Let A, B and C are `nxxn` matrices such that `|A|-2, |B|=3 and |C|=5`. If `|(2A)^(2)(3B)(5C)^(-1)|=(1728)/(125)`, then the value of n is equal to

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the problem, we need to find the value of \( n \) given the determinants of matrices \( A \), \( B \), and \( C \), and the determinant of the product of transformed matrices. Here’s the step-by-step solution:

### Step 1: Write down the given information
We know:
- \( |A| = 2 \)
- \( |B| = 3 \)
- \( |C| = 5 \)
- \( |(2A)^2 (3B) (5C)^{-1}| = \frac{1728}{125} \)

### Step 2: Use the properties of determinants
Using the properties of determinants:
1. \( |kP| = k^n |P| \) for any scalar \( k \) and \( n \times n \) matrix \( P \).
2. \( |P^2| = |P|^2 \).
3. \( |P^{-1}| = \frac{1}{|P|} \).

### Step 3: Calculate the determinant of each term
Now we calculate each term in the expression \( |(2A)^2 (3B) (5C)^{-1}| \):

1. For \( |(2A)^2| \):
\[ |(2A)^2| = |2A|^2 = (2^n |A|)^2 = (2^n \cdot 2)^2 = 2^{2n} \cdot 4 = 4 \cdot 2^{2n} \]

2. For \( |3B| \):
\[ |3B| = 3^n |B| = 3^n \cdot 3 = 3^{n+1} \]

3. For \( |(5C)^{-1}| \):
\[ |(5C)^{-1}| = \frac{1}{|5C|} = \frac{1}{5^n |C|} = \frac{1}{5^n \cdot 5} = \frac{1}{5^{n+1}} \]

### Step 4: Combine the determinants
Now we can combine these results:
\[ |(2A)^2 (3B) (5C)^{-1}| = |(2A)^2| \cdot |3B| \cdot |(5C)^{-1}| \]
Substituting the values we calculated:
\[ = (4 \cdot 2^{2n}) \cdot (3^{n+1}) \cdot \left(\frac{1}{5^{n+1}}\right) \]
This simplifies to:
\[ = \frac{4 \cdot 2^{2n} \cdot 3^{n+1}}{5^{n+1}} \]

### Step 5: Set the equation
We know this equals \( \frac{1728}{125} \):
\[ \frac{4 \cdot 2^{2n} \cdot 3^{n+1}}{5^{n+1}} = \frac{1728}{125} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4 \cdot 2^{2n} \cdot 3^{n+1} \cdot 125 = 1728 \cdot 5^{n+1} \]
Now simplify \( 1728 \):
\[ 1728 = 12^3 = (3 \cdot 2^2)^3 = 3^3 \cdot 2^6 \]
Thus, we have:
\[ 4 \cdot 2^{2n} \cdot 3^{n+1} \cdot 125 = 3^3 \cdot 2^6 \cdot 5^{n+1} \]

### Step 7: Factor and equate powers
Now we can equate the coefficients:
- The left-hand side has \( 4 = 2^2 \), so we have \( 2^{2 + 2n} \).
- The right-hand side has \( 2^6 \).

Equating powers of 2:
\[ 2 + 2n = 6 \implies 2n = 4 \implies n = 2 \]

### Step 8: Conclusion
Thus, the value of \( n \) is:
\[ \boxed{2} \]