Question

If $A$ and $B$ are invertible square matrices of order $n$, then which of the following isnot correct?

• A. $\det(AB) = \det(A) \cdot \det(B)$
• B. $\det(kA) = k^n \det(A)$
• C. $\det(A + B) = \det(A) + \det(B)$
• D. $\det(A^{-1}) = \frac{1}{\det(A)}$

Hint:

A common pitfall is assuming that operations like determinant or trace are fully linear. Remember: Determinants distribute over multiplication but NOT over addition. (Conversely, the Trace operator distributes over addition but not multiplication).

Answer 1

The Correct Option is C

Step 1: Recall determinant properties

• $\det(AB) = \det(A)\det(B)$
• $\det(kA) = k^n \det(A)$
• $\det(A^{-1}) = \dfrac{1}{\det(A)}$

Step 2: Check each option
(1) \[ \det(AB) = \det(A)\det(B) \] Correct property. (2) \[ \det(kA) = k^n \det(A) \] Correct for $n \times n$ matrix. (3) \[ \det(A+B) \neq \det(A) + \det(B) \] This is not a valid general property. (4) \[ \det(A^{-1}) = \frac{1}{\det(A)} \] Correct property.
Step 3: Conclusion
Only option (3) is incorrect. Final Answer:
\[ \boxed{\text{Option (3)}} \]