Algebra Examples

$4 x - y = - 4$ , $6 x - y = 0$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = \frac{1}{4} R_{1}$ 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 $R_{1} = \frac{1}{4} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{1} = \frac{1}{4} R_{1}$

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

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

$R_{1} = \frac{1}{4} R_{1}$

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

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

Perform the row operation $R_{2} = - 6 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 6 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ - 6 \cdot R_{1} + R_{2} & - 6 \cdot R_{1} + R_{2} & - 6 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 6 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 6 \cdot R_{1} + R_{2}$ .

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

$R_{2} = - 6 \cdot R_{1} + R_{2}$

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

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

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & - \frac{1}{4} & - 1 \\ 0 & 1 & 12 \end{array}]$

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

$[\begin{array}{c} \frac{1}{4} R_{2} + R_{1} & \frac{1}{4} R_{2} + R_{1} & \frac{1}{4} R_{2} + R_{1} \\ 0 & 1 & 12 \end{array}]$

$R_{1} = \frac{1}{4} R_{2} + R_{1}$

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

$[\begin{array}{c} (\frac{1}{4}) \cdot (0) + 1 & (\frac{1}{4}) \cdot (1) - \frac{1}{4} & (\frac{1}{4}) \cdot (12) - 1 \\ 0 & 1 & 12 \end{array}]$

$R_{1} = \frac{1}{4} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & 2 \\ 0 & 1 & 12 \end{array}]$

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

$x = 2$

$y = 12$

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

$(2, 12)$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 2 \\ 12 \end{array}]$

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