## Elementary Linear Algebra 7th Edition

Write the augmented matrix of the system of linear equations. $\begin{bmatrix} 1 & 2 & 6 & 1\\ 2 & 5 & 15 & 4\\ 3 & 1 & 3 & -6 \end{bmatrix}$ Add -2 times the 1st row to the 2nd row to produce a new 2nd row. Add -3 times the 1st row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & 2 & 6 & 1\\ 0 & 1 & 3 & 2\\ 0& -5 & -15 & -9 \end{bmatrix}$ Add 5 times the 2nd row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & 2 & 6 & 1\\ 0 & 1 & 3 & 2\\ 0& 0 & 0 & 1 \end{bmatrix}$ Use back-substitution to find the solution. $0z=1 \rightarrow$ this equation is not possible $\rightarrow$ the system has no solution.