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 solve the problem, we need to analyze the given system of equations and determine the conditions under which it has a unique solution. The system of equations is:

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

### Step 1: Write the Matrix of Coefficients
We can represent the system of equations in matrix form \( A \mathbf{x} = \mathbf{0} \), where \( A \) is the matrix of coefficients:

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

### Step 2: Find the Determinant of the Matrix
To determine if the system has a unique solution, we need to calculate the determinant of matrix \( A \) and set the condition that it should not equal zero.

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

### Step 3: Calculate the Determinant
Using the determinant formula for a 3x3 matrix:

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

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

Now substituting back into the determinant:

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

### Step 4: Set the Condition for Unique Solution
For the system to have a unique solution, the determinant must not be equal to zero:

\[ 1 - \lambda^2 \neq 0 \]
\[ \lambda^2 \neq 1 \]
\[ \lambda \neq 1 \quad \text{and} \quad \lambda \neq -1 \]

### Step 5: Identify Values of \( a \) and \( b \)
From the conditions derived, we see that \( a = 1 \) and \( b = -1 \).

### Step 6: Calculate \( a^2 + b^2 \)
Now we compute \( a^2 + b^2 \):

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

### Final Answer
Thus, the value of \( a^2 + b^2 \) is \( \boxed{2} \).