Let A and B be invertible square matrices of order 3. Consider the following statements:
(A) A square matrix is invertible if and only if its determinant is non-zero.
(B) \((AB)^{-1} = A^{-1}B^{-1}\) for all such A and B.
(C) \((A^{-1})^{-1} = A\).
(D) \(|\mathrm{adj}\,A| = |A|^{3}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
For a square matrix A of order n:
\( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A \ \text{ exists exactly when } |A| \ne 0 \)
\( (AB)^{-1} = B^{-1}A^{-1} \)
\( |\mathrm{adj}\,A| = |A|^{n - 1} \)
So the inverse exists only for a non-singular matrix, the inverse of a product reverses the order, and for order 3, \(|\mathrm{adj}\,A| = |A|^{2}\).
Checking the statements
(A) A square matrix is invertible if and only if its determinant is non-zero. — Correct. This is the condition for \(\frac{1}{|A|}\,\mathrm{adj}\,A\) to exist.
(B) \((AB)^{-1} = A^{-1}B^{-1}\) for all such A and B. — Incorrect. The order reverses: \((AB)^{-1} = B^{-1}A^{-1}\), which differs from \(A^{-1}B^{-1}\) when \(A^{-1}\) and \(B^{-1}\) do not commute.
(C) \((A^{-1})^{-1} = A\). — Correct. \(A^{-1}A = AA^{-1} = I\), so A is the inverse of \(A^{-1}\).
(D) \(|\mathrm{adj}\,A| = |A|^{3}\). — Incorrect. For order 3 it is \(|A|^{3 - 1} = |A|^{2}\).