Reduce to reduced row echelon form. \( \left[\begin{array}{cccc}-5 & -15 & 6 & 40 \\ 3 & 9 & -7 & -41 \\ 1 & 3 & -3 & -17 \\ -15 & -45 & 18 & 120\end{array}\right] \)

To reduce the given matrix to reduced row echelon form, we can perform a series of row operations. Let's denote the matrix as \( A \): \[ A = \left[\begin{array}{cccc}-5 & -15 & 6 & 40 \\ 3 & 9 & -7 & -41 \\ 1 & 3 & -3 & -17 \\ -15 & -45 & 18 & 120\end{array}\right] \] We will perform the following steps: 1. **Row Swaps**: Start by swapping \( R_1 \) and \( R_3 \) to get a leading 1 at the top: \[ \left[\begin{array}{cccc} 1 & 3 & -3 & -17 \\ 3 & 9 & -7 & -41 \\ -5 & -15 & 6 & 40 \\ -15 & -45 & 18 & 120\end{array}\right] \] 2. **Eliminate below the leading 1 in the first column**: - Replace \( R_2 \) with \( R_2 - 3R_1 \) - Replace \( R_3 \) with \( R_3 + 5R_1 \) - Replace \( R_4 \) with \( R_4 + 15R_1 \) Calculating those gives: \[ R_2: 3 - 3(1) = 0, \quad 9 - 3(3) = 0, \quad -7 + 3(3) = 2, \quad -41 + 3(17) = 10 \] \[ R_3: -5 + 5(1) = 0, \quad -15 + 5(3) = 0, \quad 6 - 5(3) = -9, \quad 40 - 5(17) = -45 \] \[ R_4: -15 + 15(1) = 0, \quad -45 + 15(3) = 0, \quad 18 - 15(3) = -27, \quad 120 - 15(17) = -15 \] Thus, our matrix changes to: \[ \left[\begin{array}{cccc} 1 & 3 & -3 & -17 \\ 0 & 0 & 2 & 10 \\ 0 & 0 & -9 & -45 \\ 0 & 0 & -27 & -15 \end{array}\right] \] 3. **Normalize the second row** by multiplying \( R_2 \) by \( \frac{1}{2} \): \[ R_2: \left[ 0, 0, 1, 5 \right] \] Our matrix now looks like this: \[ \left[\begin{array}{cccc} 1 & 3 & -3 & -17 \\ 0 & 0 & 1 & 5 \\ 0 & 0 & -9 & -45 \\ 0 & 0 & -27 & -15 \end{array}\right] \] 4. **Eliminate the third column** from \( R_3 \) and \( R_4 \): - For \( R_3 \), replace it with \( R_3 + 9R_2 \) - For \( R_4 \), replace it with \( R_4 + 27R_2 \) This results in: \[ R_3: 0 + 9(0) = 0, \quad 0 + 9(0) = 0, \quad -9 + 9(1) = 0, \quad -45 + 9(5) = 0 \] \[ R_4: 0 + 27(0) = 0, \quad 0 + 27(0) = 0, \quad -27 + 27(1) = 0, \quad -15 + 27(5) = 120 \] So, we update our matrix to: \[ \left[\begin{array}{cccc} 1 & 3 & -3 & -17 \\ 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \] 5. **Now eliminate elements above the pivot in the second column based on R2**: - Replace \( R_1 \) using \( R_1 - 3R_2 \): \[ R_1: 1 - 0,