Solve Using a Matrix by Elimination x+4y-z=4 , 2x+5y-8z=15 , x+3y-3z=1

$x + 4 y - z = 4$ , $2 x + 5 y - 8 z = 15$ , $x + 3 y - 3 z = 1$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ 2 & 5 & - 8 & 15 \\ 1 & 3 & - 3 & 1 \end{array}]$

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \\ 1 & 3 & - 3 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (4) + 5 & (- 2) \cdot (- 1) - 8 & (- 2) \cdot (4) + 15 \\ 1 & 3 & - 3 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ 0 & - 3 & - 6 & 7 \\ 1 & 3 & - 3 & 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ 0 & - 3 & - 6 & 7 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 1 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ 0 & - 3 & - 6 & 7 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (4) + 3 & (- 1) \cdot (- 1) - 3 & (- 1) \cdot (4) + 1 \end{array}]$

$R_{3} = - 1 \cdot R_{1} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ 0 & - 3 & - 6 & 7 \\ 0 & - 1 & - 2 & - 3 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{3} R_{2}$ 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 $R_{2} = - \frac{1}{3} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ - \frac{1}{3} R_{2} & - \frac{1}{3} R_{2} & - \frac{1}{3} R_{2} & - \frac{1}{3} R_{2} \\ 0 & - 1 & - 2 & - 3 \end{array}]$

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

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ (- \frac{1}{3}) \cdot (0) & (- \frac{1}{3}) \cdot (- 3) & (- \frac{1}{3}) \cdot (- 6) & (- \frac{1}{3}) \cdot (7) \\ 0 & - 1 & - 2 & - 3 \end{array}]$

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

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

$[\begin{array}{c} 1 & 4 & - 1 & 4 \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & - 1 & - 2 & - 3 \end{array}]$

Perform the row operation $R_{1} = - 4 \cdot R_{2} + R_{1}$ 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 $R_{1} = - 4 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - 4 \cdot R_{2} + R_{1} & - 4 \cdot R_{2} + R_{1} & - 4 \cdot R_{2} + R_{1} & - 4 \cdot R_{2} + R_{1} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & - 1 & - 2 & - 3 \end{array}]$

$R_{1} = - 4 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - 4 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (- 4) \cdot (0) + 1 & (- 4) \cdot (1) + 4 & (- 4) \cdot (2) - 1 & (- 4) \cdot (- \frac{7}{3}) + 4 \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & - 1 & - 2 & - 3 \end{array}]$

$R_{1} = - 4 \cdot R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & - 1 & - 2 & - 3 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ R_{2} + R_{3} & R_{2} + R_{3} & R_{2} + R_{3} & R_{2} + R_{3} \end{array}]$

$R_{3} = R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 + 0 & 1 - 1 & 2 - 2 & - \frac{7}{3} - 3 \end{array}]$

$R_{3} = R_{2} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & 0 & 0 & - \frac{16}{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ - \frac{3}{16} R_{3} & - \frac{3}{16} R_{3} & - \frac{3}{16} R_{3} & - \frac{3}{16} R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ (- \frac{3}{16}) \cdot (0) & (- \frac{3}{16}) \cdot (0) & (- \frac{3}{16}) \cdot (0) & (- \frac{3}{16}) \cdot (- \frac{16}{3}) \end{array}]$

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

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & - 9 & \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{1} = - \frac{40}{3} R_{3} + R_{1}$ 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 $R_{1} = - \frac{40}{3} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{40}{3} R_{3} + R_{1} & - \frac{40}{3} R_{3} + R_{1} & - \frac{40}{3} R_{3} + R_{1} & - \frac{40}{3} R_{3} + R_{1} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & 0 & 0 & 1 \end{array}]$

$R_{1} = - \frac{40}{3} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{40}{3} R_{3} + R_{1}$ .

$[\begin{array}{c} (- \frac{40}{3}) \cdot (0) + 1 & (- \frac{40}{3}) \cdot (0) + 0 & (- \frac{40}{3}) \cdot (0) - 9 & (- \frac{40}{3}) \cdot (1) + \frac{40}{3} \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & 0 & 0 & 1 \end{array}]$

$R_{1} = - \frac{40}{3} R_{3} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & - 9 & 0 \\ 0 & 1 & 2 & - \frac{7}{3} \\ 0 & 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{2} = \frac{7}{3} R_{3} + R_{2}$ 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 $R_{2} = \frac{7}{3} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & - 9 & 0 \\ \frac{7}{3} R_{3} + R_{2} & \frac{7}{3} R_{3} + R_{2} & \frac{7}{3} R_{3} + R_{2} & \frac{7}{3} R_{3} + R_{2} \\ 0 & 0 & 0 & 1 \end{array}]$

$R_{2} = \frac{7}{3} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{7}{3} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & - 9 & 0 \\ (\frac{7}{3}) \cdot (0) + 0 & (\frac{7}{3}) \cdot (0) + 1 & (\frac{7}{3}) \cdot (0) + 2 & (\frac{7}{3}) \cdot (1) - \frac{7}{3} \\ 0 & 0 & 0 & 1 \end{array}]$

$R_{2} = \frac{7}{3} R_{3} + R_{2}$

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

$[\begin{array}{c} 1 & 0 & - 9 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

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

$x - 9 z = 0$

$y + 2 z = 0$

$0 = 1$

Add $9 z$ to both sides of the equation.

$x = 9 z$

$y + 2 z = 0$

$0 = 1$

Subtract $2 z$ from both sides of the equation.

$y = - 2 z$

$x = 9 z$

$0 = 1$

Since $0 \neq 1$ , there are no solutions.

No solution

Solve Using a Matrix by Elimination x+4y-z=4 , 2x+5y-8z=15 , x+3y-3z=1

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