Question

\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]

• A. \( \frac{1}{x} + \frac{1}{y} \)
• B. \( xy \)
• C. \( \frac{1}{xy} \)
• D. \( x + y \)

Hint:

When dealing with adjugate matrices, use the property \( |Adj(M)| = |M|^{n-1} \) and apply determinant rules carefully for matrix multiplication.

Answer 1

The Correct Option is C

We are given the properties of the adjugate matrices of \( A \) and \( B \), and we need to determine \( \left( |Adj(AB)| \right)^{-1} \).

 Step 1: Property of Determinant of Adjugate 
For any square matrix \( M \), the determinant of the adjugate is given by: \[ |Adj(M)| = |M|^{n-1} \] where \( n \) is the order of the matrix. 

Step 2: Using the Determinant Multiplication Rule 
Since \( AB \) is the product of two matrices, we use the property: \[ |Adj(AB)| = |AB|^{n-1} \] Applying the determinant property: \[ |AB| = |A| \cdot |B| \] Thus, \[ |Adj(AB)| = (|A| \cdot |B|)^{n-1} \] 

Step 3: Expressing in Terms of Given Values 
We know that: \[ |Adj(A)| = |A|^{n-1} = x, \quad |Adj(B)| = |B|^{n-1} = y. \] Multiplying these equations: \[ |Adj(A)| \cdot |Adj(B)| = (|A|^{n-1}) \cdot (|B|^{n-1}) = |AB|^{n-1}. \] So, \[ |Adj(AB)| = |Adj(A)| \cdot |Adj(B)| = x \cdot y. \] 

Step 4: Finding the Inverse 
\[ \left( |Adj(AB)| \right)^{-1} = \frac{1}{|Adj(AB)|} = \frac{1}{xy}. \]