For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.2x1 + 3x2 − x3 = 14x1 + 2x2 + x3 = 45x1 + 9x2 + 2x3 = 71 0 -5 160 1 3 -60 0 0 -19Incorrect

Answer:The reduced row-echelon form of the linear system is [tex]\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right][/tex]Step-by-step explanation:We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:Interchange two rowsMultiply one row by a nonzero numberAdd a multiple of one row to a different rowTo find the reduced row-echelon form of this augmented matrix[tex]\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right][/tex]You need to follow these steps:Divide row 1 by 2 [tex]\left(R_1=\frac{R_1}{2}\right)[/tex][tex]\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right][/tex]Subtract row 1 from row 2 [tex]\left(R_2=R_2-R_1\right)[/tex][tex]\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right][/tex]Subtract row 1 multiplied by 5 from row 3 [tex]\left(R_3=R_3-\left(5\right)R_1\right)[/tex][tex]\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right][/tex]Subtract row 2 multiplied by 3 from row 1 [tex]\left(R_1=R_1-\left(3\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right][/tex]Subtract row 2 multiplied by 3 from row 3 [tex]\left(R_3=R_3-\left(3\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right][/tex]Multiply row 2 by 2 [tex]\left(R_2=\left(2\right)R_2\right)[/tex][tex]\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right][/tex]Divide row 3 by −19 [tex]\left(R_3=\frac{R_3}{-19}\right)[/tex][tex]\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right][/tex]Subtract row 3 multiplied by 16 from row 1 [tex]\left(R_1=R_1-\left(16\right)R_3\right)[/tex][tex]\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right][/tex]Add row 3 multiplied by 6 to row 2 [tex]\left(R_2=R_2+\left(6\right)R_3\right)[/tex][tex]\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right][/tex]