Question

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations $ a x+2 a y-3 a z=1$ $ (2 a+1) x+(2 a+3) y+(a+1) z=2$ $(3 a+5) x+(a+5) y+(a+2) z=3$ has unique solution and infinitely many solutions Then

• A. $S_1=\Phi$ and $S_2=R-\{0\}$
• B. $S_1$ is an infinite set and $n\left(S_2\right)=2$
• C. $S_1=R-\{0\}$ and $S_2=\Phi$
• D. $n \left( S _1\right)=2$ and $S _2$ is an infinite set

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the given system of linear equations and determine the conditions under which it has a unique solution or infinitely many solutions.

The system of equations given is:

• \(a x + 2a y - 3a z = 1\)
• \((2a + 1) x + (2a + 3) y + (a + 1) z = 2\)
• \((3a + 5) x + (a + 5) y + (a + 2) z = 3\)

We first write the system in matrix form:

\(\begin{bmatrix} a & 2a & -3a \\ 2a + 1 & 2a + 3 & a + 1 \\ 3a + 5 & a + 5 & a + 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\)

For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. Calculate the determinant:

\(\text{det} = \begin{vmatrix} a & 2a & -3a \\ 2a + 1 & 2a + 3 & a + 1 \\ 3a + 5 & a + 5 & a + 2 \end{vmatrix}\)

To find the determinant, use cofactor expansion along the first row:

\(\text{det} = a \begin{vmatrix} 2a + 3 & a + 1 \\ a + 5 & a + 2 \end{vmatrix} - 2a \begin{vmatrix} 2a + 1 & a + 1 \\ 3a + 5 & a + 2 \end{vmatrix} - 3a \begin{vmatrix} 2a + 1 & 2a + 3 \\ 3a + 5 & a + 5 \end{vmatrix}\)

Calculate each of the 2x2 determinants:

• \(\begin{vmatrix} 2a + 3 & a + 1 \\ a + 5 & a + 2 \end{vmatrix} = (2a + 3)(a + 2) - (a + 1)(a + 5) = a^2 + 4a + 1\)
• \(\begin{vmatrix} 2a + 1 & a + 1 \\ 3a + 5 & a + 2 \end{vmatrix} = (2a + 1)(a + 2) - (a + 1)(3a + 5) = -a^2 - 3a - 3\)
• \(\begin{vmatrix} 2a + 1 & 2a + 3 \\ 3a + 5 & a + 5 \end{vmatrix} = (2a + 1)(a + 5) - (2a + 3)(3a + 5) = -4a - 7\)

Then substitute these back into the determinant expression:

\(\text{det} = a(a^2 + 4a + 1) + 2a(a^2 + 3a + 3) + 3a(4a + 7)\)

Simplify:

\(= a^3 + 4a^2 + a + 2a^3 + 6a^2 + 6a + 12a^2 + 21a = 3a^3 + 22a^2 + 28a\)

Factor out the common term \(a\):

\(= a(3a^2 + 22a + 28)\)

The system will have a unique solution if \(a(3a^2 + 22a + 28) \neq 0\). As \(a \neq 0\) is given, we must have \(3a^2 + 22a + 28 \neq 0\).

Solving \(3a^2 + 22a + 28 = 0\) gives:

\(D = 22^2 - 4 \cdot 3 \cdot 28\) \(= 484 - 336 = 148\\)\(> 0,\)

Therefore, it has two distinct roots:

\(a = \frac{-22 \pm \sqrt{148}}{6}\)

Thus, \(a\) cannot be these two roots for the system to have a unique solution. Hence, \(S_1 = R - \{0\}\) because outside these roots \(0\) is already restricted by the set definition.

The set \(S_2\) is empty because for \(a\) such that \(3a^2 + 22a + 28 = 0\) (the degree of three equations = number of rows of matrix), the system cannot have infinitely many solutions. Points resulting from zero determinant typically yield no solutions or a contradiction.

We conclude:

Correct Answer: \(S_1 = R - \{0\}\) and \(S_2 = \Phi\)