For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.x1 − 2x2 + 3x3 + 2x4 + x5 = 102x1 − 4x2 + 8x3 + 3x4 + 10x5 = 73x1 − 6x2 + 10x3 + 6x4 + 5x5 = 27

Accepted Solution

Answer:This is the reduced row-echelon form [tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&0&2&-3\\0&0&0&1&-4&7\end{array}\right][/tex] from the augmented matrix[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\2&-4&8&3&10&7\\3&-6&10&6&5&27\end{array}\right][/tex] Step-by-step explanation:To transform an augmented matrix to the reduced row-echelon form we need to follow this steps:1. Write the system of equations as an augmented matrix.The augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.[tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\2&-4&8&3&10&7\\3&-6&10&6&5&27\end{array}\right][/tex]2. Make zeros in column 1 except the entry at row 1, column 1 (this is the pivot entry). Subtract row 1 multiplied by 2 from row 2 [tex]\left(R_2=R_2-\left(2\right)R_1\right)[/tex][tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\0&0&2&-1&8&-13\\3&-6&10&6&5&27\end{array}\right][/tex]3. Subtract row 1 multiplied by 3 from row 3 [tex]\left(R_3=R_3-\left(3\right)R_1\right)[/tex][tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\0&0&2&-1&8&-13\\0&0&1&0&2&-3\end{array}\right][/tex]4. Since element at row 2 and column 2 (pivot element) equals 0, we need to swap rows. Find the first non-zero element in the column 2 under the pivot entry. As can be seen, there are no such entries. So, move to the next column. Make zeros in column 3 except the entry at row 2, column 3 (pivot entry). Divide row 2 by 2 [tex]\left(R_2=\frac{R_2}{2}\right)[/tex][tex]\left[\begin{array}{cccccc}1&-2&3&2&1&10\\0&0&1&-1/2&4&-13/2\\0&0&1&0&2&-3\end{array}\right][/tex]5. 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}{cccccc}1&-2&0&7/2&-11&59/2\\0&0&1&-1/2&4&-13/2\\0&0&1&0&2&-3\end{array}\right][/tex]6. Subtract row 2 from row 3 [tex]\left(R_3=R_3-R_2\right)[/tex][tex]\left[\begin{array}{cccccc}1&-2&0&7/2&-11&59/2\\0&0&1&-1/2&4&-13/2\\0&0&0&1/2&-2&7/2\end{array}\right][/tex]7. Make zeros in column 4 except the entry at row 3, column 4 (pivot entry). Subtract row 3 multiplied by 7 from row 1 [tex]\left(R_1=R_1-\left(7\right)R_3\right)[/tex][tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&-1/2&4&-13/2\\0&0&0&1/2&-2&7/2\end{array}\right][/tex]8. Add row 3 to row 2 [tex]\left(R_2=R_2+R_3\righ)[/tex][tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&0&2&-3\\0&0&0&1/2&-2&7/2\end{array}\right][/tex]9. Multiply row 3 by 2 [tex]\left(R_3=\left(2\right)R_3\right)[/tex][tex]\left[\begin{array}{cccccc}1&-2&0&0&3&5\\0&0&1&0&2&-3\\0&0&0&1&-4&7\end{array}\right][/tex]