Answer
$$45$$
Work Step by Step
Given: $$\begin{array}{c}{A=\left[\begin{array}{ccc}{2} & {1} & {3} \\ {-1} & {2} & {6} \\ {4} & {1} & {12} \end{array}\right] }\end{array} $$
Perform row operations $R_2 \rightarrow R_2-\dfrac{R_1}{2}$ and $R_3 \rightarrow R_3-2R_1$ to obtain:
$$A=\begin{vmatrix}{2} & {1} & {3} \\ {0} & {5/2} & {15/2} \\ {0} & {-1} & {6}\end{vmatrix}$$
Again, perform row operations $R_3 \rightarrow R_3+\dfrac{2R_2}{5}$ to obtain:
$$A=\begin{vmatrix}{2} & {1} & {3} \\ {0} & {5/2} & {15/2} \\ {0} & {0} & {9}\end{vmatrix}$$
Determinant of triangular matrix is the product of its diagonal elements, so: $det (A)=(2)(\dfrac{5}{2})(9)=45$
