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 \( S = I - A + A^2 - A^3 + \cdots \), which is an infinite series. It can be expressed as:

\[ S = (I - A)^{-1} \]

if the matrix \( (I - A) \) is invertible.

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

The identity matrix \( I \) is:

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

So,

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

The determinant of \( I - A \) is:

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

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

Step 4: Verify the series sum:

The inverse of \( I - A \) is:

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

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

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

The correct option is:

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