Question

For a given square matrix \(A\) of order n, if there exists another square matrix \(B\) of the same order n, such that \(AB = BA = I\), then \(A\) is said to be invertible and \(B\) is called the inverse matrix of \(A\).

Which of the following statements are not TRUE ?

A. Inverse of a matrix, if it exists, is unique.
B. For two invertible matrices of same order, say \(A\) and \(B\) , then \((AB)^{-1} = A^{-1}B^{-1}\)
C. For an invertible matrix \(A\), \((A^{-1})^{-1} = A\)
D. For an invertible matrix \(A\), \((A^{T})^{-1} = A^{T}\)

Choose the correct answer from the options given below:

• A. A and C only
• B. B and D only
• C. A, C and D only
• D. A and D only

Hint:

Use \((AB)^{-1} = B^{-1}A^{-1}\) and \((A^T)^{-1} = (A^{-1})^T\).

Answer 1

The Correct Option is B

Step 1: Plan.
Use two known laws of inverses and test each statement with them or with a quick example.

Step 2: Test with a concrete pair.
Take \(A = \begin{bmatrix}1 & 2\\3 & 5\end{bmatrix}\) and \(B = \begin{bmatrix}2 & 1\\1 & 1\end{bmatrix}\). Their determinants are -1 and 1, so both are invertible. Direct computation shows \((AB)^{-1} = B^{-1}A^{-1}\), and this is not equal to \(A^{-1}B^{-1}\). So statement B fails.

Step 3: Transpose test.
The inverse of a transpose is the transpose of the inverse. For the same \(A\), \((A^T)^{-1} = (A^{-1})^T\), which is not \(A^T\). So statement D fails.

Step 4: Remaining statements.
Uniqueness of the inverse (A) and \((A^{-1})^{-1} = A\) (C) are standard theorems, so both hold.

Step 5: Match.
The false statements are B and D, so the choice is option 2.

Final Answer:
Not true: B and D. \[ \boxed{\text{Option 2}} \]