If A is a square matrix of order \( n \) and \( k \) is a scalar such that \( |kA|=\alpha|A| \) then the value of α is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( k^n \)
Multiplying A by k multiplies every row by k. A factor common to one row can be taken outside the determinant, and here each of the n rows gives one factor k:
\( \displaystyle |kA|=k\cdot k\cdots k\ (n\text{ factors})\cdot|A| \)
\( \displaystyle =k^n|A| \)
So α = kn.
Why the other options are incorrect
Option (1) \( k\,n \) — Incorrect. kn adds up the factors instead of multiplying them; each of the n rows contributes a factor k. For a 2 × 2 matrix and k = 3, \( |3A|=9|A| \), not \( 6|A| \).
Option (2) \( n^k \) — Incorrect. nk has the base and the exponent interchanged.
Option (4) \( k/n \) — Incorrect. The factor is kn, not k/n; for a 2 × 2 matrix and k = 3, \( |3A|=9|A| \).
CBSE Applied Mathematics (Class XII):
Properties of a Determinant
