Compute the inverse using row reduction. \( \left[\begin{array}{cccc}6 & 7 & -1 & 7 \\ 12 & 15 & 5 & 13 \\ 12 & 16 & 10 & 11 \\ 18 & 24 & 12 & 22\end{array}\right] \)

To find the inverse of the matrix \( A = \begin{bmatrix} 6 & 7 & -1 & 7 \\ 12 & 15 & 5 & 13 \\ 12 & 16 & 10 & 11 \\ 18 & 24 & 12 & 22 \end{bmatrix} \) using row reduction, we will augment the matrix \( A \) with the identity matrix \( I \) of the same size: \[ \left[ \begin{array}{cccc|cccc} 6 & 7 & -1 & 7 & 1 & 0 & 0 & 0 \\ 12 & 15 & 5 & 13 & 0 & 1 & 0 & 0 \\ 12 & 16 & 10 & 11 & 0 & 0 & 1 & 0 \\ 18 & 24 & 12 & 22 & 0 & 0 & 0 & 1 \end{array} \right] \] We will perform row operations to transform the left side into the identity matrix. ### Step 1: Normalize the first row Divide the first row by 6: \[ R_1 \leftarrow \frac{1}{6} R_1 \implies \left[ \begin{array}{cccc|cccc} 1 & \frac{7}{6} & -\frac{1}{6} & \frac{7}{6} & \frac{1}{6} & 0 & 0 & 0 \\ 12 & 15 & 5 & 13 & 0 & 1 & 0 & 0 \\ 12 & 16 & 10 & 11 & 0 & 0 & 1 & 0 \\ 18 & 24 & 12 & 22 & 0 & 0 & 0 & 1 \end{array} \right] \] ### Step 2: Eliminate the first column below the first row - For \( R_2 \): \( R_2 \leftarrow R_2 - 12R_1 \) - For \( R_3 \): \( R_3 \leftarrow R_3 - 12R_1 \) - For \( R_4 \): \( R_4 \leftarrow R_4 - 18R_1 \) Calculating these: \[ R_2: 12 - 12 \cdot 1 = 0, \quad 15 - 12 \cdot \frac{7}{6} = 15 - 14 = 1, \quad 5 + 2 = 7, \quad 13 - 14 = -1 \] \[ R_3: 12 - 12 \cdot 1 = 0, \quad 16 - 12 \cdot \frac{7}{6} = 16 - 14 = 2, \quad 10 + 2 = 12, \quad 11 - 14 = -3 \] \[ R_4: 18 - 18 \cdot 1 = 0, \quad 24 - 18 \cdot \frac{7}{6} = 24 - 21 = 3, \quad 12 + 3 = 15, \quad 22 - 21 = 1 \] The augmented matrix now looks like: \[ \left[ \begin{array}{cccc|cccc} 1 & \frac{7}{6} & -\frac{1}{6} & \frac{7}{6} & \frac{1}{6} & 0 & 0 & 0 \\ 0 & 1 & 7 & -1 & -2 & 1 & 0 & 0 \\ 0 & 2 & 12 & -3 & -4 & 0 & 1 & 0 \\ 0 & 3 & 15 & 1 & -3 & 0 & 0 & 1 \end{array} \right] \] ### Step 3: Normalize the second row The second row is already normalized. Now we will eliminate the second column below the second row. - For \( R_3 \): \( R_3 \leftarrow R_3 - 2R_2 \) - For \( R_4 \): \( R_4 \leftarrow R_4 - 3R_2 \) Calculating these: \[ R_3: 0, \quad 2 - 2 \cdot 1 = 0, \quad 12 - 2 \cdot 7 = -2, \quad -3 +