Question

Let \(A\) and \(B\) be two invertible matrices, then which of the following is true ?

• A. \(A^{-1}\) may not invertible
• B. \((A^{-1})^{-1} = A\)
• C. \((AB)^{-1} = A^{-1} B\)
• D. \((A^{-1})^{-1} = A^{-1}\)

Hint:

The inverse operation is an involution, meaning that applying it twice returns the original object:
\[ (A^{-1})^{-1} = A \]
This is analogous to double negation (\(-(-x) = x\)) or double transpose (\((A^T)^T = A\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question tests our understanding of the properties of invertible matrices and their inverses.
We need to determine the correct identity regarding the double inverse of an invertible matrix.
Key Formula or Approach:
For an invertible matrix \(A\), there exists a unique matrix \(A^{-1}\) such that:
\[ A A^{-1} = A^{-1} A = I \]
By the definition of the inverse, the inverse of any matrix \(B\) is a matrix \(C\) such that \(B C = C B = I\).

Step 2: Detailed Explanation:

• Let us evaluate each option:
- Option (A) states "\(A^{-1}\) may not be invertible".
This is false because the inverse of an invertible matrix is always invertible, with its inverse being the original matrix \(A\).

• Evaluate Option (B): \((A^{-1})^{-1} = A\).
Let \(B = A^{-1}\). We want to find the inverse of \(B\), which is denoted by \(B^{-1} = (A^{-1})^{-1}\).
By definition:
\[ B \cdot A = A^{-1} A = I \] and \[ A \cdot B = A A^{-1} = I \] Since multiplying \(B\) by \(A\) yields the identity matrix, \(A\) must be the inverse of \(B\).
Thus, \(B^{-1} = A \implies (A^{-1})^{-1} = A\). This option is correct.

• Evaluate Option (C): \((AB)^{-1} = A^{-1} B\).
This is incorrect because the inverse of a product is the product of the inverses in reverse order: \((AB)^{-1} = B^{-1} A^{-1}\).

• Option (D) states \((A^{-1})^{-1} = A^{-1}\), which is incorrect as the double inverse returns \(A\).

Step 3: Final Answer:
The correct statement is \((A^{-1})^{-1} = A\).