Question

If the system of equations
\( 3x + y + 4z = 3 \)
\( 2x + \alpha y - z = -3 \)
\( x + 2y + z = 4 \)
has no solution, then the value of \( \alpha \) is equal to :

• A. 19
• B. 13
• C. 4
• D. 23

Hint:

For a \( 3 \times 3 \) system, \( \Delta = 0 \) is a necessary condition for both "no solution" and "infinitely many solutions".

Answer 1

The Correct Option is A

To determine the value of \(\alpha\) for which the given system of equations has no solution, we need to analyze the condition for inconsistency in a system of linear equations. The system of equations provided is:

1. \(3x + y + 4z = 3\)
2. \(2x + \alpha y - z = -3\) 
3. \(x + 2y + z = 4\)

For a system of linear equations to have no solution, the determinant of the coefficient matrix must be zero and the rank of the coefficient matrix should not be equal to the augmented matrix.

The coefficient matrix \(A\) and its determinant \(\text{det}(A)\) are:

Coefficient Matrix \( A \)
\(\begin{pmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{pmatrix}\)

The determinant of matrix \(A\) is given by:

\(\begin{align*} \text{det}(A) &= 3\left(\alpha \cdot 1 - (-1) \cdot 2\right) - 1\left(2 \cdot 1 - (-1) \cdot 1\right) + 4\left(2 \cdot 2 - \alpha \cdot 1\right) \\ &= 3(\alpha + 2) - (2 + 1) + 4(4 - \alpha) \\ &= 3\alpha + 6 - 3 + 16 - 4\alpha \\ &= -\alpha + 19 \end{align*}\)

For the system to have no solution, set the determinant to zero:

\(-\alpha + 19 = 0\)

Solving for \(\alpha\) gives:

\(\alpha = 19\)

Thus, the value of \(\alpha\) for which the system of equations has no solution is 19.

The correct option is: 19.