Linear Algebra Examples

$[\begin{array}{c} - 9 & - 6 \\ 1 & - 2 \end{array}]$

Step 1

Perform the row operation $R_{1} = - \frac{1}{9} R_{1}$ 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 $R_{1} = - \frac{1}{9} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{1}{9} R_{1} & - \frac{1}{9} R_{1} \\ 1 & - 2 \end{array}]$

$R_{1} = - \frac{1}{9} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{9} R_{1}$ .

$[\begin{array}{c} (- \frac{1}{9}) \cdot (- 9) & (- \frac{1}{9}) \cdot (- 6) \\ 1 & - 2 \end{array}]$

$R_{1} = - \frac{1}{9} R_{1}$

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

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

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 $0$ .

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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 $0$ .

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

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

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

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

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

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

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

Step 3

Perform the row operation $R_{2} = - \frac{3}{8} R_{2}$ 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 $R_{2} = - \frac{3}{8} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{2} = - \frac{3}{8} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{3}{8} R_{2}$ .

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

$R_{2} = - \frac{3}{8} R_{2}$

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

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

Step 4

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

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

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

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

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

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

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

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

Step 5

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

$1, 2$

Step 6

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 $1, 2$ are the pivot positions.

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

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