Question

Let \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} , \ \alpha > 0 \), such that \( \det(A) = 0 \) and \( \alpha + \beta = 1. \) If \( I \) denotes the \( 2 \times 2 \) identity matrix, then the matrix \( (I + A)^8 \) is:

• A. \( \begin{bmatrix} 4 & -1 \\6 & -1 \end{bmatrix} \)
• B. \( \begin{bmatrix} 257 & -64 \\514 & -127 \end{bmatrix} \)
• C. \( \begin{bmatrix} 1025 & -511 \\2024 & -1024 \end{bmatrix} \)
• D. \( \begin{bmatrix} 766 & -255 \\1530 & -509 \end{bmatrix} \)

Hint:

When a matrix satisfies \( A^2 = A \), powers simplify: \( A^n = A \). Use this in binomial expansions.

Answer 1

The Correct Option is D

This problem requires us to first find the values of \( \alpha \) and \( \beta \) using the given conditions on the matrix \( A \). Once matrix \( A \) is determined, we need to compute the matrix \( (I + A)^8 \).

Concept Used:

1. Determinant of a 2x2 Matrix: For a matrix \( M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the determinant is \( \det(M) = ad - bc \).

2. System of Equations: Solving for two variables using two given equations.

3. Matrix Exponentiation using Diagonalization: If a matrix \( B \) can be written as \( B = PDP^{-1} \), where \( D \) is a diagonal matrix of eigenvalues and \( P \) is the matrix of corresponding eigenvectors, then \( B^n = PD^n P^{-1} \). The eigenvalues \( \lambda \) are found by solving the characteristic equation \( \det(B - \lambda I) = 0 \).

Step-by-Step Solution:

We are given the matrix \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} \) with the conditions \( \alpha > 0 \), \( \det(A) = 0 \), and \( \alpha + \beta = 1 \).

First, we calculate the determinant of \( A \):

\[ \det(A) = (\alpha)(\beta) - (-1)(6) = \alpha\beta + 6 \]

Using the condition \( \det(A) = 0 \), we get:

\[ \alpha\beta + 6 = 0 \implies \alpha\beta = -6 \]

We now have a system of two equations with two variables:

\[ \alpha + \beta = 1 \] \[ \alpha\beta = -6 \]

We can solve for \( \alpha \) and \( \beta \) by considering a quadratic equation \( t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0 \), where the roots are \( \alpha \) and \( \beta \). This gives:

\[ t^2 - (1)t + (-6) = 0 \implies t^2 - t - 6 = 0 \]

Factoring the quadratic equation, we get:

\[ (t - 3)(t + 2) = 0 \]

The possible values for \( \alpha \) and \( \beta \) are \( 3 \) and \( -2 \). Since we are given that \( \alpha > 0 \), we must have \( \alpha = 3 \) and \( \beta = -2 \).

Now, we can write the matrix \( A \):

\[ A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \]

We need to find \( (I + A)^8 \). Let's first compute the matrix \( B = I + A \):

\[ B = I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix} \]

To compute \( B^8 \), we use diagonalization. First, we find the eigenvalues of \( B \) by solving the characteristic equation \( \det(B - \lambda I) = 0 \):

\[ \det\left(\begin{bmatrix} 4-\lambda & -1 \\ 6 & -1-\lambda \end{bmatrix}\right) = (4-\lambda)(-1-\lambda) - (-1)(6) = 0 \] \[ -4 - 4\lambda + \lambda + \lambda^2 + 6 = 0 \] \[ \lambda^2 - 3\lambda + 2 = 0 \] \[ (\lambda - 1)(\lambda - 2) = 0 \]

The eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \).

Next, we find the corresponding eigenvectors. For \( \lambda_1 = 1 \):

\[ (B - 1I)\mathbf{v}_1 = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 3x - y = 0 \]

An eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \).

For \( \lambda_2 = 2 \):

\[ (B - 2I)\mathbf{v}_2 = \begin{bmatrix} 2 & -1 \\ 6 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 2x - y = 0 \]

An eigenvector is \( \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \).

Now we form the matrices \( P \), \( D \), and \( P^{-1} \):

\[ P = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}, \quad D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \] \[ P^{-1} = \frac{1}{(1)(2) - (1)(3)} \begin{bmatrix} 2 & -1 \\ -3 & 1 \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 2 & -1 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} \]

Final Computation & Result:

We use the formula \( B^8 = PD^8P^{-1} \). First, we calculate \( D^8 \):

\[ D^8 = \begin{bmatrix} 1^8 & 0 \\ 0 & 2^8 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 256 \end{bmatrix} \]

Now, we compute \( B^8 \):

\[ B^8 = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 256 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} \]

First, multiply the first two matrices:

\[ \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 256 \end{bmatrix} = \begin{bmatrix} 1 & 256 \\ 3 & 512 \end{bmatrix} \]

Now, multiply the result by \( P^{-1} \):

\[ B^8 = \begin{bmatrix} 1 & 256 \\ 3 & 512 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} (1)(-2) + (256)(3) & (1)(1) + (256)(-1) \\ (3)(-2) + (512)(3) & (3)(1) + (512)(-1) \end{bmatrix} \] \[ B^8 = \begin{bmatrix} -2 + 768 & 1 - 256 \\ -6 + 1536 & 3 - 512 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix} \]

Thus, the matrix \( (I+A)^8 \) is \( \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix} \).

Answer 2

From \( \det(A) = \alpha \beta + 6 = 0 \), we get \( \alpha \beta = -6 \), and \( \alpha + \beta = 1 \). Solving: \[ \alpha = 3,\quad \beta = -2 \Rightarrow A = \begin{bmatrix} 3 & -1 6 & -2 \end{bmatrix} \] Check powers: \[ A^2 = A \Rightarrow A^n = A,\ \forall n \geq 1 \] Use binomial expansion: \[ (1 + A)^5 = I + 5A + 10A^2 + 10A^3 + 5A^4 + A^5 = I + 31A \] \[ (1 + A)^5 = \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix} + 31 \cdot \begin{bmatrix} 3 & -1 \\6 & -2 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\1530 & -509 \end{bmatrix} \]