If the system of equations `{:(x-lamday-z=0),(lamdax-y-z=0),(x+y-z=0):}}` has unique solution then the range of `lamda` is `R-{a,b}` Then the value of `(a^(2)+b^(2))` is :

Options

A. 1
B. 2
C. 4
D. 9

Explanation

To determine the values of \( a \) and \( b \) such that the system of equations has a unique solution, we need to analyze the determinant of the coefficient matrix. The system of equations is given as follows:

1. \( x - \lambda y - z = 0 \)
2. \( \lambda x - y - z = 0 \)
3. \( x + y - z = 0 \)

We can represent this system in matrix form \( A \mathbf{x} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \).

The coefficient matrix \( A \) is:

\[ A = \begin{pmatrix} 1 & -\lambda & -1 \\ \lambda & -1 & -1 \\ 1 & 1 & -1 \end{pmatrix} \]

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

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

Calculating the determinant using cofactor expansion:

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

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} -1 & -1 \\ 1 & -1 \end{vmatrix} = (-1)(-1) - (-1)(1) = 1 + 1 = 2 \)

2. \( \begin{vmatrix} \lambda & -1 \\ 1 & -1 \end{vmatrix} = \lambda(-1) - (-1)(1) = -\lambda + 1 = 1 - \lambda \)

3. \( \begin{vmatrix} \lambda & -1 \\ 1 & 1 \end{vmatrix} = \lambda(1) - (-1)(1) = \lambda + 1 \)

Substituting these back into the determinant calculation:

\[ \text{det}(A) = 1 \cdot 2 + \lambda(1 - \lambda) - (\lambda + 1) \]
\[ = 2 + \lambda - \lambda^2 - \lambda - 1 \]
\[ = 2 - \lambda^2 - 1 \]
\[ = 1 - \lambda^2 \]

For the system to have a unique solution, we require:

\[ 1 - \lambda^2 \neq 0 \]

This implies:

\[ \lambda^2 \neq 1 \]

Thus, the values of \( \lambda \) that make the determinant zero are \( \lambda = 1 \) and \( \lambda = -1 \). Therefore, the range of \( \lambda \) for which the system has a unique solution is \( \mathbb{R} - \{-1, 1\} \).

Now, since \( a = -1 \) and \( b = 1 \), we need to find \( a^2 + b^2 \):

\[ a^2 + b^2 = (-1)^2 + (1)^2 = 1 + 1 = 2 \]

Thus, the final answer is:

\[ \boxed{2} \]