Question

If \( A = \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix} \), then the value of \( I - A + A^2 - A^3 + \ldots \) is:

• A. \( \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \)
• B. \( \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \)
• C. \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
• D. \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

Hint:

For infinite geometric series \( I - A + A^2 - A^3 + \ldots \), check if \( A^2 = 0 \).

Answer 1

The Correct Option is A

Step 1: Define the series \( S \): The given infinite series is \( S = I - A + A^2 - A^3 + \cdots \). This series converges to \( S = (I - A)^{-1} \) provided that the matrix \( (I - A) \) is invertible.

Step 2: Calculate \( I - A \): Given the identity matrix \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and the matrix \( A = \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix} \), we compute \( I - A \) as follows: \[ I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \].

Step 3: Assess invertibility of \( I - A \): To determine if \( I - A \) is invertible, we calculate its determinant: \[ \text{det}(I - A) = (-1)(3) - (-1)(4) = -3 + 4 = 1 \]. Since \( \text{det}(I - A) = 1 eq 0 \), the matrix \( I - A \) is invertible.

Step 4: Determine the sum of the series: The sum of the series \( S \) is equal to the inverse of \( I - A \). We find the inverse of \( I - A = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \). The inverse is given by: \[ (I - A)^{-1} = \frac{1}{\text{det}(I - A)} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \frac{1}{1} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \] Oh, there was a calculation error in the original text. Let's recompute the inverse of \( \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \). The determinant is \( (-1)(3) - (-1)(4) = -3 + 4 = 1 \). The inverse is \( \frac{1}{1} \begin{bmatrix} 3 & -(-1) \\ -(4) & -1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \). Let me recheck the provided answer. The provided answer is \( \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \). This is the matrix \( I-A \), not its inverse.

Corrected Step 4: Determine the sum of the series: The sum of the series \( S \) is equal to the inverse of \( I - A \). We find the inverse of \( I - A = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \). The determinant is \( 1 \). The inverse is \( (I - A)^{-1} = \frac{1}{1} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \). Thus, the sum of the series \( S = I - A + A^2 - A^3 + \cdots \) is: \[ S = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \).

Conclusion: The correct option, based on the calculation of \( (I - A)^{-1} \), is \( \mathbf{(B)} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \). The original text incorrectly stated the sum.