Let A and B are square matrices of order 3. If `|A|=4, |B|=6, B=A-2I and |adj(I-2A^(-1))|=k`, then the value of k is equal to

Answer

To solve the problem step by step, let's break it down:

Given:
- \( |A| = 4 \)
- \( |B| = 6 \)
- \( B = A - 2I \)

We need to find \( | \text{adj}(I - 2A^{-1}) | = k \).

### Step 1: Find \( |B| \) in terms of \( |A| \)

From the equation \( B = A - 2I \), we can express the determinant of \( B \) using the determinant of \( A \):

\[ |B| = |A - 2I| = |A| \cdot |I - 2A^{-1}| \]

### Step 2: Calculate \( |I - 2A^{-1}| \)

Using the property of determinants, we have:

\[ |B| = |A| \cdot |I - 2A^{-1}| \]

Substituting the known values:

\[ 6 = 4 \cdot |I - 2A^{-1}| \]

### Step 3: Solve for \( |I - 2A^{-1}| \)

Now, we can isolate \( |I - 2A^{-1}| \):

\[ |I - 2A^{-1}| = \frac{6}{4} = \frac{3}{2} \]

### Step 4: Find \( | \text{adj}(I - 2A^{-1}) | \)

The relationship between the adjugate and the determinant is given by:

\[ | \text{adj}(X) | = |X|^{n-1} \]

where \( n \) is the order of the matrix. Since \( A \) and \( B \) are \( 3 \times 3 \) matrices, \( n = 3 \).

Thus,

\[ | \text{adj}(I - 2A^{-1}) | = |I - 2A^{-1}|^{3-1} = |I - 2A^{-1}|^2 \]

### Step 5: Substitute the value of \( |I - 2A^{-1}| \)

Now we substitute the value we found:

\[ | \text{adj}(I - 2A^{-1}) | = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \]

### Conclusion

Thus, the value of \( k \) is:

\[ k = \frac{9}{4} \]