Linear Algebra Examples

$[\begin{array}{c} 6 & 7 \\ 9 & 7 \end{array}]$

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

$A = [\begin{array}{c} 6 & 7 \\ 9 & 7 \end{array}]$

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} \frac{1}{6} R_{1} & \frac{1}{6} R_{1} \\ 9 & 7 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{6}) \cdot (6) & (\frac{1}{6}) \cdot (7) \\ 9 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{7}{6} \\ 9 & 7 \end{array}]$

Perform the row operation $R_{2} = - 9 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 9 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{7}{6} \\ - 9 \cdot R_{1} + R_{2} & - 9 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 9 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 9 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{6} \\ (- 9) \cdot (1) + 9 & (- 9) \cdot (\frac{7}{6}) + 7 \end{array}]$

$R_{2} = - 9 \cdot R_{1} + R_{2}$

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

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

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 $1$ .

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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 $1$ .

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

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

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

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

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

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

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

Perform the row operation $R_{1} = - \frac{7}{6} R_{2} + R_{1}$ 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 $R_{1} = - \frac{7}{6} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{1} = - \frac{7}{6} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{7}{6} R_{2} + R_{1}$ .

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

$R_{1} = - \frac{7}{6} R_{2} + R_{1}$

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

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

$D$ 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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Linear Algebra Examples

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