Algebra Examples

$3 x + 3 y + 3 z = 6$ , $x - y = - 3$ , $- 4 x + y - z = - 1$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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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 $1$ .

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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 $1$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & - 2 & - 1 & - 5 \\ - 4 & 1 & - 1 & - 1 \end{array}]$

Perform the row operation $R_{3} = 4 \cdot R_{1} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = 4 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & - 2 & - 1 & - 5 \\ 4 \cdot R_{1} + R_{3} & 4 \cdot R_{1} + R_{3} & 4 \cdot R_{1} + R_{3} & 4 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = 4 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = 4 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & - 2 & - 1 & - 5 \\ (4) \cdot (1) - 4 & (4) \cdot (1) + 1 & (4) \cdot (1) - 1 & (4) \cdot (2) - 1 \end{array}]$

$R_{3} = 4 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & - 2 & - 1 & - 5 \\ 0 & 5 & 3 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ - \frac{1}{2} R_{2} & - \frac{1}{2} R_{2} & - \frac{1}{2} R_{2} & - \frac{1}{2} R_{2} \\ 0 & 5 & 3 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ (- \frac{1}{2}) \cdot (0) & (- \frac{1}{2}) \cdot (- 2) & (- \frac{1}{2}) \cdot (- 1) & (- \frac{1}{2}) \cdot (- 5) \\ 0 & 5 & 3 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & 1 & 1 & 2 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 5 & 3 & 7 \end{array}]$

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

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

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 5 & 3 & 7 \end{array}]$

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (\frac{1}{2}) + 1 & (- 1) \cdot (\frac{5}{2}) + 2 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 5 & 3 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 5 & 3 & 7 \end{array}]$

Perform the row operation $R_{3} = - 5 \cdot R_{2} + R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = - 5 \cdot R_{2} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} & - 5 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 5 \cdot R_{2} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 5 \cdot R_{2} + R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ (- 5) \cdot (0) + 0 & (- 5) \cdot (1) + 5 & (- 5) \cdot (\frac{1}{2}) + 3 & (- 5) \cdot (\frac{5}{2}) + 7 \end{array}]$

$R_{3} = - 5 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} & - \frac{11}{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 2 \cdot R_{3} & 2 \cdot R_{3} & 2 \cdot R_{3} & 2 \cdot R_{3} \end{array}]$

$R_{3} = 2 \cdot R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ (2) \cdot (0) & (2) \cdot (0) & (2) \cdot (\frac{1}{2}) & (2) \cdot (- \frac{11}{2}) \end{array}]$

$R_{3} = 2 \cdot R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & 1 & - 11 \end{array}]$

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

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

$[\begin{array}{c} - \frac{1}{2} R_{3} + R_{1} & - \frac{1}{2} R_{3} + R_{1} & - \frac{1}{2} R_{3} + R_{1} & - \frac{1}{2} R_{3} + R_{1} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & 1 & - 11 \end{array}]$

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

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

$[\begin{array}{c} (- \frac{1}{2}) \cdot (0) + 1 & (- \frac{1}{2}) \cdot (0) + 0 & (- \frac{1}{2}) \cdot (1) + \frac{1}{2} & (- \frac{1}{2}) \cdot (- 11) - \frac{1}{2} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & 1 & - 11 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 5 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & 1 & - 11 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 5 \\ - \frac{1}{2} R_{3} + R_{2} & - \frac{1}{2} R_{3} + R_{2} & - \frac{1}{2} R_{3} + R_{2} & - \frac{1}{2} R_{3} + R_{2} \\ 0 & 0 & 1 & - 11 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 5 \\ (- \frac{1}{2}) \cdot (0) + 0 & (- \frac{1}{2}) \cdot (0) + 1 & (- \frac{1}{2}) \cdot (1) + \frac{1}{2} & (- \frac{1}{2}) \cdot (- 11) + \frac{5}{2} \\ 0 & 0 & 1 & - 11 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

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

$x = 5$

$y = 8$

$z = - 11$

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

$(5, 8, - 11)$

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 5 \\ 8 \\ - 11 \end{array}]$

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