# Chapter 1 - Linear Equations in Linear Algebra - 1.2 Exercises: 9

Row Reduced Echelon Form: $\left[ {\begin{array}{cc} 1 & 0 & 7 & -6\\ 0& 1 & -6 & 5 \\ \end{array} } \right]$ Parametric Form Solutions: $x_1 = 4+5x_3$ $x_2 = 5+6x_3$ $x_3$ is free

#### Work Step by Step

Find the general solutions of the systems $\left[ {\begin{array}{cc} 0 & 1 & -6 & 5\\ 1& -2 & 7 & -6 \\ \end{array} } \right]$ Swap the first row with the second row $\left[ {\begin{array}{cc} 1 & -2 & 7 & -6\\ 0& 1 & -6 & 5 \\ \end{array} } \right]$ Pivots are 1 in row 1 and 1 in row 2. Pivot columns in column 1 and 2 Create a zero for the second row pivot by multiplying the second pivot by 2 and adding it to -2 $\left[ {\begin{array}{cc} 1 & 0 & 7 & -6\\ 0& 1 & -6 & 5 \\ \end{array} } \right]$ The matrix is now in reduced echelon form Define the variables $x_1 ....+7x_3 = -6$ $..+x_2 -6x_3 = 5$ Solve the first equation for $x_1$ and the second equation for $x_2$. $x_3$ has no restrictions therefore is the free variable. $x_1 -2(5+6x_3)+7x_3 = -6$ $x_1 - 10 - 12x_3 + 7x_3 = -6$ $x_1 -5x_3 = 4$ $x_1 = 4+5x_3$ $x_1 = 4+5x_3$ $x_2 = 5+6x_3$ $x_3$ is free

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