The reduced row echelon form of \( \left[\begin{array}{ccc}4 & 24 & -3 \\ -8 & -48 & -6\end{array}\right] \) is

To find the reduced row echelon form (RREF) of the matrix \( \left[\begin{array}{ccc}4 & 24 & -3 \\ -8 & -48 & -6\end{array}\right] \), we can start by performing row operations. 1. First, let's simplify the first row by dividing it by 4: \[ \left[\begin{array}{ccc}1 & 6 & -\frac{3}{4} \\ -8 & -48 & -6\end{array}\right] \] 2. Next, we'll eliminate the entry below the leading 1 in the first column by adding 8 times the first row to the second row: \[ R_2' = R_2 + 8R_1 \] This gives us: \[ \left[\begin{array}{ccc}1 & 6 & -\frac{3}{4} \\ 0 & 0 & -\frac{24}{4}\end{array}\right] = \left[\begin{array}{ccc}1 & 6 & -\frac{3}{4} \\ 0 & 0 & -6\end{array}\right] \] 3. Next, we can simplify the second row by dividing it by -6: \[ R_2'' = \frac{1}{-6} R_2 = \left[\begin{array}{ccc}1 & 6 & -\frac{3}{4} \\ 0 & 0 & 1\end{array}\right] \] 4. Then we eliminate the entry in the third column of the first row by adding \(\frac{3}{4}\) times the second row to the first row: \[ R_1'' = R_1 + \frac{3}{4} R_2 = \left[\begin{array}{ccc}1 & 6 & 0 \\ 0 & 0 & 1\end{array}\right] \] 5. Finally, we want the leading entry in the second row to be 0 in the first column, resulting in: \[ \left[\begin{array}{ccc} 1 & 6 & 0 \\ 0 & 0 & 1 \end{array}\right] \] Thus, the reduced row echelon form of the given matrix is \[ \left[\begin{array}{ccc} 1 & 6 & 0 \\ 0 & 0 & 1 \end{array}\right]. \]