Answer
$\begin{bmatrix} 26 & 7 & -25\\ -3 & -1 & 3\\-27 & -7 & 26 \end{bmatrix}$
Work Step by Step
Step 1. Write the original matrix together with an identity matrix.
$\begin{array}(\\B=A|I= \\ \\ \end{array} \begin{bmatrix} 5 & 7 & 4 & | &1 & 0 & 0\\ 3 & -1 & 3 & | &0 & 1 & 0\\6 & 7 & 5 & | &0 & 0 & 1 \end{bmatrix} \begin{array}( 6R_1\to R_1 \\10R_2\to R_2 \\5R_3\to R_3 \\ \end{array}$
Step 2. Use row operations given on the right side of the matrix to transform the left half into reduced row-echelon form towards $B=IA^{-1}$
$\begin{bmatrix} 30 & 42 & 24 & | &6 & 0 & 0\\ 30 & -10 & 30 & | &0 & 10 & 0\\30 & 35 & 25 & | &0 & 0 & 5 \end{bmatrix} \begin{array}( \\R_1-R_2\to R_2 \\R_1-R_3\to R_3 \\ \end{array}$
$\begin{bmatrix} 30 & 42 & 24 & | &6 & 0 & 0\\ 0 & 52 & -6 & | &6 & -10 & 0\\0 & 7 & -1 & | &6 & 0 & -5 \end{bmatrix} \begin{array}( \\7R_2/2\to R_2 \\26R_3\to R_3 \\ \end{array}$
$\begin{bmatrix} 30 & 42 & 24 & | &6 & 0 & 0\\ 0 & 182 & -21 & | &21 & -35 & 0\\0 & 182 & -26 & | &156 & 0 & -130 \end{bmatrix} \begin{array}( \\ \\R_2-R_3\to R_3 \\ \end{array}$
$\begin{bmatrix} 30 & 42 & 24 & | &6 & 0 & 0\\ 0 & 182 & -21 & | &21 & -35 & 0\\0 & 0 & 5 & | &-135 & -35 & 130 \end{bmatrix} \begin{array}( R_1/30 \\R_2/182 \\R_3/5 \\ \end{array}$
$\begin{bmatrix} 1 & 7/5 & 4/5 & | &1/5 & 0 & 0\\ 0 & 1 & -3/26 & | &3/26 & -5/26 & 0\\0 & 0 & 1 & | &-27 & -7 & 26 \end{bmatrix} \begin{array}( \\R_2+3R_3/26\to R_2 \\ \\ \end{array}$
$\begin{bmatrix} 1 & 7/5 & 4/5 & | &1/5 & 0 & 0\\ 0 & 1 & 0 & | &-3 & -1 & 3\\0 & 0 & 1 & | &-27 & -7 & 26 \end{bmatrix} \begin{array}( R_1-7R_2/5-4R_3/5\to R_1\\ \\ \\ \end{array}$
$\begin{bmatrix} 1 & 0 &0 & | &26 & 7 & -25\\ 0 & 1 & 0 & | &-3 & -1 & 3\\0 & 0 & 1 & | &-27 & -7 & 26 \end{bmatrix}$
Step 3. Conclusion: the inverse of the original matrix is
$\begin{bmatrix} 26 & 7 & -25\\ -3 & -1 & 3\\-27 & -7 & 26 \end{bmatrix}$
