Question

Let A and B be two square matrices of order 3 such that $\text{det} = 3$ and $\text{det} = -4$. Find the value of $\text{det}(-6AB)$.

Hint:

When multiplying matrices and calculating determinants, always remember that the determinant of a product is the product of the determinants, i.e., $\text{det}(AB) = \text{det} \cdot \text{det}$.

Answer 1

Given that $\text{det} = 3$ and $\text{det} = -4$. We aim to determine the value of $\text{det}(-6AB)$.

Applying the determinant property $\text{det}(cA) = c^n \cdot \text{det}$ for an $n \times n$ matrix, and the property for matrix products $\text{det}(AB) = \text{det} \cdot \text{det}$, we can express $\text{det}(-6AB)$ as:

\[\text{det}(-6AB) = \text{det}(-6) \cdot \text{det} \cdot \text{det}.\]

Assuming the matrix is of order 3, we have:

\[\text{det}(-6AB) = (-6)^3 \cdot \text{det} \cdot \text{det} = (-216) \cdot 3 \cdot (-4).\]

Thus:

\[\text{det}(-6AB) = 2592.\]