Question

If \(A\) and \(B\) are square matrices of order \(3\) such that \(|A|=-1\) and \(|B|=3\), then \(|3AB|\) equals:

• A. \(-81\)
• B. \(-27\)
• C. \(-9\)
• D. \(81\)

Hint:

For \(n\times n\) matrices, a single global scalar \(k\) outside a product contributes exactly a factor \(k^n\) to the determinant, regardless of how you parenthesize the product.

Answer 1

The Correct Option is A

Step 1: Recall two fundamental determinant properties for an \(n\times n\) matrix.
(1) Multiplicativity: \(|AB|=|A|\,|B|\).
(2) Scalar pull-out: \(|kA|=k^n|A|\) (because multiplying a single row by \(k\) scales the determinant by \(k\); doing this to all \(n\) rows scales it by \(k^n\)). For \(n=3\), \(|kA|=k^3|A|\).

Step 2: Rewrite \(|3AB|\) so the scalar can be extracted cleanly.
A convenient trick is to absorb the scalar into one factor: \[ |3AB|=\big|(3A)B\big|=|3A|\;|B| \text{(by multiplicativity).} \]

Step 3: Apply the scalar rule to \(|3A|\) for a \(3\times3\) matrix.
\[ |3A|=3^3|A|=27\,|A|. \]

Step 4: Substitute the given determinant values and compute.
\[ |3AB|=\underbrace{27|A|}_{=27(-1)=-27}\cdot |B| =(-27)\cdot 3 =-81. \]

Step 5: Quick consistency checks.
\(\bullet\) Since \(|A||B|=(-1)\cdot 3=-3\), scaling the product \(AB\) by \(3\) should multiply the determinant by \(3^3=27\) (order \(3\)), giving \(27\cdot(-3)=-81\).
\(\bullet\) Beware of the common error \(|3AB|=(3^3)(|A||B|)(3^3)\); the scalar \(3\) appears once (on \(AB\)), not on both factors separately.

Final Answer:
\[ \boxed{-81} \]