Let A be a square matrix of order n, then which of the following are TRUE?
(A) \(|\mathrm{adj}\, A| = |A|^{n-1}\)
(B) \(|A \cdot \mathrm{adj}\, A| = |A|^n\)
(C) \(A (\mathrm{adj}\, A)= |A|\)
(D) \(|KA| = K|A|\)
(E) \(|A^{-1}| = \displaystyle\frac{1}{|A|}, |A| \ne 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (E) only
The statements rest on the adjoint identity, where I is the identity matrix of order n:
\( \displaystyle A(\mathrm{adj}\,A) = (\mathrm{adj}\,A)A = |A|\,I \)
Checking the statements
(A) \(|\mathrm{adj}\, A| = |A|^{n-1}\) — Correct. Taking determinants in the identity, with \(|kI| = k^n\) for a scalar k:
\(|A|\,|\mathrm{adj}\,A| = |A|^n\)
Dividing by \(|A|\) when \(|A| \ne 0\) gives \(|\mathrm{adj}\,A| = |A|^{n-1}\). This standard result also holds when \(|A| = 0\), since adj A is then singular too.
(B) \(|A \cdot \mathrm{adj}\, A| = |A|^n\) — Correct. \(|A \cdot \mathrm{adj}\,A| = \big|\,|A|\,I\,\big| = |A|^n\).
(C) \(A (\mathrm{adj}\, A)= |A|\) — Incorrect. \(A(\mathrm{adj}\,A) = |A|\,I\) is a matrix, not the number \(|A|\).
(D) \(|KA| = K|A|\) — Incorrect. Multiplying A by K multiplies each of its n rows by K, and each row contributes a factor K to the determinant:
\(|KA| = K^n|A|\)
(E) \(|A^{-1}| = \frac{1}{|A|}, |A| \ne 0\) — Correct. From \(AA^{-1} = I\):
\(|A|\,|A^{-1}| = |I| = 1\)
\(\displaystyle |A^{-1}| = \frac{1}{|A|}\)
So the true statements are (A), (B) and (E).