Question

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

The given series is \( S = I - A + A^2 - A^3 + \cdots \). This infinite geometric series converges to \( (I - A)^{-1} \) if \( (I - A) \) is invertible.

Step 2: Compute \( I - A \):

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 \( I - A \):

\[ I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \] Step 3: Check if \( I - A \) is invertible:

To determine invertibility, we calculate the determinant of \( I - A \):

\[ \det(I - A) = (-1)(3) - (-1)(4) = -3 + 4 = 1 \]

Since \( \det(I - A) = 1 eq 0 \), the matrix \( I - A \) is invertible.

Step 4: Calculate the series sum \( S \):

The sum of the series \( S \) is equal to the inverse of \( (I - A) \). We compute the inverse:

\[ (I - A)^{-1} = \frac{1}{\det(I - A)} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \frac{1}{1} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \]

Correction: Recomputing the inverse. The formula for a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\). For \( I - A = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \), the inverse is:

\[ (I - A)^{-1} = \frac{1}{(-1)(3) - (-1)(4)} \begin{bmatrix} 3 & -(-1) \\ -(4) & -1 \end{bmatrix} = \frac{1}{1} \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix} \]

Wait, the provided example in Step 4 seems to have the original \( I - A \) matrix. Let me re-evaluate the calculation for the sum of the series \( S \).

The series \( S = I - A + A^2 - A^3 + \cdots \) is a geometric series with first term \( I \) and common ratio \( -A \). The sum is \( (I - (-A))^{-1} = (I + A)^{-1} \). Let's recalculate based on this understanding.

Recalculation based on \( S = (I+A)^{-1} \):

Step 2 (Revised): Compute \( I + A \):

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

Step 3 (Revised): Check if \( I + A \) is invertible:

\[ \det(I + A) = (3)(-1) - (1)(-4) = -3 + 4 = 1 eq 0 \]

Since the determinant is non-zero, \( I + A \) is invertible.

Step 4 (Revised): Compute the inverse of \( I + A \):

\[ (I + A)^{-1} = \frac{1}{\det(I + A)} \begin{bmatrix} -1 & -1 \\ -(-4) & 3 \end{bmatrix} = \frac{1}{1} \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \]

Thus, the sum of the series \( S = I - A + A^2 - A^3 + \cdots \) is \( (I + A)^{-1} \).

Conclusion:

The sum of the series is:

\[ S = \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \]

The correct option is:

\[ \mathbf{(A)} \quad \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix} \]