1. **State the problem:** Find the determinant of the matrix $$\begin{bmatrix} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix}$$ using the row reduction method. 2. **Recall the rules for determinant and row operations:** - Swapping two rows multiplies the determinant by $-1$. - Multiplying a row by a scalar $k$ multiplies the determinant by $k$. - Adding a multiple of one row to another does not change the determinant. 3. **Start with the original matrix:** $$A = \begin{bmatrix} 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ 4 & 5 & 6 & 7 \\ 9 & 10 & 11 & 12 \end{bmatrix}$$ 4. **Make zeros below the pivot in column 1:** - Replace $R_2$ by $R_2 - \frac{3}{2} R_1$: $$R_2 = \begin{bmatrix} 3 & 4 & 5 & 6 \end{bmatrix} - \frac{3}{2} \times \begin{bmatrix} 2 & 3 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 3 - 3 & 4 - \frac{9}{2} & 5 - 6 & 6 - \frac{15}{2} \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{2} & -1 & -\frac{3}{2} \end{bmatrix}$$ - Replace $R_3$ by $R_3 - 2 R_1$: $$R_3 = \begin{bmatrix} 4 & 5 & 6 & 7 \end{bmatrix} - 2 \times \begin{bmatrix} 2 & 3 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 - 4 & 5 - 6 & 6 - 8 & 7 - 10 \end{bmatrix} = \begin{bmatrix} 0 & -1 & -2 & -3 \end{bmatrix}$$ - Replace $R_4$ by $R_4 - \frac{9}{2} R_1$: $$R_4 = \begin{bmatrix} 9 & 10 & 11 & 12 \end{bmatrix} - \frac{9}{2} \times \begin{bmatrix} 2 & 3 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 9 - 9 & 10 - \frac{27}{2} & 11 - 18 & 12 - \frac{45}{2} \end{bmatrix} = \begin{bmatrix} 0 & -\frac{7}{2} & -7 & -\frac{21}{2} \end{bmatrix}$$ Matrix now: $$\begin{bmatrix} 2 & 3 & 4 & 5 \\ 0 & -\frac{1}{2} & -1 & -\frac{3}{2} \\ 0 & -1 & -2 & -3 \\ 0 & -\frac{7}{2} & -7 & -\frac{21}{2} \end{bmatrix}$$ 5. **Make zeros below the pivot in column 2:** - Pivot is $-\frac{1}{2}$ at $R_2, C_2$. - Replace $R_3$ by $R_3 - 2 R_2$: $$R_3 = \begin{bmatrix} 0 & -1 & -2 & -3 \end{bmatrix} - 2 \times \begin{bmatrix} 0 & -\frac{1}{2} & -1 & -\frac{3}{2} \end{bmatrix} = \begin{bmatrix} 0 & -1 + 1 & -2 + 2 & -3 + 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \end{bmatrix}$$ - Replace $R_4$ by $R_4 - 7 R_2$: $$R_4 = \begin{bmatrix} 0 & -\frac{7}{2} & -7 & -\frac{21}{2} \end{bmatrix} - 7 \times \begin{bmatrix} 0 & -\frac{1}{2} & -1 & -\frac{3}{2} \end{bmatrix} = \begin{bmatrix} 0 & -\frac{7}{2} + \frac{7}{2} & -7 + 7 & -\frac{21}{2} + \frac{21}{2} \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \end{bmatrix}$$ Matrix now: $$\begin{bmatrix} 2 & 3 & 4 & 5 \\ 0 & -\frac{1}{2} & -1 & -\frac{3}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ 6. **Since rows 3 and 4 are zero rows, the matrix is singular and determinant is zero.** 7. **Final answer:** $$\boxed{0}$$ The determinant of the given matrix is zero because the matrix is not full rank after row reduction.