Solve Using a Matrix by Elimination 2x+4y=16 -4x-8y=32

$2 x + 4 y = 16$ $- 4 x - 8 y = 32$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & 4 & 16 \\ - 4 & - 8 & 32 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

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 $1$ .

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 $1$ .

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ - 4 & - 8 & 32 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (4) & (\frac{1}{2}) \cdot (16) \\ - 4 & - 8 & 32 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 8 \\ - 4 & - 8 & 32 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 2 & 8 \\ 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 8 \\ (4) \cdot (1) - 4 & (4) \cdot (2) - 8 & (4) \cdot (8) + 32 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 8 \\ 0 & 0 & 64 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{64} R_{2}$ 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 $R_{2} = \frac{1}{64} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 2 & 8 \\ \frac{1}{64} R_{2} & \frac{1}{64} R_{2} & \frac{1}{64} R_{2} \end{array}]$

$R_{2} = \frac{1}{64} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{64} R_{2}$ .

$[\begin{array}{c} 1 & 2 & 8 \\ (\frac{1}{64}) \cdot (0) & (\frac{1}{64}) \cdot (0) & (\frac{1}{64}) \cdot (64) \end{array}]$

$R_{2} = \frac{1}{64} R_{2}$

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

$[\begin{array}{c} 1 & 2 & 8 \\ 0 & 0 & 1 \end{array}]$

Perform the row operation $R_{1} = - 8 \cdot R_{2} + R_{1}$ 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 $R_{1} = - 8 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - 8 \cdot R_{2} + R_{1} & - 8 \cdot R_{2} + R_{1} & - 8 \cdot R_{2} + R_{1} \\ 0 & 0 & 1 \end{array}]$

$R_{1} = - 8 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - 8 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (- 8) \cdot (0) + 1 & (- 8) \cdot (0) + 2 & (- 8) \cdot (1) + 8 \\ 0 & 0 & 1 \end{array}]$

$R_{1} = - 8 \cdot R_{2} + R_{1}$

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

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

Step 3

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

$x + 2 y = 0$

$0 = 1$

Step 4

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

$x = - 2 y$

$0 = 1$

Step 5

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

No solution

Solve Using a Matrix by Elimination 2x+4y=16 -4x-8y=32

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