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.

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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

  1. 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:

$$ A = \begin{bmatrix} \lambda & 2 & -2 \ 4 & 2\lambda & -1 \ 6 & 6 & \lambda \end{bmatrix} $$

  1. 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).

  1. 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).

  1. Set the Determinant Not Equal to Zero

Set (\text{det}(A) ≠ 0) and solve for the values of (\lambda).

  1. 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).

  1. 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).

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