Elementary Linear Algebra 7th Edition

$$\left|\begin{array}{rrr}{4} & {3} & {-2} \\ {5} & {4} & {1} \\ {-2} & {3} & {4}\end{array}\right|=6\left|\begin{array}{rrr}{2} & {-3} & {-2} \\ {7} & {-10} & {0} \\ {1} & {0} & {0}\end{array}\right|=-60.$$
Given $$\left|\begin{array}{rrr}{4} & {3} & {-2} \\ {5} & {4} & {1} \\ {-2} & {3} & {4}\end{array}\right|.$$ Adding the first row to $2$ times the second row, adding $2$ times the first row to the third row, we get $$\left|\begin{array}{rrr}{4} & {3} & {-2} \\ {14} & {11} & {0} \\ {6} & {9} & {0}\end{array}\right|.$$ Factor $2$ out of the first column and $3$ out of the third row, we have $$\left|\begin{array}{rrr}{4} & {3} & {-2} \\ {14} & {11} & {0} \\ {6} & {9} & {0}\end{array}\right|=6\left|\begin{array}{rrr}{2} & {3} & {-2} \\ {7} & {11} & {0} \\ {1} & {3} & {0}\end{array}\right|.$$ Adding $-3$ times the first column to the second column, we have $$6\left|\begin{array}{rrr}{2} & {-3} & {-2} \\ {7} & {-10} & {0} \\ {1} & {0} & {0}\end{array}\right|.$$ Finally, we have $$\left|\begin{array}{rrr}{4} & {3} & {-2} \\ {5} & {4} & {1} \\ {-2} & {3} & {4}\end{array}\right|=6\left|\begin{array}{rrr}{2} & {-3} & {-2} \\ {7} & {-10} & {0} \\ {1} & {0} & {0}\end{array}\right|=-60.$$