Linear Algebra Examples

$[\begin{array}{c} 9 & 6 \\ 5 & 2 \end{array}]$

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 $1$ .

Tap for more steps...

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 $1$ .

$[\begin{array}{c} \frac{1}{9} R_{1} & \frac{1}{9} R_{1} \\ 5 & 2 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{9}) \cdot (9) & (\frac{1}{9}) \cdot (6) \\ 5 & 2 \end{array}]$

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

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

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

Step 2

Perform the row operation $R_{2} = - 5 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 5 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = - 5 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 5 \cdot R_{1} + R_{2}$ .

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

$R_{2} = - 5 \cdot R_{1} + R_{2}$

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

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

Step 3

Perform the row operation $R_{2} = - \frac{3}{4} R_{2}$ 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 $R_{2} = - \frac{3}{4} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{2} = - \frac{3}{4} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{3}{4} R_{2}$ .

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

$R_{2} = - \frac{3}{4} R_{2}$

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 $0$ .

Tap for more steps...

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

Enter YOUR Problem