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Question

Mathematics Question on Solution of a Linear Equation

Let S1S_1 and S2S_2 be respectively the sets of all aR0a \in R -\\{0\\} for which the system of linear equations
ax+2ay3az=1a x+2 a y-3 a z=1
(2a+1)x+(2a+3)y+(a+1)z=2(2 a+1) x+(2 a+3) y+(a+1) z=2
(3a+5)x+(a+5)y+(a+2)z=3(3 a+5) x+(a+5) y+(a+2) z=3
has unique solution and infinitely many solutions. Then

A

S1=ΦS_1=\Phi and S2=R0S_2=R-\\{0\\}

B

S1S_1 is an infinite set and n(S2)=2n\left(S_2\right)=2

C

S1=R0S_1=R-\\{0\\} and S2=ΦS_2=\Phi

D

n(S1)=2n \left( S _1\right)=2 and S2S _2 is an infinite set

Answer

S1=R0S_1=R-\\{0\\} and S2=ΦS_2=\Phi

Explanation

Solution

The correct answer is (C) : S1=R0S_1=R-\\{0\\} and S2=ΦS_2=\Phi
Δ=∣∣​a2a+13a+5​2a2a+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