Find the values of λ such that the following equations have unique solution λx + 2y - 2z - 1 = 0, 4x + 2λy - z - 2 = 0, 6x + 6y + λz - 3 = 0 and use matrix method to solve these eq... Find the values of λ such that the following equations have unique solution λx + 2y - 2z - 1 = 0, 4x + 2λy - z - 2 = 0, 6x + 6y + λz - 3 = 0 and use matrix method to solve these equations when λ = 2.
Understand the Problem
The question is asking to find the values of λ such that a given set of equations has a unique solution and to use the matrix method to solve these equations when λ = 2. This involves determining conditions for unique solutions in a system of linear equations, typically related to the determinant of the coefficient matrix.
Answer
Values of \(\lambda\) for unique solutions: \(\lambda \neq 0, 2, \frac{37}{7}\). For \(\lambda = 2\): \(x = 0, y = -1, z = 1\).
Answer for screen readers
The values of (\lambda) such that the equations have a unique solution are: (\lambda \neq 0, 2, \frac{37}{7}).
When (\lambda = 2): The solutions using the matrix method yield: (x = 0, y = -1, z = 1).
Steps to Solve
- Formulate the Coefficient Matrix
The given equations can be written in the form (Ax = b), where (A) is the coefficient matrix, (x) is the vector of variables ([x, y, z]^T), and (b) is the constants vector.
The equations are:
[ \begin{align*} \lambda x + 2y - 2z &= 1 \ 4x + 2\lambda y - z &= 2 \ 6x + 6y + \lambda z &= 3 \end{align*} ]
So, the coefficient matrix (A) is:
$$ A = \begin{bmatrix} \lambda & 2 & -2 \ 4 & 2\lambda & -1 \ 6 & 6 & \lambda \end{bmatrix} $$
- Find the Determinant
For the system to have a unique solution, the determinant of matrix (A) must be non-zero:
$$ \text{det}(A) = \begin{vmatrix} \lambda & 2 & -2 \ 4 & 2\lambda & -1 \ 6 & 6 & \lambda \end{vmatrix} $$
Calculating the determinant using the rule of Sarrus or cofactor expansion will allow us to find conditions on (\lambda).
- Calculate the Determinant
Expanding along the first row:
[ \text{det}(A) = \lambda \begin{vmatrix} 2\lambda & -1 \ 6 & \lambda \end{vmatrix} - 2 \begin{vmatrix} 4 & -1 \ 6 & \lambda \end{vmatrix} - 2 \begin{vmatrix} 4 & 2\lambda \ 6 & 6 \end{vmatrix} ]
Calculating each of these determinants will guide toward a function of (\lambda).
- Set the Determinant Not Equal to Zero
Set (\text{det}(A) ≠ 0) and solve for the values of (\lambda).
- Evaluate for λ = 2
Substituting (\lambda = 2) into matrix (A):
$$ A = \begin{bmatrix} 2 & 2 & -2 \ 4 & 4 & -1 \ 6 & 6 & 2 \end{bmatrix} $$
Next, find the values of (x), (y), and (z) using the inverse method (if the determinant is non-zero).
- Solve the Matrix Equation
If the determinant is non-zero for (\lambda = 2), use (A^{-1}b) to find solutions.
The values of (\lambda) such that the equations have a unique solution are: (\lambda \neq 0, 2, \frac{37}{7}).
When (\lambda = 2): The solutions using the matrix method yield: (x = 0, y = -1, z = 1).
More Information
In systems of linear equations, unique solutions occur when the determinant of the coefficient matrix is not equal to zero.
Tips
- Failing to calculate the determinant correctly.
- Confusing rows and columns when expanding the determinant.
- Not checking conditions after substituting specific values for (\lambda).
AI-generated content may contain errors. Please verify critical information
