Linear Algebra Examples

$[\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (4) & (\frac{1}{2}) \cdot (6) \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 3 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

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

$[\begin{array}{c} 1 & 2 & 3 \\ - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} & - 8 \cdot R_{1} + R_{2} \\ 1 & 2 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 3 \\ (- 8) \cdot (1) + 8 & (- 8) \cdot (2) + 10 & (- 8) \cdot (3) + 12 \\ 1 & 2 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 1 & 2 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

$R_{3} = - 1 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (2) + 2 & (- 1) \cdot (3) + 0 \end{array}]$

$R_{3} = - 1 \cdot R_{1} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & - 6 & - 12 \\ 0 & 0 & - 3 \end{array}]$

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

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

$R_{2} = - \frac{1}{6} R_{2}$

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

$[\begin{array}{c} 1 & 2 & 3 \\ (- \frac{1}{6}) \cdot (0) & (- \frac{1}{6}) \cdot (- 6) & (- \frac{1}{6}) \cdot (- 12) \\ 0 & 0 & - 3 \end{array}]$

$R_{2} = - \frac{1}{6} R_{2}$

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

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

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

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

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

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

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

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

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

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

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

Tap for more steps...

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

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

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

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

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

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

Simplify $R_{3}$ (row $3$ ).

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

Perform the row operation $R_{1} = R_{3} + R_{1}$ 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 $R_{1} = R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{1} = R_{3} + R_{1}$

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

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

$R_{1} = R_{3} + R_{1}$

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

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

Perform the row operation $R_{2} = - 2 \cdot R_{3} + R_{2}$ 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 $R_{2} = - 2 \cdot R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = - 2 \cdot R_{3} + R_{2}$

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

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

$R_{2} = - 2 \cdot R_{3} + R_{2}$

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

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

Pivot columns are the columns, which contains pivot positions, so those pivot columns are $1, 2, 3$ .

$1, 2, 3$

A pivot position in a matrix is a position that after row reduction contains a leading $1$ . Thus, the leading one in the pivot columns $1, 2, 3$ are the pivot positions.

The leading $1 s$ in the pivot columns $1, 2, 3$ are the pivot positions

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