Find the determinant by row reduction to echelon form. \( \left|\begin{array}{rrrr}1 & -1 & -2 & 0 \\ 3 & -2 & 5 & 4 \\ 1 & 2 & 2 & 1 \\ -2 & 5 & 8 & 1\end{array}\right| \) Use row operations to reduce the matrix to echelon form. \( \left[\begin{array}{rrrr}1 & -1 & -2 & 0 \\ 3 & -2 & 5 & 4 \\ 1 & 2 & 2 & 1 \\ -2 & 8 & 1\end{array}\right] \sim \square \)

To find the determinant of the given matrix through row reduction to echelon form, we can follow a systematic approach. Start with the matrix: \[ \left[\begin{array}{rrrr} 1 & -1 & -2 & 0 \\ 3 & -2 & 5 & 4 \\ 1 & 2 & 2 & 1 \\ -2 & 5 & 8 & 1 \end{array}\right] \] ### Row Operations: 1. **Eliminate below the first pivot (1st row, 1st column)**: - Row 2: \( R_2 - 3R_1 \) - Row 3: \( R_3 - R_1 \) - Row 4: \( R_4 + 2R_1 \) Updating the rows gives: \[ \left[\begin{array}{rrrr} 1 & -1 & -2 & 0 \\ 0 & 1 & 11 & 4 \\ 0 & 3 & 4 & 1 \\ 0 & 3 & 4 & 1 \end{array}\right] \] 2. **Next, eliminate below the second pivot (2nd row, 2nd column)**: - Row 3: \( R_3 - 3R_2 \) - Row 4: \( R_4 - 3R_2 \) So, we perform: Row 3 becomes \( 0, 0, -29, -11 \) Row 4 becomes \( 0, 0, -29, -11 \) Now we have: \[ \left[\begin{array}{rrrr} 1 & -1 & -2 & 0 \\ 0 & 1 & 11 & 4 \\ 0 & 0 & -29 & -11 \\ 0 & 0 & -29 & -11 \end{array}\right] \] 3. **Final step, simplify**: - The last two rows are identical. We can say they lead to a determinant of 0 since their presence indicates linearly dependent rows. ### Final Matrix Form: Therefore, the matrix in echelon form becomes: \[ \left[\begin{array}{rrrr} 1 & -1 & -2 & 0 \\ 0 & 1 & 11 & 4 \\ 0 & 0 & -29 & -11 \\ 0 & 0 & 0 & 0 \end{array}\right] \] ### Determinant: The determinant of the original matrix is 0, because of the row of zeros, which indicates that the matrix is singular. In box notation for education: \[ \left[\begin{array}{rrrr} 1 & -1 & -2 & 0 \\ 0 & 1 & 11 & 4 \\ 0 & 0 & -29 & -11 \\ 0 & 0 & 0 & 0 \end{array}\right] \sim \boxed{0} \]