Finite Math Examples

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

Step 1

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 $1$ .

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

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (3) & (\frac{1}{2}) \cdot (3) & (\frac{1}{2}) \cdot (- 1) & (\frac{1}{2}) \cdot (- 4) & (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (5) & (\frac{1}{2}) \cdot (0) & (\frac{1}{2}) \cdot (1) & (\frac{1}{2}) \cdot (4) \\ 1 & 5 & 2 & 0 & 0 & 0 & 1 & - 2 & - 3 & - 4 \end{array}]$

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

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

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

Step 2

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 $0$ .

Tap for more steps...

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 $0$ .

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

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

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

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

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

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

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 0 & \frac{7}{2} & \frac{1}{2} & \frac{1}{2} & 2 & - 1 & - \frac{3}{2} & - 2 & - \frac{7}{2} & - 6 \end{array}]$

Step 3

Perform the row operation $R_{2} = \frac{2}{7} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Tap for more steps...

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{2}{7} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{2} = \frac{2}{7} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{2}{7} R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ (\frac{2}{7}) \cdot (0) & (\frac{2}{7}) \cdot (\frac{7}{2}) & (\frac{2}{7}) \cdot (\frac{1}{2}) & (\frac{2}{7}) \cdot (\frac{1}{2}) & (\frac{2}{7}) \cdot (2) & (\frac{2}{7}) \cdot (- 1) & (\frac{2}{7}) \cdot (- \frac{3}{2}) & (\frac{2}{7}) \cdot (- 2) & (\frac{2}{7}) \cdot (- \frac{7}{2}) & (\frac{2}{7}) \cdot (- 6) \end{array}]$

$R_{2} = \frac{2}{7} R_{2}$

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

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{3}{2} & - \frac{1}{2} & - 2 & 1 & \frac{5}{2} & 0 & \frac{1}{2} & 2 \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

Step 4

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

Tap for more steps...

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

$[\begin{array}{c} - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

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

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

$[\begin{array}{c} (- \frac{3}{2}) \cdot (0) + 1 & (- \frac{3}{2}) \cdot (1) + \frac{3}{2} & (- \frac{3}{2}) \cdot (\frac{1}{7}) + \frac{3}{2} & (- \frac{3}{2}) \cdot (\frac{1}{7}) - \frac{1}{2} & (- \frac{3}{2}) \cdot (\frac{4}{7}) - 2 & (- \frac{3}{2}) \cdot (- \frac{2}{7}) + 1 & (- \frac{3}{2}) \cdot (- \frac{3}{7}) + \frac{5}{2} & (- \frac{3}{2}) \cdot (- \frac{4}{7}) + 0 & (- \frac{3}{2}) \cdot (- 1) + \frac{1}{2} & (- \frac{3}{2}) \cdot (- \frac{12}{7}) + 2 \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{9}{7} & - \frac{5}{7} & - \frac{20}{7} & \frac{10}{7} & \frac{22}{7} & \frac{6}{7} & 2 & \frac{32}{7} \\ 0 & 1 & \frac{1}{7} & \frac{1}{7} & \frac{4}{7} & - \frac{2}{7} & - \frac{3}{7} & - \frac{4}{7} & - 1 & - \frac{12}{7} \end{array}]$

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