# Chapter 3 - Section 3.4 - Matrix Solutions to Linear Systems - Exercise Set - Page 229: 29

$\{(1,2,-1)\}$.

#### Work Step by Step

The given system of equations is $2x+2y+7z=-1$ $2x+y+2z=2$ $4x+6y+z=15$ The augmented matrix is $\Rightarrow \left[\begin{array}{ccc|c} 2 & 2 & 7& -1\\ 2 &1 & 2& 2 \\ 4&6&1&15 \end{array}\right]$ Perform $R_1\Rightarrow R_1/(2)$ $\Rightarrow \left[\begin{array}{ccc|c} 2/(2) & 2/(2) & 7/(2)& -1/(2)\\ 2 &1 & 2& 2 \\ 4&6&1&15 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & 7/2& -1/2\\ 2 &1 & 2& 2 \\ 4&6&1&15 \end{array}\right]$ Perform $R_2\rightarrow R_2-2 R_1$ and $R_3\rightarrow R_3-4 R_1$. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & 7/2& -1/2\\ 2-2(1) &1-2(1) & 2-2(7/2)& 2-2(-1/2) \\ 4-4(1)&6-4(1)&1-4(7/2)&15-4(-1/2) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & 7/2& -1/2\\ 0 &-1 & -5& 3 \\ 0&2&-13&17 \end{array}\right]$ Perform $R_2\Rightarrow R_2/(-1)$ Change the sign of the second row and swap the 3rd and 2nd row. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & 7/2& -1/2\\ 0/(-1) &-1/(-1) & -5/(-1)& 3/(-1) \\ 0&2&-13&17 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 1 & 7/2& -1/2\\ 0 &1 & 5& -3 \\ 0&2&-13&17 \end{array}\right]$ Perform $R_1\rightarrow R_1- R_2$ and $R_3\rightarrow R_3-2 R_2$. $\Rightarrow \left[\begin{array}{ccc|c} 1-0 & 1-1 & 7/2-5& -1/2-(-3)\\ 0 &1 & 5& -3 \\ 0-2(0)&2-2(1)&-13-2(5)&17-2(-3) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & -3/2& 5/2\\ 0 &1 & 5& -3 \\ 0&0&-23&23 \end{array}\right]$ Perform $R_3\rightarrow \frac{R_3}{(-23)}$. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & -3/2& 5/2\\ 0 &1 & 5& -3 \\ 0/(-23)&0/(-23)&-23/(-23)&23/(-23) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & -3/2& 5/2\\ 0 &1 & 5& -3 \\ 0&0&1&-1 \end{array}\right]$ Perform $R_1\rightarrow R_1+(3/2)R_3$ and $R_2\rightarrow R_2-5 R_3$. $\Rightarrow \left[\begin{array}{ccc|c} 1+(3/2)(0) & 0+(3/2)(0) & -3/2+(3/2)(1)& 5/2+(3/2)(-1)\\ 0-5(0) &1-5(0) & 5-5(1) & -3-5(-1) \\ 0&0&1&-1 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0& 1\\ 0&1 & 0 & 2 \\ 0&0&1&-1 \end{array}\right]$ Use back substitution to solve the linear system. $\Rightarrow x=1$. and $\Rightarrow y=2$. and $\Rightarrow z=-1$. The solution set is $\{(x,y,z)\}=\{(1,2,-1)\}$.

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