Question

If A and B are invertible matrices of order $3 \times 3$ such that $\text{det} = 4$ and $\text{det}([AB]^{-1}) = \frac{1}{20}$, then $\text{det}$ is equal to:

• A. $\frac{1}{20}$
• B. $\frac{1}{5}$
• C. 20
• D. 5

Hint:

To solve for the determinant of a matrix product, use the property: $\text{det}(AB) = \text{det} \cdot \text{det}$. Also, for the inverse, $\text{det}(A^{-1}) = \frac{1}{\text{det}}$.

Answer 1

The Correct Option is B

We are given that: \[ \text{det} = 4 \quad \text{and} \quad \text{det}([AB]^{-1}) = \frac{1}{20} \] Using the determinant property for matrix inverse: \[ \text{det}([AB]^{-1}) = \frac{1}{\text{det}(AB)} \] Since $\text{det}(AB) = \text{det} \cdot \text{det}$, we have: \[ \frac{1}{20} = \frac{1}{\text{det} \cdot \text{det}} \] Substituting $\text{det} = 4$: \[ \frac{1}{20} = \frac{1}{4 \cdot \text{det}} \] Solving for $\text{det}$: \[ 4 \cdot \text{det} = 20 \] \[ \text{det} = \frac{20}{4} = 5 \] Therefore, $\text{det} = \frac{1}{5}$.