Question

Consider the matrix \[ M= \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \] Let \(p,q,r,s,a,b,c,d\) be integers such that \[ M^{26}= \begin{bmatrix} p & q \\ r & s \end{bmatrix} \] and \[ \sum_{k=1}^{26} M^k= \begin{bmatrix} a & b \\ c & d \end{bmatrix}. \] Then which of the following statements is (are) TRUE?

• A. There exists a \(2\times2\) invertible matrix \(N\) with real entries such that \[ MN= N \begin{bmatrix} 1 & 1 0 & 1 \end{bmatrix} \]
• B. The value of \(a\) is \(378\)
• C. For any two given integers \(m\) and \(n\), there exist unique integers \(x\) and \(y\) such that \[ px+qy=m \] and \[ rx+sy=n \]
• D. For each positive real number \(t\), the system of linear equations \[ (a+t)x+by=1 \] \[ cx+(d+t)y=-1 \] has a unique solution

Hint:

If: \[ A^2=0 \] then: \[ (I+A)^n=I+nA \] using binomial expansion.

Answer 1

The Correct Option is A

To determine which statement is true, we need to analyze the properties of the matrix \( M \) and its powers. The matrix \( M \) is given by:

\[M= \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}\]

First, let's calculate the characteristic polynomial of \( M \). This will help us determine the nature of its eigenvalues and eigenvectors:

The characteristic polynomial is given by:

\[\det(M - \lambda I) = \begin{vmatrix} 2 - \lambda & -1 \\ 1 & -\lambda \end{vmatrix} = (2-\lambda)(-\lambda) - (-1)(1) = \lambda^2 - 2\lambda + 1\]

Factoring, we find:

\[\lambda^2 - 2\lambda + 1 = (\lambda - 1)^2\]

This means the eigenvalue of \( M \) is \( \lambda = 1 \) with algebraic multiplicity 2. However, we need to find a generalized eigenvector or matrix \( N \) that can simultaneously diagonalize \( M \) and another matrix.

Now, consider the possibility of transforming \( M \) to the form:

\[\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\]

This transformation is associated with a Jordan form when all eigenvalues are identical and there's a defect in the geometric multiplicity.

Since the problem states there exists a \( 2 \times 2 \) invertible matrix \( N \), we can indeed transform \( M \) to its Jordan form using some \( N \) such that:

\(MN = N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)

This statement is true as a result of \( M \) having the eigenvalue 1 and leveraging generalized eigenvectors. Thus, the correct option is:

There exists a \(2 \times 2\) invertible matrix \(N\) with real entries such that

\[MN= N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\]