Given `A=[(4,2,5),(2,0,3),(-1,1,0)]` write the value of det. `("2AA"^(-1))` .

Answer

To find the value of \( \text{det}(2AA^{-1}) \), we can follow these steps:

### Step 1: Understand the properties of determinants
We know that for any square matrix \( A \):
\[ \text{det}(AA^{-1}) = \text{det}(I) = 1 \]
where \( I \) is the identity matrix.

### Step 2: Use the property of determinants with scalar multiplication
The determinant of a scalar multiple of a matrix can be expressed as:
\[ \text{det}(kA) = k^n \cdot \text{det}(A) \]
where \( n \) is the order (dimension) of the square matrix \( A \) and \( k \) is a scalar.

### Step 3: Apply the properties to our problem
In our case, we need to find:
\[ \text{det}(2AA^{-1}) = \text{det}(2I) \]
Since \( AA^{-1} = I \).

### Step 4: Calculate the determinant of \( 2I \)
For a \( 3 \times 3 \) identity matrix \( I \):
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Multiplying by 2 gives:
\[ 2I = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \]

### Step 5: Calculate the determinant of \( 2I \)
Using the property of determinants:
\[ \text{det}(2I) = 2^3 \cdot \text{det}(I) = 2^3 \cdot 1 = 8 \]

### Final Answer
Thus, the value of \( \text{det}(2AA^{-1}) \) is:
\[ \boxed{8} \]