Question

For α, β ∈ R , α ≠ β, if −2 and 5 are the eigenvalues of the matrix 𝑀=[1−𝛼1+𝛽𝛽 𝛼+𝛽] and 𝑋 = [𝑥1 𝑥2 ] is an eigenvector of 𝑀 associated to −2, then\(M=\begin{bmatrix} 1-α& 1+β\\ β & a+B& \end{bmatrix}\)
and 𝑋 = \(\begin{bmatrix}  X_1\\ X_2 \end{bmatrix}\) is an eigenvector of 𝑀 associated to −2, then

• A. 2𝑥₁ + 𝑥₂ = 0
• B. 𝛽 − 𝛼 = 5
• C. 𝛼 ² − 𝛽 ² = 5
• D. 𝑥₁ + 3𝑥₂ = 0

Answer 1

The Correct Option is A, B, C

To solve the given problem, we need to follow a step-by-step approach, focusing on the properties of eigenvalues and eigenvectors.

Given the matrix: 

\(M=\begin{bmatrix} 1-α & 1+β \\ β & α+β \end{bmatrix}\)

It is provided that the eigenvalues of this matrix are -2 and 5. This information gives us essential conditions about the matrix. The eigenvalues of a 2x2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) are given by the solution to the characteristic equation:

\(λ^2 - (a+d)λ + (ad-bc) = 0\)

For this problem, equating the trace (sum of eigenvalues) and determinant (product of eigenvalues) to the matrix properties:

• Trace Condition: The sum of eigenvalues equals the sum of the diagonal elements.

\(-2 + 5 = (1-α) + (α+β) = 1 + β\)

From the above, we find:

\(1 + β = 3 \Rightarrow β = 2\)

• Determinant Condition: The product of eigenvalues equals the determinant of the matrix.

\(-2 \cdot 5 = (1-α)(α+β) - (1+β)β\)

\(-10 = (1-α)(α+2) - (1+2) \cdot 2\)

Expanding and simplifying, we get:

\(-10 = (α^2 +2α - α^2 +2) - 6\)

\(-10 = 2α - 4\)

\(2α = -6 \Rightarrow α = -3\)

Substituting the values of \(α\) and \(β\) back, verify:

\(β − α = 2 - (-3) = 5\)

\(α^2 - β^2 = (-3)^2 - 2^2 = 9 - 4 = 5\)

Verification of eigenvector \(X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) associated with eigenvalue -2:

Using the eigenvalue equation \(MX = -2X\), we have:

\(\begin{bmatrix} 1-(-3) & 1+2 \\ 2 & -3+2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = -2\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\)

Which results in:

\(\begin{bmatrix}4 & 3 \\ 2 & -1\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -2x_1 \\ -2x_2 \end{bmatrix}\)

This gives two equations:

• \(4x_1 + 3x_2 = -2x_1\) → simplifying gives \(6x_1 + 3x_2 = 0\) → \(2x_1 + x_2 = 0\)

Thus, the correct options are:

• \(2x_1 + x_2 = 0\)
• \(\beta - \alpha = 5\)
• \(\alpha^2 - \beta^2 = 5\)

These match the conditions given by the options.