Question
Mathematics Question on Solution of a Linear Equation
Let S1 and S2 be respectively the sets of all a∈R−0 for which the system of linear equations
ax+2ay−3az=1
(2a+1)x+(2a+3)y+(a+1)z=2
(3a+5)x+(a+5)y+(a+2)z=3
has unique solution and infinitely many solutions. Then
A
S1=Φ and S2=R−0
B
S1 is an infinite set and n(S2)=2
C
S1=R−0 and S2=Φ
D
n(S1)=2 and S2 is an infinite set
Answer
S1=R−0 and S2=Φ
Explanation
Solution
The correct answer is (C) : S1=R−0 and S2=Φ
Δ=∣∣a2a+13a+52a2a+3a+5−3aa+1a+2∣∣
=a(15a2+31a+36)=0⇒a=0
Δ=0 for all a∈R−{0}
Hence S1=R−{0}S2=Φ