## College Algebra (6th Edition)

Start with the augmented matrix. Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$ (row-echelon form). $\left[\begin{array}{lllll} 2 & -4 & 1 & | & 3\\ 1 & -3 & 1 & | & 5\\ 3 & -7 & 2 & | & 12 \end{array}\right]$ Swapping R1 and R2 brings a 1 into the first diagonal position $\left[\begin{array}{lllll} 1 & -3 & 1 & | & 5\\ 2 & -4 & 1 & | & 3\\ 3 & -7 & 2 & | & 12 \end{array}\right]\left\{\begin{array}{l} .\\ -2R1+R2\rightarrow R2.\\ -3R1+R3\rightarrow R3. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -3 & 1 & | & 5\\ 0 & 2 & -1 & | & -7\\ 0 & 2 & -1 & | & -3 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ -R2+R3\rightarrow R3. \end{array}\right.$ $\left[\begin{array}{lllll} 1 & -3 & 1 & | & 5\\ 0 & 2 & -1 & | & -7\\ 0 & 0 & 0 & | & 4 \end{array}\right]$ The last row represents the equation 0=4 which is impossible. The system has no solution.