Algebra Examples

$2 x + y = - 2$ , $x + 2 y = 2$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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

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

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

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

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

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

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

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

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 $0$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$x = - 2$

$y = 2$

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

$(- 2, 2)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 2 \end{array}]$

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