Let S1 and S2 be respectively the sets of all \[a \in \mathbb{R} - \{ 0\} \] 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
Solutions · 1
Official solution
Given system of equations
\[
\begin{aligned}
& a x+2 a y-3 a z=1 \\\\
& (2 a+1) x+(2 a+3) y+(a+1) z=2 \\\\
& (3 a+5) x+(a+5) y+(a+2) z=3 \\\\
& \text { Let } A=\left|\[\begin{array}{|ccccccc|}a & 2 a & -3 a \\ \\ 2 a+1 & 2 a+3 & a+1 \\ \\ 3 a+5 & a+5 & a+2\end{array}\end{aligned}\]\right| \\\\
\& =a\left|\[12a+13a+501−2a−5a−507a+410a+17\]\right| \\\\
\& =a\left(15 a\^{2}+31 a+37\right) \\\\
\& \text { Now } A=0 \\\\
\& \Rightarrow a=0
\]
So, S1=R−{0} and at a=0
System has infinite solution but a∈R−{0}∴S2=Φ