Question

The value of \( \lambda \) for which the following system of equations has unique solution: \[ \lambda x+3y-z=1 \] \[ x+2y+z=2 \] \[ -\lambda x+y+2z=-1 \] are:

• A. \( \lambda\neq \frac{5}{2} \)
• B. \( \lambda\neq \frac{3}{2} \)
• C. \( \lambda\neq \frac{7}{2} \)
• D. \( \lambda\neq -\frac{7}{2} \)

Hint:

For a system of linear equations: \[ AX=B \]
• Unique solution exists when: \[ |A|\neq 0 \]
• No unique solution occurs when: \[ |A|=0 \] In determinant expansion: \[ \begin{vmatrix} a & b & c \end{vmatrix} \] always remember the sign pattern: \[ + \quad - \quad + \] while expanding along a row or column.

Answer 1

The Correct Option is A

To determine the value of \( \lambda \) for which the given system of equations has a unique solution, we must ensure the determinant of the coefficient matrix is non-zero.

The system of equations is:

\[\begin{align*} \lambda x + 3y - z &= 1 \\ x + 2y + z &= 2 \\ -\lambda x + y + 2z &= -1 \end{align*}\]

The coefficient matrix \( A \) is:

\[A = \begin{bmatrix} \lambda & 3 & -1 \\ 1 & 2 & 1 \\ -\lambda & 1 & 2 \end{bmatrix}\]

To find the values of \( \lambda \) for which this system has a unique solution, we need to compute the determinant of matrix \( A \) and set it not equal to zero:

\[\text{det}(A) = \begin{vmatrix} \lambda & 3 & -1 \\ 1 & 2 & 1 \\ -\lambda & 1 & 2 \end{vmatrix}\]

Expanding this determinant along the first row gives:

\[\text{det}(A) = \lambda \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} - 3 \begin{vmatrix} 1 & 1 \\ -\lambda & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ -\lambda & 1 \end{vmatrix}\]

Calculating the 2x2 determinants:

\[\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 2(2) - 1(1) = 4 - 1 = 3 \]