1. **State the problem:** Find the inverse of the matrix $$\begin{bmatrix}4 & -3 \\ 12 & 3\end{bmatrix}$$ given that its determinant is 38. 2. **Recall the formula for the inverse of a 2x2 matrix:** For a matrix $$A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$, the inverse is given by $$ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix} $$ where $$\det(A) = ad - bc$$. 3. **Apply the formula:** Here, $$a=4$$, $$b=-3$$, $$c=12$$, $$d=3$$, and $$\det(A) = 38$$. 4. **Calculate the adjugate matrix:** $$ \begin{bmatrix}d & -b \\ -c & a\end{bmatrix} = \begin{bmatrix}3 & 3 \\ -12 & 4\end{bmatrix} $$ 5. **Write the inverse matrix:** $$ A^{-1} = \frac{1}{38} \begin{bmatrix}3 & 3 \\ -12 & 4\end{bmatrix} = \begin{bmatrix}\frac{3}{38} & \frac{3}{38} \\ -\frac{12}{38} & \frac{4}{38}\end{bmatrix} $$ 6. **Simplify fractions where possible:** $$ -\frac{12}{38} = -\frac{6}{19}, \quad \frac{4}{38} = \frac{2}{19} $$ 7. **Final inverse matrix:** $$ A^{-1} = \begin{bmatrix}\frac{3}{38} & \frac{3}{38} \\ -\frac{6}{19} & \frac{2}{19}\end{bmatrix} $$ This matches the bottom-right colored box matrix given in the problem. **Answer:** The inverse matrix is $$\begin{bmatrix}\frac{3}{38} & \frac{3}{38} \\ -\frac{6}{19} & \frac{2}{19}\end{bmatrix}$$.