Question

For \( a, b \in \mathbb{R} \), consider the system of linear equations \[ x + y + az = 2 \] \[ 2y + 2z = 1 \] \[ ax + 2z = b \] If the system has infinitely many solutions, then which of the following statements is/are correct?

• A. \( a = 2, b = 3 \)
• B. \( a = 2, b = 5 \)
• C. \( a = -1, b = -\frac{3}{2} \)
• D. \( a = 3, b = 5 \)

Hint:

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent and dependent.

Answer 1

The Correct Option is A, C

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, indicating that the system is consistent but dependent.

The coefficient matrix of the system is:

\[ \begin{bmatrix} 1 & 1 & a \\ 0 & 2 & 2 \\ a & 0 & 2 \end{bmatrix} \]

We can compute the determinant of this matrix to find the condition for infinitely many solutions:

\[ \text{det} = \begin{vmatrix} 1 & 1 & a \\ 0 & 2 & 2 \\ a & 0 & 2 \end{vmatrix} \]

Expanding the determinant:

\[ \text{det} = 1 \cdot \begin{vmatrix} 2 & 2 \\ 0 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 2 \\ a & 2 \end{vmatrix} + a \cdot \begin{vmatrix} 0 & 2 \\ a & 0 \end{vmatrix} \]

\[ \text{det} = 1 \cdot (2 \cdot 2 - 2 \cdot 0) - 1 \cdot (0 \cdot 2 - 2 \cdot a) + a \cdot (0 \cdot 0 - 2 \cdot a) \]

\[ \text{det} = 1 \cdot 4 - 1 \cdot (-2a) + a \cdot (-2a) \]

\[ \text{det} = 4 + 2a - 2a^2 \]

Setting \( \text{det} = 0 \) for infinitely many solutions:

\[ 4 + 2a - 2a^2 = 0 \]

\[ 2a^2 - 2a - 4 = 0 \]

\[ a^2 - a - 2 = 0 \]

Solving this quadratic equation:

\[ a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \]

Thus, \( a = 2 \) or \( a = -1 \).

Now, substituting these values of \( a \) back into the system, we find that for \( a = 2 \), \( b = 3 \), and for \( a = -1 \), \( b = -\frac{3}{2} \).

Thus, the correct answers are \( a = 2, b = 3 \) and \( a = -1, b = -\frac{3}{2} \).