If A =  then det(A^-1) is _____________ (correct to two decimal places).

The correct answer is 0.25.

To find the determinant of the inverse matrix, we use the property of determinants for invertible matrices, which states that:
$det\left(A^{-1}\right)=\frac{1}{det\left(A\right)}$

From the given image, the matrix $A$ is:
$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 0& 0& 1\end{array}\right]$

Notice that matrix $A$ is an upper triangular matrix (all entries below the main diagonal are zero). The determinant of any triangular matrix is simply the product of its diagonal elements:
$det\left(A\right)=1\times 4\times 1=4$

Alternatively, we can compute the determinant by expanding along the first column:
$det\left(A\right)=1\times (4\times 1-5\times 0)-0+0=4$

Now, using the relationship between the determinant of a matrix and its inverse:
$det\left(A^{-1}\right)=\frac{1}{4}=0.25$