a + b + c = 0
b – c + a = 0
b + c – a = 0
Explanation
P1 : 2x + 2y + 3z = a
P2 : 3x $$-$$ y + 5z = b
P3 : x $$-$$ 3y + 2z = c
We find
P1 + P3 = P2 $$ \Rightarrow $$ a + c = b
P2 : 3x $$-$$ y + 5z = b
P3 : x $$-$$ 3y + 2z = c
We find
P1 + P3 = P2 $$ \Rightarrow $$ a + c = b
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