If A is an invertible matrix of order n and k is any positive real number, then the value of `[det(kA)]^(1) det A` is

Options

A. `k^(-n)`
B. `k^(-1)`
C. `k^(n)`
D. nk

Explanation

To solve the problem, we need to find the value of \([ \det(kA) ]^{-1} \cdot \det(A)\), where \(A\) is an invertible matrix of order \(n\) and \(k\) is a positive real number.

### Step-by-Step Solution:

1. **Understanding the Determinant of a Scalar Multiple of a Matrix**:
The determinant of a scalar multiplied by a matrix can be expressed as:
\[ \det(kA) = k^n \det(A) \]
where \(n\) is the order of the matrix \(A\).

2. **Substituting into the Expression**:
We need to substitute \(\det(kA)\) into our expression:
\[ [\det(kA)]^{-1} \cdot \det(A) = [k^n \det(A)]^{-1} \cdot \det(A) \]

3. **Calculating the Inverse of the Determinant**:
The inverse of \(\det(kA)\) is:
\[ [k^n \det(A)]^{-1} = \frac{1}{k^n \det(A)} \]

4. **Combining the Terms**:
Now, we can combine this with \(\det(A)\):
\[ [\det(kA)]^{-1} \cdot \det(A) = \frac{1}{k^n \det(A)} \cdot \det(A) \]

5. **Simplifying the Expression**:
The \(\det(A)\) in the numerator and denominator cancels out:
\[ = \frac{1}{k^n} \]

6. **Final Result**:
Therefore, the final value of the expression \([ \det(kA) ]^{-1} \cdot \det(A)\) is:
\[ \frac{1}{k^n} \]

### Conclusion:
The value of \([ \det(kA) ]^{-1} \cdot \det(A)\) is \(\frac{1}{k^n}\).