Question

Consider a 2x2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\). If \(a+d=1\) and \(ad-bc=1\), then \(A^3\) is equal to

• A. 0
• B. \(-I\)
• C. \(3I\)
• D. \(I\)

Hint:

The Cayley-Hamilton theorem is a very powerful tool for finding powers of a matrix. For any 2x2 matrix, always remember the formula \(A^2 - \text{tr}(A)A + \det(A)I = 0\). This allows you to express higher powers of \(A\) in terms of \(A\) and \(I\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. For a 2x2 matrix, the characteristic equation is \(\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0\). The matrix \(A\) will satisfy the equation \(A^2 - \text{tr}(A)A + \det(A)I = 0\).

Step 2: Key Formula or Approach:
1. Identify the trace (\(\text{tr}(A) = a+d\)) and determinant (\(\det(A) = ad-bc\)) from the given information.
2. Write the Cayley-Hamilton equation for matrix \(A\).
3. Manipulate this equation algebraically to find an expression for \(A^3\).

Step 3: Detailed Explanation:
We are given a 2x2 matrix \(A\) with:

\(\text{tr}(A) = a+d = 1\)
\(\det(A) = ad-bc = 1\) \end{itemize} According to the Cayley-Hamilton theorem for a 2x2 matrix: \[ A^2 - (\text{tr}(A))A + (\det(A))I = 0 \] Substitute the given values into this equation: \[ A^2 - (1)A + (1)I = 0 \] \[ A^2 - A + I = 0 \] From this equation, we can express \(A^2\) as: \[ A^2 = A - I \] To find \(A^3\), we multiply the entire equation by \(A\): \[ A(A^2) = A(A - I) \] \[ A^3 = A^2 - AI \] \[ A^3 = A^2 - A \] Now, we can substitute the expression for \(A^2\) back into this equation: \[ A^3 = (A - I) - A \] \[ A^3 = A - I - A \] \[ A^3 = -I \]
Step 4: Final Answer:
The matrix \(A^3\) is equal to \(-I\).