Let A and B be invertible matrices of the same order and let k be a non-zero real number. Consider the following statements:
(A) \((A')^{-1} = (A^{-1})'\)
(B) \((kA)^{-1} = kA^{-1}\)
(C) \((A + B)^{-1} = A^{-1} + B^{-1}\)
(D) If A is symmetric, then \(A^{-1}\) is symmetric.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (D) only
A matrix M is the inverse of P when \(PM = MP = I\). Using \((PQ)' = Q'P'\):
\(A'(A^{-1})' = (A^{-1}A)' = I' = I\), and similarly \((A^{-1})'A' = I\)
\( \displaystyle (kA)\left(\frac{1}{k}A^{-1}\right) = \frac{k}{k}AA^{-1} = I \)
For \(A = B = I\): \((A + B)^{-1} = (2I)^{-1} = \displaystyle\frac{1}{2}I\), while \(A^{-1} + B^{-1} = 2I\).
So \((A')^{-1} = (A^{-1})'\), \((kA)^{-1} = \displaystyle\frac{1}{k}A^{-1}\), and \((A + B)^{-1}\) need not equal \(A^{-1} + B^{-1}\).
Checking the statements
(A) \((A')^{-1} = (A^{-1})'\) — Correct. \(A'(A^{-1})' = (A^{-1}A)' = I\).
(B) \((kA)^{-1} = kA^{-1}\) — Incorrect. The inverse of kA is \(\displaystyle\frac{1}{k}A^{-1}\); for \(k = 2\), \(A = I\) it is \(\displaystyle\frac{1}{2}I\), not 2I.
(C) \((A + B)^{-1} = A^{-1} + B^{-1}\) — Incorrect. For \(A = B = I\) the left side is \(\displaystyle\frac{1}{2}I\) and the right side is 2I.
(D) If A is symmetric, then \(A^{-1}\) is symmetric. — Correct. \((A^{-1})' = (A')^{-1} = A^{-1}\).