Question

The system of linear equations \( 3x + y - z = 2, x - z = 1 \) and \( 2x + 2y + az = 5 \) has unique solution when:

• A. \( a \neq 3 \)
• B. \( a \neq 4 \)
• C. \( a \neq 5 \)
• D. \( a \neq 2 \)
• E. \( a \neq 1 \)

Hint:

For a system \(AX = B\), a unique solution exists if and only if \(|A| \neq 0\). If \(|A| = 0\), the system may have either no solution or infinitely many solutions.

Answer 1

The Correct Option is D

To determine when the given system of linear equations has a unique solution, we need to analyze the conditions for the existence and uniqueness of solutions in a linear system. The system of equations given is:

1. \(3x + y - z = 2\) 
2. \(x - z = 1\)
3. \(2x + 2y + az = 5\)

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

The coefficient matrix \(A\) for the given system is:

\(A = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 0 & -1 \\ 2 & 2 & a \end{bmatrix}\)

To find the determinant of matrix \(A\), use the formula for the determinant of a 3x3 matrix:

\(\text{det}(A) = 3(0 \cdot a - (-1) \cdot 2) - 1(1 \cdot a - (-1) \cdot 2) + (-1)(1 \cdot 2 - 0 \cdot 2)\)

Simplifying, we get:

\(\text{det}(A) = 3(2) - 1(a + 2) - 2\)

\(\text{det}(A) = 6 - a - 2 - 2\)

\(\text{det}(A) = 6 - a - 4\)

\(\text{det}(A) = 2 - a\)

The system will have a unique solution if \(\text{det}(A) \neq 0\).

Thus, \(2 - a \neq 0 \implies a \neq 2\).

Therefore, the system of equations has a unique solution when \(a \neq 2\). This matches the option:

Correct Answer: \(a \neq 2\)