Question

The system of linear equations in x₁, x₂, x₃
\(\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & \alpha \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 3 \\ 1 \\ \beta \end{pmatrix}\)
where α, β ∈ R, has

• A. at least one solution for any α and β
• B. a unique solution for any β when α ≠ 1
• C. NO solution for any α when β ≠ 5
• D. infinitely many solutions for any α when β = 5

Answer 1

The Correct Option is B

To determine the solution nature of the given system of linear equations with the matrix form: 

\(\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & \alpha \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ \beta \end{pmatrix}\)

we will analyze the possible scenarios based on the values of \( \alpha \) and \( \beta \).

1. The matrix of coefficients is:

\(\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 2 & 3 & \alpha \end{pmatrix}\)

1. The augmented matrix is:

\(\begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 0 & -1 & 1 & | & 1 \\ 2 & 3 & \alpha & | & \beta \end{pmatrix}\)

1. We perform row operations to reduce the augmented matrix to row-echelon form:
   1. Subtract 2 times the first row from the third row:

\(\begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 0 & -1 & 1 & | & 1 \\ 0 & 1 & \alpha - 2 & | & \beta - 6 \end{pmatrix}\)

1. Add the second row to the third row:

\(\begin{pmatrix} 1 & 1 & 1 & | & 3 \\ 0 & -1 & 1 & | & 1 \\ 0 & 0 & \alpha - 1 & | & \beta - 5 \end{pmatrix}\)

1. For the system to have a unique solution, the last row of the matrix must not create a contradiction, i.e., \(\alpha - 1 \neq 0\). Therefore, the condition is \(\alpha \neq 1\).
2. If \(\alpha = 1\), the last row becomes \(0 = \beta - 5\), implying \(\beta = 5\). In this case, there are infinitely many solutions when \(\alpha = 1\) and \(\beta = 5\).
3. Conclusion: Given the analysis, the correct statement is: "A unique solution for any \(\beta\) when \(\alpha \neq 1\)" due to the independence of rows for \(\alpha \neq 1\).