JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 13)
Let S$$_1$$ and S$$_2$$ 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
$$\mathrm{n({S_1}) = 2}$$ and S$$_2$$ is an infinite set
$$\mathrm{{S_1} = \Phi} $$ and $$\mathrm{{S_2} = \mathbb{R} - \{ 0\}}$$
$$\mathrm{{S_1} = \mathbb{R} - \{ 0\}}$$ and $$\mathrm{{S_2} = \Phi} $$
S$$_1$$ is an infinite set and n(S$$_2$$) = 2
Explanation
Given system of equations
$$ \begin{aligned} & a x+2 a y-3 a z=1 \\\\ & (2 a+1) x+(2 a+3) y+(a+1) z=2 \\\\ & (3 a+5) x+(a+5) y+(a+2) z=3 \\\\ & \text { Let } A=\left|\begin{array}{ccc} a & 2 a & -3 a \\\\ 2 a+1 & 2 a+3 & a+1 \\\\ 3 a+5 & a+5 & a+2 \end{array}\right| \\\\ & =a\left|\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 a+1 & 1-2 a & 7 a+4 \\\\ 3 a+5 & -5 a-5 & 10 a+17 \end{array}\right| \\\\ & =a\left(15 a^{2}+31 a+37\right) \\\\ & \text { Now } A=0 \\\\ & \Rightarrow a=0 \end{aligned} $$
So, $S_{1}=R-\{0\}$ and at $a=0$
System has infinite solution but $a \in R-\{0\}$
$\therefore S_{2}=\Phi$
$$ \begin{aligned} & a x+2 a y-3 a z=1 \\\\ & (2 a+1) x+(2 a+3) y+(a+1) z=2 \\\\ & (3 a+5) x+(a+5) y+(a+2) z=3 \\\\ & \text { Let } A=\left|\begin{array}{ccc} a & 2 a & -3 a \\\\ 2 a+1 & 2 a+3 & a+1 \\\\ 3 a+5 & a+5 & a+2 \end{array}\right| \\\\ & =a\left|\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 a+1 & 1-2 a & 7 a+4 \\\\ 3 a+5 & -5 a-5 & 10 a+17 \end{array}\right| \\\\ & =a\left(15 a^{2}+31 a+37\right) \\\\ & \text { Now } A=0 \\\\ & \Rightarrow a=0 \end{aligned} $$
So, $S_{1}=R-\{0\}$ and at $a=0$
System has infinite solution but $a \in R-\{0\}$
$\therefore S_{2}=\Phi$
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