Solve Using a Matrix by Elimination 3x+y-z=3 , 8x+8y-7z=8

$3 x + y - z = 3$ , $8 x + 8 y - 7 z = 8$

Write the system of equations in matrix form.

$[\begin{array}{c} 3 & 1 & - 1 & 3 \\ 8 & 8 & - 7 & 8 \end{array}]$

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{1} = \frac{1}{3} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{3} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} \\ 8 & 8 & - 7 & 8 \end{array}]$

$R_{1} = \frac{1}{3} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{3} R_{1}$ .

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (1) & (\frac{1}{3}) \cdot (- 1) & (\frac{1}{3}) \cdot (3) \\ 8 & 8 & - 7 & 8 \end{array}]$

$R_{1} = \frac{1}{3} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ 8 & 8 & - 7 & 8 \end{array}]$

Perform the row operation $R_{2} = - 8 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 8 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 8 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 8 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ (- 8) \cdot (1) + 8 & (- 8) \cdot (\frac{1}{3}) + 8 & (- 8) \cdot (- \frac{1}{3}) - 7 & (- 8) \cdot (1) + 8 \end{array}]$

$R_{2} = - 8 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ 0 & \frac{16}{3} & - \frac{13}{3} & 0 \end{array}]$

Perform the row operation $R_{2} = \frac{3}{16} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{3}{16} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ \frac{3}{16} R_{2} & \frac{3}{16} R_{2} & \frac{3}{16} R_{2} & \frac{3}{16} R_{2} \end{array}]$

$R_{2} = \frac{3}{16} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{3}{16} R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ (\frac{3}{16}) \cdot (0) & (\frac{3}{16}) \cdot (\frac{16}{3}) & (\frac{3}{16}) \cdot (- \frac{13}{3}) & (\frac{3}{16}) \cdot (0) \end{array}]$

$R_{2} = \frac{3}{16} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & \frac{1}{3} & - \frac{1}{3} & 1 \\ 0 & 1 & - \frac{13}{16} & 0 \end{array}]$

Perform the row operation $R_{1} = - \frac{1}{3} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{1}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{1}{3} R_{2} + R_{1} & - \frac{1}{3} R_{2} + R_{1} & - \frac{1}{3} R_{2} + R_{1} & - \frac{1}{3} R_{2} + R_{1} \\ 0 & 1 & - \frac{13}{16} & 0 \end{array}]$

$R_{1} = - \frac{1}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{1}{3}) \cdot (0) + 1 & (- \frac{1}{3}) \cdot (1) + \frac{1}{3} & (- \frac{1}{3}) \cdot (- \frac{13}{16}) - \frac{1}{3} & (- \frac{1}{3}) \cdot (0) + 1 \\ 0 & 1 & - \frac{13}{16} & 0 \end{array}]$

$R_{1} = - \frac{1}{3} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & - \frac{1}{16} & 1 \\ 0 & 1 & - \frac{13}{16} & 0 \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x - \frac{z}{16} = 1$

$y - \frac{13 z}{16} = 0$

Add $\frac{z}{16}$ to both sides of the equation.

$x = 1 + \frac{z}{16}$

$y - \frac{13 z}{16} = 0$

Add $\frac{13 z}{16}$ to both sides of the equation.

$y = \frac{13 z}{16}$

$x = 1 + \frac{z}{16}$

The solution is the set of ordered pairs that makes the system true.

$(1 + \frac{z}{16}, \frac{13 z}{16}, z)$

Solve Using a Matrix by Elimination 3x+y-z=3 , 8x+8y-7z=8

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