Find A of the matrix

Question

Find $$(2A)^T = \left[\begin{array}{cc}1&-1\\2&3\end{array}\right]^{-1}$$

My solution

$$2A^T = \left[\begin{array}{cc}3&-2\\1&1\end{array}\right] = \frac12\left[\begin{array}{cc}3&-2\\1&1\end{array}\right], A = \left[\begin{array}{cc}3&-2\\1&1\end{array}\right]$$

Apparently this is completely wrong and the answer is

$$\frac{1}{10}\left[\begin{array}{cc}3&-2\\1&1\end{array}\right]$$ I do not understand where they got the $1/10$.

Answer 1

Determinant = ad - bc = (3 $\times$ 1) - (-2 $\times$ 1) = 5

The inverse of the matrix = $\frac{1}{determinant}$adj(matrix) = $\frac {1}{5}$adj(matrix)

Bring the 2 over, it becomes $\frac{1}{10}$.

Answer 2

Hint: You may have forgotten to include the determinant $$\det\left[\begin{array}{cc}1&-1\\2&3\end{array}\right]$$ in your computation.