For every square matrix A of order n x n, which of the following statements are correct?
(A) (adj A)A = A(adj A) = |A|I, where I is identity matrix of order n.
(B) |kA| = kn |A|, k is any scalar.
(C) |adj A| = |A|n
(D) A2 = A
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
The basic property of the adjoint holds for every square matrix of order n: the (i, i) entry of \( \mathrm{A}(\mathrm{adj}\,\mathrm{A}) \) is the expansion of |A| along row i, and every other entry is a sum of elements of one row times cofactors of another row, which is 0. So
\( \displaystyle (\mathrm{adj}\,\mathrm{A})\mathrm{A}=\mathrm{A}(\mathrm{adj}\,\mathrm{A})=|\mathrm{A}|\mathrm{I} \)
Multiplying A by k multiplies each of its n rows by k, and each row gives one factor k:
\( \displaystyle |k\mathrm{A}|=k^{n}|\mathrm{A}| \)
Taking determinants in the first result:
\( \displaystyle |\mathrm{A}|\,|\mathrm{adj}\,\mathrm{A}|=\big|\,|\mathrm{A}|\mathrm{I}\,\big| \)
\( \displaystyle =|\mathrm{A}|^{n} \)
\( \displaystyle |\mathrm{adj}\,\mathrm{A}|=|\mathrm{A}|^{n-1} \)
So for every square matrix of order n, (adj A)A = A(adj A) = |A|I, |kA| = kn|A| and |adj A| = |A|n−1.
Checking the statements
(A) (adj A)A = A(adj A) = |A|I, where I is identity matrix of order n. — Correct. This is the basic property of the adjoint.
(B) |kA| = kn |A|, k is any scalar. — Correct. Each of the n rows gives one factor k.
(C) |adj A| = |A|n — Incorrect. The power is n − 1: |adj A| = |A|n−1.
(D) A2 = A — Incorrect. This holds only for special (idempotent) matrices such as I or O; for example \( \mathrm{A}=2\mathrm{I} \) gives \( \mathrm{A}^2=4\mathrm{I}\ne\mathrm{A} \).