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

We are given that $\text{det} = 3$ and $\text{det} = -4$. We need to find the value of $\text{det}(-6AB)$.
Using the property of determinants: \[ \text{det}(cA) = c^n \cdot \text{det} \quad \text{for an } n \times n \text{ matrix}. \] For the product of two matrices, the determinant is the product of the determinants: \[ \text{det}(AB) = \text{det} \cdot \text{det}. \] Now, for $-6AB$, we use the following: \[ \text{det}(-6AB) = \text{det}(-6) \cdot \text{det} \cdot \text{det}. \] Since the matrix is of order 3, we get: \[ \text{det}(-6AB) = (-6)^3 \cdot \text{det} \cdot \text{det} = (-216) \cdot 3 \cdot (-4). \] Therefore, \[ \text{det}(-6AB) = 2592. \]