Algebra Examples

$[\begin{array}{c} 7 & 2 & 2 \\ 2 & 2 & 7 \end{array}]$

Assign the matrix the name $A$ to simplify the descriptions throughout the problem.

$A = [\begin{array}{c} 7 & 2 & 2 \\ 2 & 2 & 7 \end{array}]$

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = \frac{1}{7} R_{1}$ 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 $R_{1} = \frac{1}{7} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{1} = \frac{1}{7} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{7} R_{1}$ .

$[\begin{array}{c} (\frac{1}{7}) \cdot (7) & (\frac{1}{7}) \cdot (2) & (\frac{1}{7}) \cdot (2) \\ 2 & 2 & 7 \end{array}]$

$R_{1} = \frac{1}{7} R_{1}$

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

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 2 & 2 & 7 \end{array}]$

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

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \end{array}]$

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

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

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

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

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

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 0 & \frac{10}{7} & \frac{45}{7} \end{array}]$

Perform the row operation $R_{2} = \frac{7}{10} R_{2}$ 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 $R_{2} = \frac{7}{10} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ \frac{7}{10} R_{2} & \frac{7}{10} R_{2} & \frac{7}{10} R_{2} \end{array}]$

$R_{2} = \frac{7}{10} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{7}{10} R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ (\frac{7}{10}) \cdot (0) & (\frac{7}{10}) \cdot (\frac{10}{7}) & (\frac{7}{10}) \cdot (\frac{45}{7}) \end{array}]$

$R_{2} = \frac{7}{10} R_{2}$

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

$[\begin{array}{c} 1 & \frac{2}{7} & \frac{2}{7} \\ 0 & 1 & \frac{9}{2} \end{array}]$

Perform the row operation $R_{1} = - \frac{2}{7} R_{2} + R_{1}$ 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 $R_{1} = - \frac{2}{7} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{2}{7} R_{2} + R_{1} & - \frac{2}{7} R_{2} + R_{1} & - \frac{2}{7} R_{2} + R_{1} \\ 0 & 1 & \frac{9}{2} \end{array}]$

$R_{1} = - \frac{2}{7} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{2}{7} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{2}{7}) \cdot (0) + 1 & (- \frac{2}{7}) \cdot (1) + \frac{2}{7} & (- \frac{2}{7}) \cdot (\frac{9}{2}) + \frac{2}{7} \\ 0 & 1 & \frac{9}{2} \end{array}]$

$R_{1} = - \frac{2}{7} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & - 1 \\ 0 & 1 & \frac{9}{2} \end{array}]$

$D$ represents the set of the pivot columns in the row-reduced form of the matrix.

$D = {1, 2}$

Rank is the dimension of the column space of $A$ .

$r (A) = d i m (D)$

The rank of $A$ is $2$ .

$2$

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