Question

Let \(A\) and \(B\) be two square matrices each of order 3. If \(|AB| = 21\) and \(|A^{-1}| = -7\), then the value of \(|B|\) is equal to

• A. 3
• B. -3
• C. 147
• D. -63
• E. -147

Hint:

Determinant properties like \(|AB| = |A||B|\) apply to square matrices of any order. The order "3" mentioned in the question is additional information but doesn't change the calculation here.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Determinants of matrices follow specific multiplicative and inverse properties. The determinant of a product of matrices is the product of their individual determinants.

Step 2: Key Formula or Approach:
1. \(|AB| = |A| \cdot |B|\).
2. \(|A^{-1}| = \frac{1}{|A|}\).

Step 3: Detailed Explanation:
Given:
\[ |A^{-1}| = -7 \]
Using the property of inverse determinants:
\[ \frac{1}{|A|} = -7 \implies |A| = -\frac{1}{7} \]
Now, using the property for the product \(|AB|\):
\[ |AB| = 21 \]
\[ |A| \cdot |B| = 21 \]
Substituting the value of \(|A|\):
\[ \left( -\frac{1}{7} \right) \cdot |B| = 21 \]
\[ |B| = 21 \times (-7) \]
\[ |B| = -147 \]

Step 4: Final Answer:
The value of \(|B|\) is -147.