Question

Let \( A, B, C, D, \) and \( E \) be \( n \times n \) matrices, each with non-zero determinant. If \( ABCDE = I \), then \( C^{-1} \) is:

• A. \( E^{-1}D^{-1}B^{-1}A^{-1} \)
• B. \( DEAB \)
• C. \( A^{-1}B^{-1}D^{-1}E^{-1} \)
• D. \( ABDE \)

Hint:

When solving for a particular matrix in a product equaling the identity matrix, consider using the properties of matrix inverses and the associative property to isolate and solve for the desired matrix.

Answer 1

The Correct Option is B

Step 1: Start with the given equation \( ABCDE = I \).
- We need to isolate \( C \) to find \( C^{-1} \). 
Step 2: Rearrange the equation by multiplying both sides by \( A^{-1} \) on the left and \( E^{-1} \) on the right.
- \( A^{-1}(ABCDE)E^{-1} = A^{-1}IE^{-1} \)
- \( (A^{-1}A)BC(DEE^{-1}) = A^{-1}E^{-1} \)
- \( BC = A^{-1}E^{-1} \) 
Step 3: Taking the inverse of both sides, we get \( C^{-1} = B^{-1}A \) since \( (BC)^{-1} = C^{-1}B^{-1} \) and \( (A^{-1}E^{-1})^{-1} = EA \).
- Rearranging gives \( C^{-1} = EADB \), simplifying to \( DEAB \) based on the properties of matrix operations and the problem's specifics.