Question

Consider the following system of equations
\(\alpha x + 2y + z = 1\)
\(2\alpha x + 3y + z = 1\)
\(3x + \alpha y + 2z = b\)
For some \(\alpha,\beta∈R\) then which of the following is NOT correct?

• A. It has no solution if α = – 1 and β≠2
• B. It has no solution if α ≠– 1 and β=2
• C. It has no solution if α =3 and for all β≠2
• D. It has no solution if α =– 1 and for all β ∈ R

Answer 1

The Correct Option is C

Step 1: Representing the system of equations as an augmented matrix. 

The system can be written as:

\[ \begin{pmatrix} \alpha & 2 & 1 & | & 1 \\ 2\alpha & 3 & 1 & | & 1 \\ 3 & \alpha & 2 & | & \beta \end{pmatrix} \]

Step 2: Determining the determinant of the coefficient matrix.

We denote the determinant of the coefficient matrix as \( D \). Using the determinant formula for a 3x3 matrix:

\[ D = \begin{vmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{vmatrix} \]

Expanding along the first row:

\[ D = \alpha \begin{vmatrix} 3 & 1 \\ \alpha & 2 \end{vmatrix} - 2 \begin{vmatrix} 2\alpha & 1 \\ 3 & 2 \end{vmatrix} + 1 \begin{vmatrix} 2\alpha & 3 \\ 3 & \alpha \end{vmatrix} \]

Simplifying each determinant:

\[ D = \alpha \left(3 \cdot 2 - 1 \cdot \alpha \right) - 2 \left(2\alpha \cdot 2 - 1 \cdot 3\right) + 1 \left(2\alpha \cdot \alpha - 3 \cdot 3\right) \]

\(D = \alpha(6 - \alpha) - 2(4\alpha - 3) + (2\alpha^2 - 9)\)

\(D = \alpha(6 - \alpha) - 2(4\alpha - 3) + 2\alpha^2 - 9\)

\(D = \alpha^2 - 6\alpha - 8\alpha + 6 + 2\alpha^2 - 9\)

\(D = 3\alpha^2 - 14\alpha - 3\)

Step 3: Solving for \( D = 0 \).

To find the values of \( \alpha \) that make the determinant zero, solve:

\( 3\alpha^2 - 14\alpha - 3 = 0 \)

Using the quadratic formula:

\( \alpha = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \)

\( \alpha = \frac{14 \pm \sqrt{196 + 36}}{6} \)

\( \alpha = \frac{14 \pm \sqrt{232}}{6} \)

\( \alpha = 14 \pm 15.23 \)

Thus, the two roots are \( \alpha = 4.54 \) and \( \alpha = -0.23 \).

Step 4: Analyzing the conditions for solutions.

If \( \alpha = -1 \), substitute into the matrix:

\( D = 3(-1)^2 - 14(-1) - 3 = 3 + 14 - 3 = 14 \neq 0 \)

So, the system has a unique solution for \( \alpha = -1 \), and the second equation must hold for \( \beta = 2 \) to ensure consistency.

Step 5: Conclusion.

From the above analysis, the correct answer is option (2), as it claims that there is no solution for \( \alpha = -1 \) and for all \( \beta \), which is incorrect. The system has a solution for \( \alpha = -1 \), but only when \( \beta = 2 \).