Solve Using an Augmented Matrix x_1-4x_2=-7 4x_1-x_2=2

$x_{1} - 4 x_{2} = - 7$ $4 x_{1} - x_{2} = 2$

Step 1

Write the system of equations in matrix form.

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

Step 2

Find the reduced row echelon form of the matrix.

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & - 4 & - 7 \\ 0 & 15 & 30 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{15} R_{2}$ 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 $R_{2} = \frac{1}{15} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - 4 & - 7 \\ \frac{1}{15} R_{2} & \frac{1}{15} R_{2} & \frac{1}{15} R_{2} \end{array}]$

$R_{2} = \frac{1}{15} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{15} R_{2}$ .

$[\begin{array}{c} 1 & - 4 & - 7 \\ (\frac{1}{15}) \cdot (0) & (\frac{1}{15}) \cdot (15) & (\frac{1}{15}) \cdot (30) \end{array}]$

$R_{2} = \frac{1}{15} R_{2}$

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

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

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 $0$ .

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

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

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

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

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

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

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

Step 3

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

$x_{1} = 1$

$x_{2} = 2$

Step 4

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

$(1, 2)$

Solve Using an Augmented Matrix x_1-4x_2=-7 4x_1-x_2=2

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